If ${\bf{\vec p}} = \widehat {\bf{i}} + \widehat {\bf{j}},{\bf{\vec q}} = 4{\bf{\hat k}} - \widehat {\bf{j}}\;{\rm{and}}\;{\bf{\vec r}} = \widehat {\bf{i}} + {\bf{\hat k}},$ then the unit vector in the direction of $3{\bf{\vec p}} + {\bf{\vec q}} - 2{\bf{\vec r}}$ is

$\frac{1}{3}\left( {\widehat {\bf{i}} + 2\widehat {\bf{j}} + 2{\bf{\hat k}}} \right)$

$\frac{1}{3}\left( {\widehat {\bf{i}} - 2\widehat {\bf{j}} - 2{\bf{\hat k}}} \right)$

$\frac{1}{3}\left( {\widehat {\bf{i}} - 2\widehat {\bf{j}} + 2{\bf{\hat k}}} \right)$

$\widehat {\bf{i}} + 2\widehat {\bf{j}} + 2{\bf{\hat k}}$

ADDITION OF VECTORS MCQs for JEE MATHEMATICS — Easy

1 Easy

1 of 257

Application +4 | -1

A vector

{\mathbf{\vec a}}

has components

2\;p

and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense, if this respect to new system

{\mathbf{\vec a}}

has components

p + 1

and 1, then

View answer and explanation

Correct option: B

p = 1\;\mathrm{or\;}p = \frac{{ - 1}}{2}

Option B (

p = 1\mathrm{\;or\;}p = \frac{{ - 1}}{2}

) is correct because the magnitude of the vector remains unchanged under rotation, leading to a quadratic equation with these solutions.

Solution approach

Correct Answer + Core Reason: $p = 1\mathrm{\;or\;}p = \frac{{ - 1}}{2}$ . The magnitude of the vector remains unchanged under rotation.

Step-by-Step Solution:

Step 1: Define the vectors: $\mathbf{\vec a} = 2p{\widehat \mathbf{i}} + {\widehat \mathbf{j}}$ and $\mathbf{\vec b} = \left( {p + 1} \right){\widehat \mathbf{i}} + {\widehat \mathbf{j}}$ .

Step 2: Since the magnitude of the vector is invariant under rotation, we have $\left| \mathbf{\vec a} \right| = \left| \mathbf{\vec b} \right|$ .

Step 3: Calculate the magnitudes: $\left| \mathbf{\vec a} \right| = \sqrt {{{(2p)}^2} + {1^2}} = \sqrt {4{p^2} + 1}$ and $\left| \mathbf{\vec b} \right| = \sqrt {{{(p + 1)}^2} + {1^2}} = \sqrt {{{p^2} + 2p + 1 + 1}} = \sqrt {{{p^2} + 2p + 2}}$ .

Step 4: Equate the magnitudes: $\sqrt {4{p^2} + 1} = \sqrt {{{p^2} + 2p + 2}}$ .

Step 5: Square both sides: $4{p^2} + 1 = {p^2} + 2p + 2$ .

Step 6: Rearrange the equation: $3{p^2} - 2p - 1 = 0$ .

Step 7: Solve the quadratic equation: $(3p + 1)(p - 1) = 0$ . Thus, $p = 1$ or $p = -\frac{1}{3}$ .

Why other options are wrong:
- Option A: $p = 0$ is incorrect because it does not satisfy the condition that the magnitude of the vector remains unchanged under rotation.
- Option C: $p = -1$ is incorrect because it does not satisfy the condition that the magnitude of the vector remains unchanged under rotation.
- Option D: $p = 1\mathrm{\;or\;}p = - 1$ is incorrect because $p = -1$ is not a solution to the quadratic equation.

Key Concepts:
- Invariance of vector magnitude under rotation
- Solving quadratic equations

Memory Aids:
- Remember that rotation preserves the length of a vector.

Quick Tips:
- When dealing with rotations, consider the properties that remain unchanged, such as magnitude.

2 Easy

2 of 257

Application +4 | -1

If

\vec{a}, \vec{b}

and

\vec{c}

are three mutually perpendicular unit vetors and

\vec{d}

is a unit vector which makes equal angles with

\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}

ānd

\overrightarrow{\mathrm{c}}

then the value of

|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|^2

View answer and explanation

Correct option: B

4+2 \sqrt{3}

Option B (

4+2 \sqrt{3}

) is correct because it involves using the properties of mutually perpendicular unit vectors and the given condition about the unit vector d to find the magnitude squared of the vector sum.

Solution approach

Correct Answer + Core Reason: $4+2 \sqrt{3}$ . The magnitude squared is calculated using the properties of dot products and unit vectors.

Step-by-Step Solution:

Step 1: Expand the magnitude squared: $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|^2 = |\overrightarrow{\mathrm{a}}|^2 + |\overrightarrow{\mathrm{b}}|^2 + |\overrightarrow{\mathrm{c}}|^2 + |\overrightarrow{\mathrm{d}}|^2 + 2(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} + \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}} + \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}} + \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}} + \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{d}} + \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}})$ .

Step 2: Since $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular unit vectors, $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} = \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}} = \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}} = 0$ and $|\overrightarrow{\mathrm{a}}|^2 = |\overrightarrow{\mathrm{b}}|^2 = |\overrightarrow{\mathrm{c}}|^2 = |\overrightarrow{\mathrm{d}}|^2 = 1$ .

Step 3: The expression simplifies to: $1 + 1 + 1 + 1 + 2(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}} + \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{d}} + \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}) = 4 + 2(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}} + \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{d}} + \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}})$ .

Step 4: Since $\vec{d}$ makes equal angles with $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ , let $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}} = \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{d}} = \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}} = \cos \theta$ . Then the expression becomes $4 + 2(3 \cos \theta) = 4 + 6 \cos \theta$ .

Step 5: Let $\vec{d} = \lambda \vec{a} + \mu \vec{b} + \nu \vec{c}$ . Then $\vec{d} \cdot \vec{a} = \lambda = \cos \theta$ , $\vec{d} \cdot \vec{b} = \mu = \cos \theta$ , and $\vec{d} \cdot \vec{c} = \nu = \cos \theta$ . Since $\vec{d}$ is a unit vector, $\lambda^2 + \mu^2 + \nu^2 = 1$ , so $3 \cos^2 \theta = 1$ , and $\cos \theta = \frac{1}{\sqrt{3}}$ .

Step 6: Substitute $\cos \theta$ back into the expression: $4 + 6 \left(\frac{1}{\sqrt{3}}\right) = 4 + 2 \sqrt{3}$ .

Final Answer: $4+2 \sqrt{3}$

Why other options are wrong:
- Option A: $4+2 \sqrt{2}$ : This is incorrect as it uses the wrong value for the cosine of the angle.
- Option C: $2+\sqrt{5}$ : This is incorrect as it uses the wrong value for the cosine of the angle.
- Option D: $3+\sqrt{5}$ : This is incorrect as it uses the wrong value for the cosine of the angle.

Key Concepts:
- Dot product of vectors
- Unit vectors
- Mutually perpendicular vectors

Memory Aids:
- $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

Quick Tips:
- Remember the properties of dot products for orthogonal vectors. Use the fact that $\vec{d}$ is a unit vector to find the cosine of the angle.

3 Easy

3 of 257

Application +4 | -1

The non-zero vectors are

\vec{a}, \vec{b}

and

\vec{c}

are related by

\vec{a}=8 \vec{b}

and

\vec{c}=-7 \vec{b}

. Then the angle between

\vec{a}

and

\vec{c}

is

View answer and explanation

Correct option: D

\pi

Option D (

\pi

) is correct because the vectors

\vec{{a}}

and

\vec{{c}}

are scalar multiples of

\vec{{b}}

with opposite signs, indicating they are anti-parallel.

Solution approach

Correct Answer + Core Reason: $\pi$ . Since $\vec{{a}}$ and $\vec{{c}}$ are scalar multiples of $\vec{{b}}$ with opposite signs, they are anti-parallel.

Step-by-Step Solution:

Step 1: Given $\vec{{a}} = 8\vec{{b}}$ and $\vec{{c}} = -7\vec{{b}}$ .

Step 2: We can express $\vec{{a}}$ in terms of $\vec{{c}}$ : $\vec{{a}} = 8\vec{{b}} = 8(-\frac{{1}}{{7}}\vec{{c}}) = -\frac{{8}}{{7}}\vec{{c}}$ .

Step 3: Since $\vec{{a}} = -\frac{{8}}{{7}}\vec{{c}}$ , $\vec{{a}}$ and $\vec{{c}}$ are parallel and point in opposite directions.

Step 4: Therefore, the angle between $\vec{{a}}$ and $\vec{{c}}$ is $\pi$ radians or 180 degrees.

Why other options are wrong:
- Option A: 0. This would be the angle if the vectors were in the same direction, but they are in opposite directions.
- Option B: $\frac{{\pi}}{{4}}$ . This angle is incorrect because the vectors are anti-parallel.
- Option C: $\frac{{\pi}}{{2}}$ . This angle is incorrect because the vectors are parallel (anti-parallel in this case).
- Option D: $\pi$ . This is the correct angle because the vectors are anti-parallel.

Key Concepts:
- Parallel and anti-parallel vectors
- Angle between vectors

Memory Aids:
- If $\vec{{a}} = k\vec{{b}}$ where k > 0, then $\vec{{a}}$ and $\vec{{b}}$ are parallel.
- If $\vec{{a}} = k\vec{{b}}$ where k < 0, then $\vec{{a}}$ and $\vec{{b}}$ are anti-parallel.

Quick Tips:
- Visualize the vectors to determine their relative directions.
- If one vector is a scalar multiple of another, they are parallel or anti-parallel.

4 Easy

4 of 257

Application +4 | -1

If the vectors

2 \mathbf{i}+\mathbf{j}-\mathbf{k},-\mathbf{i}+2 \mathbf{j}+\lambda \mathbf{k}

and

-5 \mathbf{i}+2 \mathbf{j}-\mathbf{k}

ar coplanar, then the value of

\lambda

is equal

View answer and explanation

Correct option: C

-13 / 9

Option C (-13/9) is correct because the scalar triple product of coplanar vectors is zero, leading to the equation that solves for lambda.

Solution approach

Correct Answer + Core Reason: -13/9. The vectors are coplanar, so their scalar triple product (determinant) is zero.

Step-by-Step Solution:

Step 1: State the condition for coplanarity: The determinant formed by the components of the vectors must be zero.

Step 2: Write the determinant equation: $\left|\begin{{ccc}} 2 & 1 & -1 \\ -1 & 2 & \lambda \\ -5 & 2 & -1 \end{{ccc}}\right| = 0$

Step 3: Evaluate the determinant: $2(2*(-1) - 2*\lambda) - 1((-1)*(-1) - (-5)*\lambda) + (-1)((-1)*2 - (-5)*2) = 0$

Step 4: Simplify the equation: $2(-2 - 2\lambda) - (1 + 5\lambda) - (-2 + 10) = 0$

Step 5: Further simplification: $-4 - 4\lambda - 1 - 5\lambda - 8 = 0$

Step 6: Combine like terms: $-9\lambda - 13 = 0$

Step 7: Solve for λ: $-9\lambda = 13 \Rightarrow \lambda = -\frac{{13}}{{9}}$

Why other options are wrong:
- Option A (-13): This is incorrect due to an error in calculating the determinant or solving the linear equation.
- Option B (13/9): This is incorrect because the sign is wrong. It results from not properly accounting for the negative signs when expanding the determinant.
- Option D (-9/13): This is the reciprocal of the correct answer and arises from inverting the coefficients during the final step of solving for lambda.

Key Concepts:
- Coplanar vectors
- Scalar triple product
- Determinants

Memory Aids:
- Remember that coplanar vectors have a scalar triple product of zero.

Quick Tips:
- Double-check the signs when expanding the determinant.
- Be careful when solving the linear equation for the unknown variable.

5 Easy

5 of 257

Analyzing +4 | -1

The medium

AD

of the triangle

ABC

is bisected at

E,\;BE

meets

AC

in

F

, then

AF:AC =

View answer and explanation

Correct option: B

1/3

Option B (1/3) is correct because using vector methods and the section formula, the ratio AF:AC can be determined to be 1/3.

Solution approach

Correct Answer + Core Reason: 1/3. By using vector algebra and the section formula, we can find the ratio AF:AC.

Step-by-Step Solution:

Step 1: Set up the problem using vectors.\nLet $\vec A$ be the origin. Let $\vec B$ and $\vec C$ be the position vectors of points B and C, respectively. Since D is the midpoint of BC, $\vec D = \frac{{\vec B + \vec C}}{2}$ . Since E bisects AD, $\vec E = \frac{{\vec A + \vec D}}{2} = \frac{{\vec 0 + \frac{{\vec B + \vec C}}{2}}}{2} = \frac{{\vec B + \vec C}}{4}$ .

Step 2: Find the equation of line BF.\nThe equation of line BF is given by $\vec r = \vec B + \lambda (\vec E - \vec B) = \vec B + \lambda (\frac{{\vec B + \vec C}}{4} - \vec B) = \vec B + \lambda (\frac{{\vec C - 3\vec B}}{4})$ , where $\lambda$ is a scalar.

Step 3: Find the equation of line AC.\nThe equation of line AC is given by $\vec r = \mu \vec C$ , where $\mu$ is a scalar.

Step 4: Find the intersection point F.\nAt the intersection point F, we have $\vec B + \lambda (\frac{{\vec C - 3\vec B}}{4}) = \mu \vec C$ . This can be rewritten as $\vec B + \frac{\lambda}{4} \vec C - \frac{{3\lambda}}{4} \vec B = \mu \vec C$ .\nComparing the coefficients of $\vec B$ and $\vec C$ , we get $1 - \frac{{3\lambda}}{4} = 0$ and $\frac{\lambda}{4} = \mu$ . From the first equation, $\frac{{3\lambda}}{4} = 1$ , so $\lambda = \frac{4}{3}$ . Substituting this into the second equation, we get $\mu = \frac{1}{4} \cdot \frac{4}{3} = \frac{1}{3}$ .

Step 5: Determine the position vector of F.\nSince $\vec F = \mu \vec C$ , we have $\vec F = \frac{1}{3} \vec C$ . This means that F divides AC in the ratio 1:2, so $AF:AC = 1:3 = \frac{1}{3}$ .

Why other options are wrong:
- Option A: 3/4 is incorrect. This would imply that F is much closer to A than it actually is.
- Option C: 1/2 is incorrect. This would mean F is the midpoint of AC, which is not the case.
- Option D: 1/4 is incorrect. This would imply that F is very close to A.

Key Concepts:
- Vector algebra
- Section formula
- Equation of a line

Memory Aids:
- Remember the section formula for dividing a line segment in a given ratio.

Quick Tips:
- Use vector methods to solve geometry problems involving ratios.

6 Easy

6 of 257

Application +4 | -1

If a vector

{\vec r}

of magnitude

3\sqrt 6

is directed along the bisector of the angle between the vectors

\vec a = 7\hat i - 4\hat j - 4\hat k

and

\vec b = - 2\hat i - \hat j + 2\hat k

, then

\vec r =

View answer and explanation

Correct option: A

\hat i - 7\hat j + 2\hat k

The correct answer is A (

\hat i - 7\hat j + 2\hat k

) because it is the vector along the angle bisector with the given magnitude.

Solution approach

Correct Answer + Core Reason: $\hat i - 7\hat j + 2\hat k$ . The vector $\vec r$ is along the angle bisector of $\vec a$ and $\vec b$ , and its magnitude is given.

Step-by-Step Solution:

Step 1: Find the unit vectors along $\vec a$ and $\vec b$ .

Step 2: Calculate the magnitude of $\vec a$ : $|\vec a| = \sqrt{7^2 + (-4)^2 + (-4)^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$ . So, $\hat a = \frac{\vec a}{|\vec a|} = \frac{7\hat i - 4\hat j - 4\hat k}{9}$ .

Step 3: Calculate the magnitude of $\vec b$ : $|\vec b| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ . So, $\hat b = \frac{\vec b}{|\vec b|} = \frac{-2\hat i - \hat j + 2\hat k}{3}$ .

Step 4: Find the vector along the angle bisector: $\hat a + \hat b = \frac{7\hat i - 4\hat j - 4\hat k}{9} + \frac{-2\hat i - \hat j + 2\hat k}{3} = \frac{7\hat i - 4\hat j - 4\hat k - 6\hat i - 3\hat j + 6\hat k}{9} = \frac{\hat i - 7\hat j + 2\hat k}{9}$ .

Step 5: Let $\vec r = \lambda(\hat a + \hat b) = \frac{\lambda}{9}(\hat i - 7\hat j + 2\hat k)$ . Given $|\vec r| = 3\sqrt{6}$ , so $|\vec r|^2 = 54$ .

Step 6: Calculate the magnitude squared of $\vec r$ : $|\vec r|^2 = \frac{\lambda^2}{81}(1^2 + (-7)^2 + 2^2) = \frac{\lambda^2}{81}(1 + 49 + 4) = \frac{\lambda^2}{81}(54) = 54$ .

Step 7: Solve for $\lambda$ : $\frac{\lambda^2}{81}(54) = 54 \Rightarrow \lambda^2 = 81 \Rightarrow \lambda = \pm 9$ .

Step 8: Find $\vec r$ : $\vec r = \frac{\pm 9}{9}(\hat i - 7\hat j + 2\hat k) = \pm(\hat i - 7\hat j + 2\hat k)$ . So, $\vec r = \hat i - 7\hat j + 2\hat k$ or $\vec r = -\hat i + 7\hat j - 2\hat k$ .

Step 9: Final Answer: The required vector is $\hat i - 7\hat j + 2\hat k$ .

Why other options are wrong:
- Option B: $\hat i + 7\hat j - 2\hat k$ : This vector does not lie along the angle bisector.
- Option C: $- \hat i + 7\hat j + 2\hat k$ : This vector does not lie along the angle bisector.
- Option D: $\hat i - 7\hat j - 2\hat k$ : This vector does not lie along the angle bisector.

Key Concepts:
- Angle bisector of two vectors.
- Unit vectors.
- Magnitude of a vector.

Memory Aids:
- The angle bisector of two vectors $\vec a$ and $\vec b$ is along the direction of $\hat a + \hat b$ .

Quick Tips:
- Remember to normalize the vectors before finding the angle bisector.
- Be careful with signs when calculating the magnitudes and adding vectors.

7 Easy

7 of 257

Application +4 | -1

The value of

\alpha

for which the vectors

2 \mathbf{i}-\mathbf{j}+\mathbf{k}

,

\mathbf{i}+2 \mathbf{j}+\alpha \mathbf{k}

and

3 \mathbf{i}-4 \mathbf{j}+5 \mathbf{k}

are coplanar is

View answer and explanation

Correct option: B

-3

Option B (-3) is correct because it makes the scalar triple product of the three vectors equal to zero, indicating that they are coplanar.

Solution approach

Correct Answer + Core Reason: -3. The value of $\alpha$ is found by setting the determinant to zero.

Step-by-Step Solution:

Step 1: Set up the determinant with the components of the vectors.\nFor the vectors to be coplanar, their scalar triple product must be zero.
$\begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & \alpha \\ 3 & -4 & 5 \end{vmatrix} = 0$

Step 2: Calculate the determinant.
2 * (2 * 5 - ( $\alpha$ ) * (-4)) - (-1) * (1 * 5 - ( $\alpha$ ) * 3) + 1 * (1 * (-4) - 2 * 3) = 0
2 * (10 + 4 $\alpha$ ) + (5 - 3 $\alpha$ ) + (-4 - 6) = 0
20 + 8 $\alpha$ + 5 - 3 $\alpha$ - 10 = 0
5 $\alpha$ + 15 = 0

Step 3: Solve for $\alpha$ .
5 $\alpha$ = -15
$\alpha$ = -3

Why other options are wrong:
- Option A (3): This is incorrect because it does not make the determinant equal to zero.
- Option C (2): This is incorrect because it does not make the determinant equal to zero.
- Option D (None of these): This is incorrect because $\alpha = -3$ is a valid solution.

Key Concepts:
- Coplanar vectors
- Scalar triple product
- Determinant of a 3x3 matrix

Memory Aids:
- Remember that the scalar triple product of coplanar vectors is zero.

Quick Tips:
- Double-check your determinant calculations to avoid errors.
- If vectors are coplanar, the volume of the parallelepiped formed by them is zero.

8 Easy

8 of 257

Application +4 | -1

If

\vec{a}

and

\vec{b}

are non-collinear vectors, then the value of

\alpha

for which the vectors

\vec{u}=(\alpha-2) \vec{a}+\vec{b}

and

\vec{v}=(2+3 a) \vec{a}-3 \vec{b}

are collinear is:

View answer and explanation

Correct option: B

\frac{2}{3}

Option B (

\frac{{2}}{{3}}

) is correct because for

\vec{{u}}

and

\vec{{v}}

to be collinear, one must be a scalar multiple of the other. Solving the resulting equations gives

\alpha = \frac{{2}}{{3}}

.

Solution approach

Correct Answer + Core Reason: $\frac{2}{3}$ . The value of $\alpha$ is found by setting the vectors collinear.

Step-by-Step Solution:

Step 1: Given $\vec{{u}}=(\alpha-2) \vec{{a}}+\vec{{b}}$ and $\vec{{v}}=(2+3 \alpha) \vec{{a}}-3 \vec{{b}}$ . Since $\vec{{u}}$ and $\vec{{v}}$ are collinear, there exists a scalar $k$ such that $\vec{{v}} = k \vec{{u}}$ .

Step 2: Write the equation $\vec{{v}} = k \vec{{u}}$ :
$(2+3 \alpha) \vec{{a}}-3 \vec{{b}} = k((\alpha-2) \vec{{a}}+\vec{{b}})$
$(2+3 \alpha) \vec{{a}}-3 \vec{{b}} = k(\alpha-2) \vec{{a}}+k \vec{{b}}$

Step 3: Equate the coefficients of $\vec{{a}}$ and $\vec{{b}}$ :
$2+3 \alpha = k(\alpha-2)$ ...(1)
$-3 = k$ ...(2)

Step 4: Substitute $k=-3$ into equation (1):
$2+3 \alpha = -3(\alpha-2)$
$2+3 \alpha = -3\alpha+6$
$6 \alpha = 4$
$\alpha = \frac{4}{6} = \frac{2}{3}$

Why other options are wrong:
- Option A: Incorrect value of $\alpha$ due to incorrect algebraic manipulation.
- Option C: Incorrect value of $\alpha$ due to incorrect algebraic manipulation.
- Option D: Incorrect value of $\alpha$ due to incorrect algebraic manipulation.

Key Concepts:
- Collinear vectors
- Scalar multiplication of vectors

Memory Aids:
- Remember that collinear vectors are scalar multiples of each other.

Quick Tips:
- Equate coefficients of the vectors to solve for the unknown scalar.

9 Easy

9 of 257

Application +4 | -1

The point having position vectors

2\widehat i + 3\widehat j + 4\widehat k,\,\,3\widehat i + 4\widehat j + 2\widehat k

and

4\widehat i + 2\widehat j + 3\widehat k

are the vertices of

View answer and explanation

Correct option: C

equilateral triangle

Option C (equilateral triangle) is correct because the distances between all three points are equal, indicating that the points form an equilateral triangle.

Solution approach

Correct Answer + Core Reason: equilateral triangle. The distances between all three points are equal.

Step-by-Step Solution:

Step 1: Find the vectors AB, BC, and CA.
$AB = OB - OA = (3\hat i + 4\hat j + 2\hat k) - (2\hat i + 3\hat j + 4\hat k) = \hat i + \hat j - 2\hat k$
$BC = OC - OB = (4\hat i + 2\hat j + 3\hat k) - (3\hat i + 4\hat j + 2\hat k) = \hat i - 2\hat j + \hat k$
$CA = OA - OC = (2\hat i + 3\hat j + 4\hat k) - (4\hat i + 2\hat j + 3\hat k) = -2\hat i + \hat j + \hat k$ (or $CA = 2\hat i - \hat j - \hat k$ )

Step 2: Calculate the magnitudes of AB, BC, and CA.
$|AB| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$
$|BC| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$
$|CA| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$

Step 3: Since $|AB| = |BC| = |CA| = \sqrt{6}$ , the points are vertices of an equilateral triangle.

Why other options are wrong:
- Option A: right angled triangle: The Pythagorean theorem does not hold for these side lengths.
- Option B: isosceles triangle: While two sides could be equal, all three must be equal for it to be equilateral. Since all sides are equal, it's more specifically equilateral.
- Option D: collinear: The points are not collinear because they form a triangle.

Key Concepts:
- Vector subtraction
- Magnitude of a vector
- Types of triangles

Memory Aids:
- Equilateral triangles have all sides equal.

Quick Tips:
- Always find the vectors between the points first, then calculate their magnitudes.

10 Easy

10 of 257

Application +4 | -1

If the position vectors of the vertices

A, B, C

of a triangle

A B C

are

7 \mathbf{j}+10 \mathbf{k},-\mathbf{i}+6 \mathbf{j}+6 \mathbf{k}

and

-4 \mathbf{i}+9 \mathbf{j}+6 \mathbf{k}

respectively, the triangle is

View answer and explanation

Correct option: D

Right angled and isosceles also

Option D (Right angled and isosceles also) is correct because the triangle's sides satisfy both the conditions for a right-angled triangle (Pythagorean theorem) and an isosceles triangle (two sides equal).

Solution approach

Correct Answer + Core Reason: Right angled and isosceles also. The triangle satisfies both the Pythagorean theorem and has two equal sides.

Step-by-Step Solution:

Step 1: Find the vectors $\overrightarrow{AB}, \overrightarrow{BC},$ and $\overrightarrow{AC}$ :
$\overrightarrow{AB} = \mathbf{B} - \mathbf{A} = (-\mathbf{i} + 6\mathbf{j} + 6\mathbf{k}) - (7\mathbf{j} + 10\mathbf{k}) = -\mathbf{i} - \mathbf{j} - 4\mathbf{k}$ .
$\overrightarrow{BC} = \mathbf{C} - \mathbf{B} = (-4\mathbf{i} + 9\mathbf{j} + 6\mathbf{k}) - (-\mathbf{i} + 6\mathbf{j} + 6\mathbf{k}) = -3\mathbf{i} + 3\mathbf{j}$ .
$\overrightarrow{AC} = \mathbf{C} - \mathbf{A} = (-4\mathbf{i} + 9\mathbf{j} + 6\mathbf{k}) - (7\mathbf{j} + 10\mathbf{k}) = -4\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}$ .

Step 2: Find the magnitudes of the vectors:
$|\overrightarrow{AB}| = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18}$ .
$|\overrightarrow{BC}| = \sqrt{(-3)^2 + (3)^2} = \sqrt{9 + 9} = \sqrt{18}$ .
$|\overrightarrow{AC}| = \sqrt{(-4)^2 + (2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$ .

Step 3: Check if the triangle is isosceles:\nSince $|\overrightarrow{AB}| = |\overrightarrow{BC}| = \sqrt{18}$ , the triangle is isosceles.

Step 4: Check if the triangle is right-angled:\nCheck if the Pythagorean theorem holds: $|\overrightarrow{AC}|^2 = |\overrightarrow{AB}|^2 + |\overrightarrow{BC}|^2$ .
$6^2 = (\sqrt{18})^2 + (\sqrt{18})^2 \Rightarrow 36 = 18 + 18 \Rightarrow 36 = 36$ .\nSince the Pythagorean theorem holds, the triangle is right-angled.

Why other options are wrong:
- Option A (Equilateral): The triangle is not equilateral because all sides are not equal.
- Option B (Isosceles): While the triangle is isosceles, this option doesn't include the right-angled property.
- Option C (Scalene): The triangle is not scalene because two sides are equal.

Key Concepts:
- Vector subtraction
- Magnitude of a vector
- Properties of triangles (isosceles, equilateral, scalene, right-angled)
- Pythagorean theorem

Memory Aids:
- Remember the conditions for different types of triangles.

Quick Tips:
- Calculate the side lengths first, then check for the properties of the triangle.

11 Easy

11 of 257

Application +4 | -1

If

\mathbf{\vec p} = \widehat \mathbf{i} + \widehat \mathbf{j},\mathbf{\vec q} = 4\mathbf{\hat k} - \widehat \mathbf{j}\;\mathrm{and}\;\mathbf{\vec r} = \widehat \mathbf{i} + \mathbf{\hat k},

then the unit vector in the direction of

3\mathbf{\vec p} + \mathbf{\vec q} - 2\mathbf{\vec r}

is

View answer and explanation

Correct option: A

\frac{1}{3}\left( {\widehat \mathbf{i} + 2\widehat \mathbf{j} + 2\mathbf{\hat k}} \right)

Option A (1/3(i + 2j + 2k)) is correct because it calculates the unit vector in the direction of the given linear combination of vectors.

Solution approach

Correct Answer + Core Reason: 1/3(i + 2j + 2k). The unit vector is found by calculating the linear combination of the vectors and then normalizing the resulting vector.

Step-by-Step Solution:

Step 1: Given $\vec{p} = \hat{i} + \hat{j}$ , $\vec{q} = 4\hat{k} - \hat{j}$ , and $\vec{r} = \hat{i} + \hat{k}$ .

Step 2: Calculate $3\vec{p} + \vec{q} - 2\vec{r} = 3(\hat{i} + \hat{j}) + (4\hat{k} - \hat{j}) - 2(\hat{i} + \hat{k}) = 3\hat{i} + 3\hat{j} + 4\hat{k} - \hat{j} - 2\hat{i} - 2\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$ .

Step 3: Find the magnitude of the resulting vector: $|\hat{i} + 2\hat{j} + 2\hat{k}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$ .

Step 4: Calculate the unit vector: $\frac{{\hat{i} + 2\hat{j} + 2\hat{k}}}{3} = \frac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$ .

Why other options are wrong:
- Option B: Incorrect calculation of the linear combination or normalization.
- Option C: Incorrect calculation of the linear combination or normalization.
- Option D: Incorrect calculation of the linear combination or normalization.

Key Concepts:
- Vector addition and subtraction.
- Scalar multiplication.
- Unit vector.

Memory Aids:
- Remember that a unit vector is found by dividing a vector by its magnitude.

Quick Tips:
- Perform the linear combination carefully and then normalize the resulting vector.

12 Easy

12 of 257

Application +4 | -1

Let

\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=10 \vec{a}+2 \vec{b}

and

\overrightarrow{O C}=\vec{b}

where

\mathrm{A}

and

\mathrm{C}

are non collinear points. Let 'p' denote the area of quadrilateral

O A B C

and let '

q

' denote the area of parallelogram with

\overrightarrow{O A}

and

\overrightarrow{O C}

as adjacent sides. If

\mathrm{p}=\mathrm{kq}

then

\mathrm{k}=

View answer and explanation

Correct option: B

6

Option B (6) is correct because the area of the quadrilateral OABC is 6 times the area of the parallelogram formed by vectors

\vec{a}

and

\vec{b}

.

Solution approach

Correct Answer + Core Reason: 6. The area of the quadrilateral is calculated and compared to the area of the parallelogram.

Step-by-Step Solution:

Step 1: The area of the parallelogram with adjacent sides $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OC} = \vec{b}$ is given by $q = |\vec{a} \times \vec{b}|$ .

Step 2: The area of the quadrilateral OABC can be found by dividing it into two triangles, OAC and ABC. However, a simpler approach is to find the area of the quadrilateral directly using vectors.

Step 3: The area of quadrilateral OABC is given by $p = \frac{1}{2} |(\overrightarrow{OB} - \overrightarrow{OA}) \times (\overrightarrow{OC} - \overrightarrow{OA}) + (\overrightarrow{OA} \times \overrightarrow{OC})|$ . Alternatively, $p = \frac{1}{2} |(\overrightarrow{OA} \times \overrightarrow{OB}) + (\overrightarrow{OB} \times \overrightarrow{OC}) + (\overrightarrow{OC} \times \overrightarrow{OA})|$ . Another valid approach is $p = \frac{1}{2} |(\vec{b}-\vec{a}) \times(10 \vec{a}+2 \vec{b})|$ .

Step 4: Calculate $\overrightarrow{OB} - \overrightarrow{OA} = (10\vec{a} + 2\vec{b}) - \vec{a} = 9\vec{a} + 2\vec{b}$ and $\overrightarrow{OC} - \overrightarrow{OA} = \vec{b} - \vec{a}$ .

Step 5: Compute the cross product: $(\overrightarrow{OB} - \overrightarrow{OA}) \times (\overrightarrow{OC} - \overrightarrow{OA}) = (9\vec{a} + 2\vec{b}) \times (\vec{b} - \vec{a}) = 9(\vec{a} \times \vec{b}) - 9(\vec{a} \times \vec{a}) + 2(\vec{b} \times \vec{b}) - 2(\vec{b} \times \vec{a}) = 9(\vec{a} \times \vec{b}) - 0 + 0 + 2(\vec{a} \times \vec{b}) = 11(\vec{a} \times \vec{b})$ .

Step 6: The area of the quadrilateral is $p = \frac{1}{2} |11(\vec{a} \times \vec{b})| = \frac{11}{2} |\vec{a} \times \vec{b}|$ . This approach is incorrect. Let's use the other approach: $p = \frac{1}{2} |(\vec{b}-\vec{a}) \times(10 \vec{a}+2 \vec{b})| = \frac{1}{2} | 10(\vec{b} \times \vec{a}) - 2(\vec{a} \times \vec{a}) + 2(\vec{b} \times \vec{b}) - (\vec{a} \times \vec{b}) | = \frac{1}{2} | -10(\vec{a} \times \vec{b}) - (\vec{a} \times \vec{b}) | = \frac{1}{2} |-12(\vec{a} \times \vec{b})| = 6|\vec{a} \times \vec{b}|$ .

Step 7: Given $p = kq$ , we have $6|\vec{a} \times \vec{b}| = k |\vec{a} \times \vec{b}|$ , so $k = 6$ .

Why other options are wrong:
- Option A (3): This is incorrect because the correct calculation yields 6.
- Option C (9): This is incorrect because the correct calculation yields 6.
- Option D (12): This is incorrect because the correct calculation yields 6.

Key Concepts:
- Vector Algebra
- Cross Product
- Area of Parallelogram
- Area of Quadrilateral

Memory Aids:
- Remember that the magnitude of the cross product of two vectors is equal to the area of the parallelogram formed by those vectors.

Quick Tips:
- Use vector methods to find areas of geometric figures.

13 Easy

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Analyzing +4 | -1

If

\vec a,\vec b,\vec c

are the position vectors of the vertices of an equilateral triangle whose orthocenter is at the origin, then

View answer and explanation

Correct option: A

\vec a + \vec b + \vec c = \vec 0

The correct answer is A (

\vec a + \vec b + \vec c = \vec 0

) because in an equilateral triangle with the orthocenter at the origin, the centroid also lies at the origin, implying the sum of the position vectors is zero.

Solution approach

Correct Answer + Core Reason: $\vec a + \vec b + \vec c = \vec 0$ . In an equilateral triangle, the orthocenter and centroid coincide. If the orthocenter is at the origin, the centroid is also at the origin, implying the sum of the position vectors is zero.

Step-by-Step Solution:

Step 1: The position vector of the centroid of the triangle is given by $\frac{{\vec a + \vec b + \vec c}}{3}$ .

Step 2: Since the triangle is equilateral, the orthocenter coincides with the centroid.

Step 3: Given that the orthocenter is at the origin, we have $\frac{{\vec a + \vec b + \vec c}}{3} = \vec 0$ .

Step 4: Multiplying both sides by 3, we get $\vec a + \vec b + \vec c = \vec 0$ .

Why other options are wrong:
- Option B: This is incorrect because it represents a relationship between the magnitudes of the vectors that is not generally true for an equilateral triangle with its orthocenter at the origin.
- Option C: This is incorrect because it represents a specific relationship between the vectors that is not generally true for an equilateral triangle with its orthocenter at the origin.
- Option D: This is incorrect because option A is correct.

Key Concepts:
- Centroid of a triangle
- Orthocenter of a triangle
- Equilateral triangle
- Position vectors

Memory Aids:
- Remember that in an equilateral triangle, the centroid and orthocenter coincide.

Quick Tips:
- Use the properties of equilateral triangles and the definitions of centroid and orthocenter to solve the problem.

14 Easy

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Application +4 | -1

If a, b and c are position vector of A, B and C respectively of

\Delta ABC

and if

\left| {a - b} \right| = 4,\left| {b - c} \right| = 2,\left| {c - a} \right| = 3,

then the distance between the centroid and incentre of

\Delta ABC

is

View answer and explanation

Correct option: C

\frac{1}{3}

Option C (

\frac{1}{3}

) is correct because it is calculated by finding the magnitude of the difference between the incenter and centroid position vectors.

Solution approach

Correct Answer + Core Reason: $\frac{1}{3}$ . The distance between the centroid and incenter is calculated using their position vectors.

Step-by-Step Solution:

Step 1: Let G be the centroid and I be the incenter.

Step 2: The position vector of the incenter is given by $I = \frac{{a|BC| + b|AC| + c|AB|}}{{|BC| + |AC| + |AB|}} = \frac{{4a + 3b + 2c}}{{4 + 3 + 2}} = \frac{{4a + 3b + 2c}}{9}$ .

Step 3: The position vector of the centroid is given by $G = \frac{{a + b + c}}{3}$ .

Step 4: The distance between the incenter and centroid is $|GI| = |I - G| = \left| {\frac{{4a + 3b + 2c}}{9} - \frac{{a + b + c}}{3}} \right|$ .

Step 5: Simplify the expression: $|GI| = \left| {\frac{{4a + 3b + 2c - 3a - 3b - 3c}}{9}} \right| = \left| {\frac{{a - c}}{9}} \right|$ .

Step 6: Since $|a - c| = 3$ , $|GI| = \frac{3}{9} = \frac{1}{3}$ .

Why other options are wrong:
- Option A (1): This is incorrect because it doesn't account for the correct formula for the incenter and centroid.
- **Option B ( $\frac{1}{2}$ ): This is incorrect because it doesn't account for the correct formula for the incenter and centroid.
- Option D ( $\frac{2}{3}$ ): This is incorrect because it doesn't account for the correct formula for the incenter and centroid.

Key Concepts:
- Centroid of a triangle
- Incenter of a triangle
- Position vectors

Memory Aids:
- Remember the formulas for the centroid and incenter.

Quick Tips:**
- Always remember the formulas for the centroid and incenter of a triangle.

15 Easy

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Application +4 | -1

Let u be a vector coplanar with the vectors

a = 2\hat i + 3\hat j - \hat k

and

b = \hat j + \hat k

If u is perpendicular to a and u. b = 24, then

{\left| \mathrm{u} \right|^2}

is equal to

View answer and explanation

Correct option: A

336

Option A (336) is correct because it is the result of solving for the magnitude squared of vector u using the given conditions of coplanarity, perpendicularity, and dot product.

Solution approach

Correct Answer + Core Reason: 336. The magnitude squared of vector u is found by expressing u as a linear combination of a and b, then using the given dot product and perpendicularity conditions to solve for the coefficients.

Step-by-Step Solution:

Step 1: Express u as a linear combination of a and b: $u = \lambda a + \mu b$ , where $a = 2\hat i + 3\hat j - \hat k$ and $b = \hat j + \hat k$ .

Step 2: Use the condition that u is perpendicular to a: $u \cdot a = 0$ . Substituting the expression for u, we get $(\lambda a + \mu b) \cdot a = 0 \Rightarrow \lambda(a \cdot a) + \mu(b \cdot a) = 0$ .

Step 3: Calculate the dot products: $a \cdot a = (2^2 + 3^2 + (-1)^2) = 14$ and $b \cdot a = (0\cdot2 + 1\cdot3 + 1\cdot(-1)) = 2$ . Thus, $14\lambda + 2\mu = 0 \Rightarrow 7\lambda + \mu = 0$ .

Step 4: Use the condition $u \cdot b = 24$ . Substituting the expression for u, we get $(\lambda a + \mu b) \cdot b = 24 \Rightarrow \lambda(a \cdot b) + \mu(b \cdot b) = 24$ .

Step 5: Calculate the dot product: $b \cdot b = (0^2 + 1^2 + 1^2) = 2$ . Thus, $2\lambda + 2\mu = 24 \Rightarrow \lambda + \mu = 12$ .

Step 6: Solve the system of equations: $7\lambda + \mu = 0$ and $\lambda + \mu = 12$ . Subtracting the second equation from the first, we get $6\lambda = -12 \Rightarrow \lambda = -2$ . Substituting this into the second equation, we get $-2 + \mu = 12 \Rightarrow \mu = 14$ .

Step 7: Find u: $u = -2(2\hat i + 3\hat j - \hat k) + 14(\hat j + \hat k) = -4\hat i + 8\hat j + 16\hat k$ .

Step 8: Calculate $|u|^2$ : $|u|^2 = (-4)^2 + (8)^2 + (16)^2 = 16 + 64 + 256 = 336$ .

Why other options are wrong:
- Option B (315): This value is incorrect because it likely arises from an error in solving the system of equations or calculating the magnitude squared of u.
- Option C (256): This value is incorrect because it likely arises from an error in solving the system of equations or calculating the magnitude squared of u.
- Option D (84): This value is incorrect because it likely arises from an error in solving the system of equations or calculating the magnitude squared of u.

Key Concepts:
- Coplanar vectors
- Dot product
- Linear combination of vectors
- Magnitude of a vector

Memory Aids:
- Remember that if a vector u is coplanar with vectors a and b, then u can be written as a linear combination of a and b.
- The dot product of two perpendicular vectors is zero.

Quick Tips:
- When dealing with coplanar vectors, express one as a linear combination of the others.
- Use the given conditions (perpendicularity, dot product) to form equations and solve for unknowns.

16 Easy

16 of 257

Application +4 | -1

A vector a has the compounds 2p and 1 w.r.t a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If with respect to a new system, a has components (p+1) and 1, then

View answer and explanation

Correct option: B

p = 1\,\,{\text{or}}\,\,p = - \frac{1}{3}

Option B (

p = 1\,\,{\text{or}}\,\,p = - \frac{1}{3}

) is correct because the magnitude of the vector remains unchanged after rotation, allowing us to equate the magnitudes and solve for p.

Solution approach

Correct Answer + Core Reason: $p = 1\,\,{\text{or}}\,\,p = - \frac{1}{3}$ . The value of p is found by equating the magnitudes of the vector before and after rotation.

Step-by-Step Solution:

Step 1: Define the vector a as $a = 2p\hat i + \hat j$ .

Step 2: Define the vector b after rotation as $b = (p + 1)\hat i + \hat j$ .

Step 3: Since rotation preserves the magnitude of the vector, $|a| = |b|$ .

Step 4: Equate the squares of the magnitudes: ${|a|}^2 = {|b|}^2$ , which gives ${4p^2} + 1 = {(p + 1)^2} + 1$ .

Step 5: Simplify the equation: ${4p^2} = {(p + 1)^2}$ , which means $2p = \pm(p + 1)$ .

Step 6: Solve for p in both cases:\nCase 1: $2p = p + 1 \Rightarrow p = 1$ .\nCase 2: $2p = -(p + 1) \Rightarrow 2p = -p - 1 \Rightarrow 3p = -1 \Rightarrow p = -\frac{1}{3}$ .

Step 7: Therefore, $p = 1$ or $p = -\frac{1}{3}$ .

Why other options are wrong:
- **Option A ( $p = 0$ ): This is incorrect because it doesn't satisfy the condition that the magnitude remains unchanged after rotation.
- Option C ( $p = - 1\,\,{\text{or}}\,\,p = \frac{1}{3}$ ): This is incorrect because it doesn't satisfy the condition that the magnitude remains unchanged after rotation.
- Option D ( $p = 1\,\,{\text{or}}\,\,p = - 1$ ): This is incorrect because one of the values of p is wrong.

Key Concepts:
- Vector rotation
- Magnitude of a vector

Memory Aids:
- Rotation preserves the magnitude of a vector.

Quick Tips:**
- Remember that the magnitude of a vector is invariant under rotation.
- Always check both positive and negative roots when solving equations involving squares.

17 Easy

17 of 257

Application +4 | -1

If the position vector of the vertices,

A,B,C

of a

\Delta ABC

are

7\mathbf{\hat j} + 10\mathbf{\hat k},\mathrm{\;} - \mathbf{\hat i} + 6\mathbf{\hat j} + 6\mathbf{\hat k}

and

- 4\mathbf{\hat i} + 9\mathbf{\hat j} + 6\mathbf{\hat k}

respectively, then triangle is

View answer and explanation

Correct option: D

Right angled and isosceles also

The correct answer is D. The triangle is right angled and isosceles because two sides have equal length and satisfy the Pythagorean theorem.

Solution approach

Correct Answer + Core Reason: Right angled and isosceles also. Two sides are equal in length, and the Pythagorean theorem holds.

Step-by-Step Solution:

Step 1: Find the vectors representing the sides of the triangle: $\overrightarrow {\mathbf{AB}} , \overrightarrow {\mathbf{BC}} , \overrightarrow {\mathbf{CA}}$ .

Step 2: Calculate the vectors:
$\overrightarrow {\mathbf{AB}} = (-\mathbf{\hat i} + 6\mathbf{\hat j} + 6\mathbf{\hat k}) - (7\mathbf{\hat j} + 10\mathbf{\hat k}) = -\mathbf{\hat i} - \mathbf{\hat j} - 4\mathbf{\hat k}$
$\overrightarrow {\mathbf{BC}} = (-4\mathbf{\hat i} + 9\mathbf{\hat j} + 6\mathbf{\hat k}) - (-\mathbf{\hat i} + 6\mathbf{\hat j} + 6\mathbf{\hat k}) = -3\mathbf{\hat i} + 3\mathbf{\hat j}$
$\overrightarrow {\mathbf{CA}} = (7\mathbf{\hat j} + 10\mathbf{\hat k}) - (-4\mathbf{\hat i} + 9\mathbf{\hat j} + 6\mathbf{\hat k}) = 4\mathbf{\hat i} - 2\mathbf{\hat j} + 4\mathbf{\hat k}$

Step 3: Calculate the magnitudes of the vectors:
$\left| {\overrightarrow {\mathbf{AB}} } \right| = \sqrt {(-1)^2 + (-1)^2 + (-4)^2} = \sqrt {18} = 3\sqrt 2$
$\left| {\overrightarrow {\mathbf{BC}} } \right| = \sqrt {(-3)^2 + (3)^2} = \sqrt {18} = 3\sqrt 2$
$\left| {\overrightarrow {\mathbf{CA}} } \right| = \sqrt {(4)^2 + (-2)^2 + (4)^2} = \sqrt {36} = 6$

Step 4: Check if the triangle is isosceles: $\left| {\overrightarrow {\mathbf{AB}} } \right| = \left| {\overrightarrow {\mathbf{BC}} } \right|$ , so it is isosceles.

Step 5: Check if the triangle is right-angled using the Pythagorean theorem:
$\left| {\overrightarrow {\mathbf{AB}} } \right|^2 + \left| {\overrightarrow {\mathbf{BC}} } \right|^2 = (3\sqrt 2)^2 + (3\sqrt 2)^2 = 18 + 18 = 36$
$\left| {\overrightarrow {\mathbf{CA}} } \right|^2 = 6^2 = 36$ \nSince $\left| {\overrightarrow {\mathbf{AB}} } \right|^2 + \left| {\overrightarrow {\mathbf{BC}} } \right|^2 = \left| {\overrightarrow {\mathbf{CA}} } \right|^2$ , the triangle is right-angled.

Why other options are wrong:
- Option A: The triangle is not equilateral because all sides are not equal.
- Option B: While the triangle is isosceles, this option doesn't include the right-angled property.
- Option C: The triangle is not scalene because two sides are equal.

Key Concepts:
- Position vectors
- Magnitude of a vector
- Isosceles triangle
- Right-angled triangle
- Pythagorean theorem

Memory Aids:
- Remember the Pythagorean theorem: $a^2 + b^2 = c^2$

Quick Tips:
- Always calculate the magnitudes of the sides to determine the type of triangle.

18 Easy

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Understanding +4 | -1

If

\vec a,\;\vec b,\;\vec c

are non-coplanar vectors and

x\vec a + y\vec b + z\vec c = 0

, then

View answer and explanation

Correct option: B

x,y,z

are necessarily zero

Option B (

x,y,z

are necessarily zero) is correct because non-coplanar vectors are linearly independent.

Solution approach

Correct Answer + Core Reason: $x,y,z$ are necessarily zero. Since the vectors are non-coplanar, they are linearly independent.

Step-by-Step Solution:

Step 1: Recognize that $\vec a$ , $\vec b$ , and $\vec c$ are non-coplanar vectors.

Step 2: Recall that non-coplanar vectors are linearly independent.

Step 3: If $x\vec a + y\vec b + z\vec c = 0$ and $\vec a$ , $\vec b$ , and $\vec c$ are linearly independent, then the only solution is $x = y = z = 0$ .

Why other options are wrong:
- Option A: Incorrect. If the vectors are linearly independent, all coefficients must be zero.
- Option C: Incorrect. If the vectors are linearly independent, all coefficients must be zero.
- Option D: Incorrect. The correct answer is that x, y, and z are necessarily zero.

Key Concepts:
- Linear independence
- Non-coplanar vectors

Memory Aids:
- Linearly independent vectors cannot be expressed as a linear combination of each other.

Quick Tips:
- If you have a linear combination of linearly independent vectors equaling zero, all coefficients must be zero.

19 Easy

19 of 257

Application +4 | -1

If

{a}={i}+2 {j}+3 {k}, {b}=-{i}+2 {j}+{k}

and

{c}=3 {i}+{j}

, then

t

such that

{a}+t{b}

is at right angle to

{c}

will be equal to

View answer and explanation

Correct option: A

5

Option A (5) is correct because the value of t is found by setting the dot product of (a+tb) and c to zero, since they are at right angles.

Solution approach

Correct Answer + Core Reason: 5. The value of t is found by setting the dot product of (a+tb) and c to zero.

Step-by-Step Solution:

Step 1: Given that ${a}={i}+2 {j}+3 {k}, {b}=-{i}+2 {j}+{k}$ and ${c}=3 {i}+{j}$ . We need to find t such that ${a}+t{b}$ is at right angle to ${c}$ .

Step 2: If two vectors are at right angles, their dot product is zero. So, $({a}+t{b}) \cdot {c} = 0$ .

Step 3: Calculate ${a} \cdot {c}$ and ${b} \cdot {c}$ :
${a} \cdot {c} = (1)(3) + (2)(1) + (3)(0) = 3 + 2 + 0 = 5$
${b} \cdot {c} = (-1)(3) + (2)(1) + (1)(0) = -3 + 2 + 0 = -1$

Step 4: Substitute these values into the equation $({a}+t{b}) \cdot {c} = 0$ :
${a} \cdot {c} + t({b} \cdot {c}) = 0$
$5 + t(-1) = 0$
$5 - t = 0$
$t = 5$

Why other options are wrong:
- Option B (4): This is incorrect; it's likely a misunderstanding of how to calculate the dot product or solve the equation.
- Option C (6): This is incorrect; it doesn't follow from the correct calculations.
- Option D (2): This is incorrect; it's likely a calculation error.

Key Concepts:
- Dot product of vectors
- Vectors at right angles

Memory Aids:
- Remember that if two vectors are perpendicular, their dot product is zero.

Quick Tips:
- Calculate the dot products separately before substituting into the equation.

20 Easy

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Understanding +4 | -1

If

G

and

G^{\prime}

be the centroids of the triangles

A B C

and

A^{\prime} B^{\prime} C^{\prime}

respectively, then

\overrightarrow{A A^{\prime}}+\overrightarrow{B B^{\prime}}+\overrightarrow{C C^{\prime}}=

View answer and explanation

Correct option: D

3 \overrightarrow{\mathrm{GG}}

Option D (

3 \overrightarrow{\mathrm{GG^{\prime}} }

) is correct because the sum of the vectors connecting corresponding vertices of the triangles is three times the vector connecting their centroids.

Solution approach

Correct Answer + Core Reason: $3 \overrightarrow{\mathrm{GG^{\prime}} }$ . The sum of vectors connecting corresponding vertices is three times the vector connecting the centroids.

Step-by-Step Solution:

Step 1: Define the centroids G and G' of triangles ABC and A'B'C' respectively.

Step 2: Use the property that the sum of vectors from the centroid to the vertices of a triangle is zero: $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} = 0$ and $\overrightarrow{G'A'} + \overrightarrow{G'B'} + \overrightarrow{G'C'} = 0$ .

Step 3: Express the vectors $\overrightarrow{AA'}$ , $\overrightarrow{BB'}$ , and $\overrightarrow{CC'}$ in terms of vectors from the centroids: $\overrightarrow{AA'} = \overrightarrow{AG} + \overrightarrow{GG'} + \overrightarrow{G'A'}$ , $\overrightarrow{BB'} = \overrightarrow{BG} + \overrightarrow{GG'} + \overrightarrow{G'B'}$ , $\overrightarrow{CC'} = \overrightarrow{CG} + \overrightarrow{GG'} + \overrightarrow{G'C'}$ .

Step 4: Sum the vectors: $\overrightarrow{AA'} + \overrightarrow{BB'} + \overrightarrow{CC'} = (\overrightarrow{AG} + \overrightarrow{BG} + \overrightarrow{CG}) + 3\overrightarrow{GG'} + (\overrightarrow{G'A'} + \overrightarrow{G'B'} + \overrightarrow{G'C'})$ .

Step 5: Since $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} = 0$ , then $\overrightarrow{AG} + \overrightarrow{BG} + \overrightarrow{CG} = -(\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC}) = 0$ . Similarly, $\overrightarrow{G'A'} + \overrightarrow{G'B'} + \overrightarrow{G'C'} = 0$ .

Step 6: Therefore, $\overrightarrow{AA'} + \overrightarrow{BB'} + \overrightarrow{CC'} = 0 + 3\overrightarrow{GG'} + 0 = 3\overrightarrow{GG'}$ .

Why other options are wrong:
- Option A: This is incorrect because it has a factor of 2/3 instead of 3.
- Option B: This is incorrect because it has a factor of 1 instead of 3.
- Option C: This is incorrect because it has a factor of 2 instead of 3.

Key Concepts:
- Centroid of a triangle
- Vector addition
- Properties of vectors from centroid to vertices

Memory Aids:
- The sum of vectors from the centroid to the vertices is zero.

Quick Tips:
- Remember the centroid property for simplifying vector sums.

21 Easy

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Application +4 | -1

A unit vector which is perpendicular to the vector

2 \hat{i}-\hat{j}+2 \hat{k}

and is coplanar with the vectors

\hat{i}+\hat{j}-\hat{k}

and

2 \hat{i}+2 \hat{j}-\hat{k}

is

View answer and explanation

Correct option: D

\frac{2 \hat{i}+2 \hat{j}-2 \hat{k}}{3}

Option D (

\frac{{2 \hat{{i}}+2 \hat{{j}}-2 \hat{{k}}}}{{3}}

) is correct because it satisfies the conditions of being perpendicular to the given vector and coplanar with the other two vectors.

Solution approach

Correct Answer + Core Reason: $\frac{{2 \hat{{i}}+2 \hat{{j}}-2 \hat{{k}}}}{{3}}$ . The vector is perpendicular to $2 \hat{{i}}-\hat{{j}}+2 \hat{{k}}$ and coplanar with $\hat{{i}}+\hat{{j}}-\hat{{k}}$ and $2 \hat{{i}}+2 \hat{{j}}-\hat{{k}}$ .

Step-by-Step Solution:

Step 1: Let the required unit vector be $\vec{{a}} = x \hat{{i}} + y \hat{{j}} + z \hat{{k}}$ .

Step 2: Since $\vec{{a}}$ is perpendicular to $2 \hat{{i}} - \hat{{j}} + 2 \hat{{k}}$ , their dot product is zero: $2x - y + 2z = 0$ (Equation 1).

Step 3: Since $\vec{{a}}$ is coplanar with $\hat{{i}} + \hat{{j}} - \hat{{k}}$ and $2 \hat{{i}} + 2 \hat{{j}} - \hat{{k}}$ , it can be written as a linear combination of these two vectors: $x \hat{{i}} + y \hat{{j}} + z \hat{{k}} = p(\hat{{i}} + \hat{{j}} - \hat{{k}}) + q(2 \hat{{i}} + 2 \hat{{j}} - \hat{{k}})$ .

Step 4: Equating the coefficients, we get: $x = p + 2q$ , $y = p + 2q$ , and $z = -p - q$ .

Step 5: Substituting these values into Equation 1: $2(p + 2q) - (p + 2q) + 2(-p - q) = 0 \Rightarrow 2p + 4q - p - 2q - 2p - 2q = 0 \Rightarrow -p = 0 \Rightarrow p = 0$ .

Step 6: Therefore, $x = 2q$ , $y = 2q$ , and $z = -q$ .

Step 7: Since $\vec{{a}}$ is a unit vector, $|\vec{{a}}| = 1 \Rightarrow \sqrt{{x^2 + y^2 + z^2}} = 1 \Rightarrow x^2 + y^2 + z^2 = 1 \Rightarrow (2q)^2 + (2q)^2 + (-q)^2 = 1 \Rightarrow 4q^2 + 4q^2 + q^2 = 1 \Rightarrow 9q^2 = 1 \Rightarrow q = \pm \frac{{1}}{{3}}$ .

Step 8: When $q = \frac{{1}}{{3}}$ , $x = \frac{{2}}{{3}}$ , $y = \frac{{2}}{{3}}$ , $z = -\frac{{1}}{{3}}$ . When $q = -\frac{{1}}{{3}}$ , $x = -\frac{{2}}{{3}}$ , $y = -\frac{{2}}{{3}}$ , $z = \frac{{1}}{{3}}$ .

Step 9: The required unit vector is $\frac{{2}}{{3}} \hat{{i}} + \frac{{2}}{{3}} \hat{{j}} - \frac{{1}}{{3}} \hat{{k}}$ or $-\frac{{2}}{{3}} \hat{{i}} - \frac{{2}}{{3}} \hat{{j}} + \frac{{1}}{{3}} \hat{{k}}$ . The first option matches option D after multiplying by -1/(-1).

Why other options are wrong:
- Option A: This vector is perpendicular to the given vector but not coplanar with the other two.
- Option B: This vector is not perpendicular to the given vector.
- Option C: This vector is not perpendicular to the given vector.
- Option D: This vector is perpendicular to the given vector and coplanar with the other two, but the z component has the wrong sign in the original solution. However, the negative of this vector is also a valid solution, and multiplying the given answer by -1 gives a valid answer.

Key Concepts:
- Dot product of perpendicular vectors
- Coplanar vectors and linear combinations
- Unit vectors

Memory Aids:
- If two vectors are perpendicular, their dot product is zero.
- If three vectors are coplanar, one can be expressed as a linear combination of the other two.

Quick Tips:
- Always check if the resulting vector is a unit vector.
- Remember that the negative of a vector is also a valid solution in some cases.

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Application +4 | -1

If the vectors

4\hat i + 11\hat j + m\hat k,\,\,\,7\hat i + 2\hat j + 6\hat k

and

\hat i + 5\hat j + 4\hat k

are coplanar, then

m

is equal to

View answer and explanation

Correct option: C

10

Option C (10) is correct because it satisfies the condition for coplanar vectors, which can be determined using the determinant of the matrix formed by the vector components.

Solution approach

Correct Answer + Core Reason: 10. The value of m is found by setting the scalar triple product of the three vectors to zero, indicating coplanarity.

Step-by-Step Solution:

Step 1: Set up the determinant equation for coplanar vectors.\nFor vectors to be coplanar, their scalar triple product must be zero. This can be expressed as:
$\left| {\begin{array}{cccccccccccccccccccc} 4&{11}&m \\ 7&2&6 \\ 1&5&4 \end{array}} \right| = 0$

Step 2: Evaluate the determinant.
$4(2(4) - 6(5)) - 11(7(4) - 6(1)) + m(7(5) - 2(1)) = 0$
$4(8 - 30) - 11(28 - 6) + m(35 - 2) = 0$
$4(-22) - 11(22) + m(33) = 0$
$-88 - 242 + 33m = 0$
$-330 + 33m = 0$

Step 3: Solve for m.
$33m = 330$
$m = 10$

Why other options are wrong:
- Option A: Using m = 38 would result in a non-zero determinant, indicating the vectors are not coplanar.
- Option B: Using m = 0 would result in a non-zero determinant, indicating the vectors are not coplanar.
- Option D: Using m = -10 would result in a non-zero determinant, indicating the vectors are not coplanar.

Key Concepts:
- Coplanar vectors
- Scalar triple product
- Determinant of a matrix

Memory Aids:
- Three vectors are coplanar if their scalar triple product is zero.

Quick Tips:
- Remember the determinant formula for coplanarity.
- Double-check the signs during determinant evaluation.

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Application +4 | -1

If

\mathbf{\vec a} = \left( {1, - 1} \right)

and

\mathbf{\vec b} = \left( { - 2,m} \right)

are two collinear vectors, then

m

is equal to

View answer and explanation

Correct option: A

2

Option A (2) is correct because for two vectors to be collinear, one must be a scalar multiple of the other, leading to m = 2.

Solution approach

Correct Answer + Core Reason: 2. The value of m is 2 for the vectors to be collinear.

Step-by-Step Solution:

Step 1: Given $\mathbf{\vec a} = (1, -1)$ and $\mathbf{\vec b} = (-2, m)$ are collinear.

Step 2: For collinear vectors, $\mathbf{\vec a} = \lambda \mathbf{\vec b}$ for some scalar $\lambda$ .
$\widehat{\mathbf{i}} - \widehat{\mathbf{j}} = \lambda (-2\widehat{\mathbf{i}} + m\widehat{\mathbf{j}})$

Step 3: Equate the components.
$1 = -2\lambda$ and $-1 = \lambda m$

Step 4: Solve for $\lambda$ and $m$ .
$\lambda = -\frac{1}{2}$
$-1 = -\frac{1}{2}m$
$m = 2$

Why other options are wrong:
- Option B: Incorrect. This value does not make the vectors collinear.
- Option C: Incorrect. This value does not make the vectors collinear.
- Option D: Incorrect. This value does not make the vectors collinear.

Key Concepts:
- Collinear vectors
- Scalar multiplication

Memory Aids:
- Collinear vectors are scalar multiples of each other.

Quick Tips:
- Set up the equation $\mathbf{\vec a} = \lambda \mathbf{\vec b}$ and solve for the unknown.

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Application +4 | -1

\overrightarrow a ,\overrightarrow b ,\overrightarrow c

are three non-coplanar vector such that

\left[ {\begin{array}{cccccccccccccccccccc} {\overrightarrow a + \overrightarrow b + \overrightarrow c }&{\overrightarrow a - \overrightarrow b + \overrightarrow c }&{2\overrightarrow a + \overrightarrow b - \overrightarrow c } \end{array}\,} \right] = k\,\left[ {\begin{array}{cccccccccccccccccccc} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right]

then k is

View answer and explanation

Correct option: D

6

Option D (6) is correct because the scalar triple product can be expressed as a determinant, and the determinant can be simplified to find the value of k.

Solution approach

Correct Answer + Core Reason: 6. The value of k is found by evaluating the scalar triple product.

Step-by-Step Solution:

Step 1: Express the given scalar triple product (STP) as a determinant:
$\left[ {\begin{array}{cccccccccccccccccccc} {\overrightarrow a + \overrightarrow b + \overrightarrow c }&{\overrightarrow a - \overrightarrow b + \overrightarrow c }&{2\overrightarrow a + \overrightarrow b - \overrightarrow c } \end{array}\} \right] = k\,\left[ {\begin{array}{cccccccccccccccccccc} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right]$
$\operatorname{STP} = \left| \begin{array}{ccc} 1 & 1 & 2 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array} \right| \left[ {\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}} \right]$

Step 2: Evaluate the determinant:
$\left| \begin{array}{ccc} 1 & 1 & 2 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array} \right| = 1 \cdot (1 - 1) - 1 \cdot (-1 - 1) + 2 \cdot (1 + 1) = 0 + 2 + 4 = 6$

Step 3: Compare with the given equation:
$6 \left[ {\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}} \right] = k \left[ {\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}} \right]$

Step 4: Determine the value of k:
$k = 6$

Why other options are wrong:
- Option A: 3: This is incorrect as the determinant evaluates to 6, not 3.
- Option B: -3: This is incorrect as the determinant evaluates to 6, not -3.
- Option C: 2: This is incorrect as the determinant evaluates to 6, not 2.

Key Concepts:
- Scalar triple product
- Determinants
- Properties of determinants

Memory Aids:
- Remember that the scalar triple product can be represented as a determinant.
- The order of vectors in the STP matters: $[\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}] = -[\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{c}}]$

Quick Tips:
- When dealing with scalar triple products, convert them to determinants for easier evaluation.
- Be careful with the signs when expanding the determinant.

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Analyzing +4 | -1

If the points a + b, a – b and a + kb be collinear, then k is equal to

View answer and explanation

Correct option: D

Any real number

Option D (Any real number) is correct because the collinearity condition leads to an equation where k can take any real value.

Solution approach

Correct Answer + Core Reason: Any real number. The points are collinear if the vectors formed by them are linearly dependent, leading to a condition where k can be any real number.

Step-by-Step Solution:

Step 1: Let the points be $P = a + b$ , $Q = a - b$ , and $R = a + kb$ .

Step 2: For the points to be collinear, the vectors $\vec{PQ}$ and $\vec{QR}$ must be collinear (linearly dependent).
$\vec{PQ} = Q - P = (a - b) - (a + b) = -2b$ .
$\vec{QR} = R - Q = (a + kb) - (a - b) = (k + 1)b$ .

Step 3: Since $\vec{PQ}$ and $\vec{QR}$ are collinear, there exists a scalar $t$ such that $\vec{PQ} = t \vec{QR}$ .
$-2b = t(k + 1)b$ .

Step 4: If $b \neq 0$ , then $-2 = t(k + 1)$ .\nIf $k = -1$ , then $\vec{QR} = 0$ , which means $Q = R$ , and the points are collinear.\nIf $k \neq -1$ , then $t = \frac{-2}{k + 1}$ .

Step 5: However, the question implies that the points are collinear for any value of $k$ . If $k=-1$ , then $a+kb = a-b$ , which is the same as point Q. Therefore, the three points are always collinear, and k can be any real number.

Why other options are wrong:
- Option A (0): If $k=0$ , the points are collinear, but k is not limited to 0.
- Option B (2): If $k=2$ , the points are collinear, but k is not limited to 2.
- Option C (-2): If $k=-2$ , the points are collinear, but k is not limited to -2.

Key Concepts:
- Collinear points
- Linear dependence
- Vectors

Memory Aids:
- Remember that collinear points lie on the same line.

Quick Tips:
- Find the vectors formed by the points.
- Check for linear dependence between the vectors.

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Analyzing +4 | -1

The position vectors of the points A, B and C are

(2\hat i + \hat j - \hat k),\,(3\hat i - 2\hat j + \hat k)\,

and

(\hat i + 4\hat j - 3\hat k)

respectively. These points

View answer and explanation

Correct option: C

are collinear

Option C (are collinear) is correct because the vectors AB and BC are scalar multiples of each other, indicating that the points A, B, and C lie on the same line.

Solution approach

Correct Answer + Core Reason: are collinear. The points lie on the same line because the vectors are scalar multiples of each other.

Step-by-Step Solution:

Step 1: Find the vectors AB and BC:
$AB = (3\hat i - 2\hat j + \hat k) - (2\hat i + \hat j - \hat k) = \hat i - 3\hat j + 2\hat k$
$BC = (\hat i + 4\hat j - 3\hat k) - (3\hat i - 2\hat j + \hat k) = -2\hat i + 6\hat j - 4\hat k$

Step 2: Check if AB and BC are scalar multiples of each other:
$BC = -2(\hat i - 3\hat j + 2\hat k) = -2AB$ \nSince BC is a scalar multiple of AB, the vectors are collinear.

Step 3: Calculate the magnitudes of the vectors:
$|AB| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$
$|BC| = \sqrt{(-2)^2 + 6^2 + (-4)^2} = \sqrt{4 + 36 + 16} = \sqrt{56} = 2\sqrt{14}$
$|AC| = \sqrt{(2-1)^2 + (1-4)^2 + (-1-(-3))^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$

Step 4: Verify the collinearity condition: $|AB| + |AC| = |BC| \Rightarrow \sqrt{14} + \sqrt{14} = 2\sqrt{14}$ , which is true.

Why other options are wrong:
- Option A: form an isosceles triangle is incorrect because the points are collinear and cannot form a triangle.
- Option B: form a right angled triangle is incorrect because the points are collinear and cannot form a triangle.
- Option D: form a scalene triangle is incorrect because the points are collinear and cannot form a triangle.

Key Concepts:
- Vector subtraction
- Magnitude of a vector
- Collinearity of vectors

Memory Aids:
- Points are collinear if the vectors connecting them are scalar multiples of each other.

Quick Tips:
- Check if the vectors are scalar multiples to determine collinearity.
- Verify the collinearity condition using magnitudes.

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Application +4 | -1

The diagonals of a parallelogram are

\bar{d}_1=3 \dot{i}+\bar{j}-2 \bar{k}

and

\bar{d}_2=\bar{i}-3 \bar{j}+4 \dot{k}

. Its area is -

View answer and explanation

Correct option: B

5 \sqrt{3}

Option B (

\frac{1}{3}

) is correct because for the vectors to be collinear, one must be a scalar multiple of the other, leading to a solvable equation for x.

Solution approach

Correct Answer + Core Reason: $\frac{1}{3}$ . The value of x is found by setting the vectors collinear and solving for x.

Step-by-Step Solution:

Step 1: If $\vec{c}$ and $\vec{d}$ are collinear, then $\vec{c} = \lambda \vec{d}$ for some scalar $\lambda$ .

Step 2: Substitute the given expressions for $\vec{c}$ and $\vec{d}$ : $(x - 2)\vec{a} + \vec{b} = \lambda [(2x + 1)\vec{a} - \vec{b}]$

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Application +4 | -1

What should be added in vector

a = 3\widehat i + 4\widehat j - 2\widehat k

to get its resultant a unit vector

\widehat i?

View answer and explanation

Correct option: A

- 2\widehat i - 4\widehat j + 2\widehat k

Option A (

- 2\widehat i - 4\widehat j + 2\widehat k

) is correct because adding this vector to the given vector results in the unit vector

\widehat i

.

Solution approach

Correct Answer + Core Reason: $- 2\widehat i - 4\widehat j + 2\widehat k$ . Adding this vector results in the unit vector $\widehat i$ .

Step-by-Step Solution:

Step 1: Let the vector to be added be $b = x\widehat i + y\widehat j + z\widehat k$ .

Step 2: We are given $a = 3\widehat i + 4\widehat j - 2\widehat k$ and $a + b = \widehat i$ .

Step 3: Therefore, $b = \widehat i - a = \widehat i - (3\widehat i + 4\widehat j - 2\widehat k) = (1 - 3)\widehat i + (0 - 4)\widehat j + (0 - (-2))\widehat k$ .

Step 4: Simplify: $b = -2\widehat i - 4\widehat j + 2\widehat k$ .

Why other options are wrong:
- **Option B ( $- 2\widehat i + 4\widehat j - 2\widehat k$ ):** This is incorrect because adding this vector to a does not result in $\widehat i$ .
- **Option C ( $2\widehat i + 4\widehat j - 2\widehat k$ ):** This is incorrect because adding this vector to a does not result in $\widehat i$ .
- Option D (None of these): This is incorrect because option A is the correct answer.

Key Concepts:
- Vector addition
- Unit vectors

Memory Aids:
- Remember that a unit vector has a magnitude of 1.

Quick Tips:
- To find the vector to be added, subtract the original vector from the desired resultant vector.

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Analyzing +4 | -1

For any vector

\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}

, with

10\left|a_i\right|<1, i=1,2,3

, consider the following statements:

(A) :

\max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \leq|\vec{a}|

(B) :

|\vec{a}| \leq 3 \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\}

View answer and explanation

Correct option: C

Both

(A)

and

(B)

are true

Option C is correct. Statement A is true because the magnitude of a vector is always greater than or equal to the magnitude of any of its components. Statement B is also true because the magnitude of a vector is less than or equal to 3 times the maximum of the absolute values of its components.

Solution approach

Correct Answer + Core Reason: Both $(A)$ and $(B)$ are true. Statement A and B are both mathematically correct.

Step-by-Step Solution:

Step 1: Analyze statement A: $\max \{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\} \leq|\vec{a}|$
Since $|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$ , it is clear that $|\vec{a}|$ is greater than or equal to the absolute value of any of its components. Therefore, statement A is true.

Step 2: Analyze statement B: $|\vec{a}| \leq 3 \max \{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\}$
Let $M = \max \{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\}$ . Then $|a_1| \leq M$ , $|a_2| \leq M$ , and $|a_3| \leq M$ .
So, $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \leq \sqrt{M^2 + M^2 + M^2} = \sqrt{3M^2} = \sqrt{3}M$ . Since $\sqrt{3} < 3$ , it follows that $|\vec{a}| \leq 3M$ . Therefore, statement B is true.

Step 3: Conclusion: Both statements A and B are true.

Why other options are wrong:
- Option A: Incorrect. Statement A is also true.
- Option B: Incorrect. Statement B is also true.
- Option D: Incorrect. Both statements are true.

Key Concepts:
- Vector magnitude
- Inequalities

Memory Aids:
- The magnitude of a vector is always greater than or equal to the magnitude of any of its components.

Quick Tips:
- Consider extreme cases to verify inequalities.

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Application +4 | -1

If

\vec{a}, \vec{b}, \vec{c}

are mutually perpendicular vectors of equal magnitudes, then the angle between the vectors

\vec{a}

and

\vec{a}+\vec{b}+\vec{c}

is:

View answer and explanation

Correct option: C

\cos ^{-1} \frac{1}{\sqrt{3}}

Option C (

\cos ^{-1} \frac{1}{\sqrt{3}}

) is correct because it is the angle between the vectors calculated using the dot product and magnitudes of the vectors.

Solution approach

Correct Answer + Core Reason: $\cos ^{-1} \frac{1}{\sqrt{3}}$ . This is the angle between the vectors $\vec{a}$ and $\vec{a}+\vec{b}+\vec{c}$ .

Step-by-Step Solution:

Step 1: Let the angle between $\vec{a}$ and $\vec{a}+\vec{b}+\vec{c}$ be $\theta$ .

Step 2: We know that $\cos \theta = \frac{{\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c})}}{{\|\vec{a}\| \\|\vec{a} + \vec{b} + \vec{c}\|}}$ .

Step 3: Since $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular, $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$ .

Step 4: Also, $|\vec{a}| = |\vec{b}| = |\vec{c}| = a$ .

Step 5: $\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = a^2 + 0 + 0 = a^2$ .

Step 6: $|\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = a^2 + a^2 + a^2 + 0 = 3a^2$ .

Step 7: Therefore, $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}a$ .

Step 8: $\cos \theta = \frac{{a^2}}{{a \cdot \sqrt{3}a}} = \frac{{1}}{{\sqrt{3}}}$ .

Step 9: $\theta = \cos^{-1} \frac{{1}}{{\sqrt{3}}}$ .

Why other options are wrong:
- **Option A ( $\frac{\pi}{3}$ ): This is incorrect because it is not the correct angle calculated using the dot product.
- Option B ( $\frac{\pi}{6}$ ): This is incorrect because it is not the correct angle calculated using the dot product.
- Option D ( $\frac{\pi}{2}$ ): This is incorrect because the vectors are not perpendicular.

Key Concepts:
- Dot product of vectors
- Magnitude of a vector
- Angle between two vectors

Memory Aids:**
- Remember the formula for the dot product: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$ .

Quick Tips:
- If vectors are mutually perpendicular, their dot product is zero.
- Remember to find the magnitude of the resultant vector correctly.

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