If $ABCD$ be a parallelogram and $M$ be the point of intersection of the diagonals. If $O$ is any point, then $\overrightarrow {{\bf{OA}}} + \overrightarrow {{\bf{OB}}} + \overrightarrow {{\bf{OC}}} + \overrightarrow {{\bf{OD}}} $ is

$4\overrightarrow {{\bf{OM}}} $

$3\overrightarrow {{\bf{OM}}} $

$2\overrightarrow {{\bf{OM}}} $

$ABCD$ is a quadrilateral , $P,Q$ are the mid points of $\overrightarrow {{\bf{BC}}} $ and $\overrightarrow {{\bf{AD}}} $, then $\overrightarrow {{\bf{AB}}} + \overrightarrow {{\bf{DC}}} $ is equal to

$2\overrightarrow {{\bf{QP}}} $

$3\;\overrightarrow {{\bf{QP}}} $

$4\overrightarrow {{\bf{QP}}} $

Let ${\bf{\vec a}}\;,{\bf{\vec b}},,\overrightarrow {{\bf{c}}\;} \;\;$ be three non-zero vectors such that no two of these are collinear. If the vector ${\bf{\vec a}} + 2\;{\bf{\vec b}}$ is collinear with $\overrightarrow {{\bf{c}}\;} \;,$ then $\overrightarrow {{\bf{a}}\;} + 2\overrightarrow {{\bf{b}}\;} + 6\;\overrightarrow {{\bf{c}}\;} \;\;{\rm{equals}}\;\;\;$

$\lambda \;{\bf{\vec c}}\;\left( {\lambda \ne 0,\;{\rm{a\;scalar}}} \right)$

$\lambda \;{\bf{\vec a}}\;\left( {\lambda \ne 0,\;{\rm{a\;scalar}}} \right)$

$\lambda \;{\bf{\vec b}}\;\left( {\lambda \ne 0,\;{\rm{a\;scalar}}} \right)$

ADDITION OF VECTORS MCQs for JEE MATHEMATICS — Easy, Page 3

61 Easy

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Analyzing +4 | -1

Let

a, b, c

denote the lengths of the sides of

a

triangle such that

(a-b) \mathbf{u}+(b-c) \mathbf{v}+(c-a)(\mathbf{u} \times \mathbf{v})=0

for any two non-collinear vectors

u

and

\mathbf{v}

, then the triongle is

View answer and explanation

Correct option: B

equilateral

Option B (equilateral) is correct because for the given vector equation to hold for any non-collinear vectors u and v, the coefficients must be zero, implying a=b=c.

Solution approach

Correct Answer + Core Reason: equilateral. The coefficients of the non-collinear vectors must be zero, leading to a=b=c.

Step-by-Step Solution:

Step 1: Given $(a-b)\mathbf{{u}} + (b-c)\mathbf{{v}} + (c-a)(\mathbf{{u}} \times \mathbf{{v}}) = 0$ for any two non-collinear vectors $\mathbf{{u}}$ and $\mathbf{{v}}$ .

Step 2: Since $\mathbf{{u}}$ , $\mathbf{{v}}$ , and $(\mathbf{{u}} \times \mathbf{{v}})$ are non-coplanar, their coefficients must be zero for the equation to hold.

Step 3: Therefore, $a - b = 0$ , $b - c = 0$ , and $c - a = 0$ .

Step 4: From $a - b = 0$ , we have $a = b$ .

Step 5: From $b - c = 0$ , we have $b = c$ .

Step 6: From $c - a = 0$ , we have $c = a$ .

Step 7: Thus, $a = b = c$ .

Step 8: If $a = b = c$ , the triangle is equilateral.

Why other options are wrong:
- Option A (right angled): This is incorrect because a right-angled triangle requires $a^2 + b^2 = c^2$ (or some permutation thereof), which is not implied by $a=b=c$ .
- Option C (isosceles): This is incorrect because an isosceles triangle only requires two sides to be equal, while this problem implies all three sides are equal.
- Option D (scalene): This is incorrect because a scalene triangle requires all three sides to be different, which contradicts $a=b=c$ .

Key Concepts:
- Linear Independence of Vectors
- Properties of Triangles

Memory Aids:
- If a linear combination of non-coplanar vectors is zero, then all coefficients must be zero.

Quick Tips:
- Recognize when a vector equation implies that the coefficients of linearly independent vectors must be zero.
- Remember the definitions of different types of triangles.

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Analyzing +4 | -1

If

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

are three non-coplanar vectors such that

\vec a + \;\vec b + \;\vec c = \alpha \vec d

and

\vec b + \vec c + \vec d = \beta \vec a

, then

\vec a + \vec b + \vec c + \vec d

is equal to

View answer and explanation

Correct option: A

{\vec 0}

Option A (

\vec 0

) is correct because the given conditions lead to a contradiction unless the sum of the vectors is zero.

Solution approach

Correct Answer + Core Reason: $\vec{0}$ . The given conditions imply that the vectors are coplanar, leading to a contradiction unless their sum is zero.

Step-by-Step Solution:

Step 1: Write down the given equations:
$\vec a + \vec b + \vec c = \alpha \vec d$ (1)
$\vec b + \vec c + \vec d = \beta \vec a$ (2)

Step 2: Add $\vec d$ to equation (1) and $\vec a$ to equation (2):
$\vec a + \vec b + \vec c + \vec d = (\alpha + 1)\vec d$ (3)
$\vec a + \vec b + \vec c + \vec d = (\beta + 1)\vec a$ (4)

Step 3: Equate equations (3) and (4):
$(\alpha + 1)\vec d = (\beta + 1)\vec a$ (5)

Step 4: If $\alpha \neq -1$ , then $\vec d = \frac{\beta + 1}{\alpha + 1} \vec a$ .\nSubstitute this into equation (1): $\vec a + \vec b + \vec c = \alpha \frac{\beta + 1}{\alpha + 1} \vec a$
$\Rightarrow (1 - \frac{\alpha(\beta + 1)}{\alpha + 1}) \vec a + \vec b + \vec c = 0$

Step 5: This implies that $\vec a, \vec b, \vec c$ are coplanar, which contradicts the given condition that they are non-coplanar. Therefore, the assumption $\alpha \neq -1$ must be false.\nHence, $\alpha = -1$ .

Step 6: Substitute $\alpha = -1$ into equation (1):
$\vec a + \vec b + \vec c = - \vec d$
$\Rightarrow \vec a + \vec b + \vec c + \vec d = \vec 0$

Why other options are wrong:
- Option B: This is incorrect because it only considers one of the given equations and doesn't account for the non-coplanarity condition.
- Option C: This is incorrect because it only considers one of the given equations and doesn't account for the non-coplanarity condition.
- Option D: This is incorrect because it doesn't account for the non-coplanarity condition and incorrectly combines the equations.

Key Concepts:
- Coplanar vectors
- Linear dependence
- Vector addition

Memory Aids:
- If three vectors are non-coplanar, any linear combination equaling zero implies all coefficients are zero.

Quick Tips:
- Look for contradictions when dealing with coplanarity or non-coplanarity conditions.
- Try to express one vector in terms of the others to simplify the equations.

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Understanding +4 | -1

For two particular vectors

\mathbf{A}

and

\mathbf{B}

, it is known that

A \times B=B \times A

. What must be true about the two vectors?

View answer and explanation

Correct option: C

One of the two vectors is a scalar multiple of the other vector

The correct answer is C (One of the two vectors is a scalar multiple of the other vector) because if

\mathbf{A} \times \mathbf{B} = \mathbf{B} \times \mathbf{A}

, then

\mathbf{A}

and

\mathbf{B}

must be parallel or one of them is a zero vector, meaning one is a scalar multiple of the other.

Solution approach

Correct Answer + Core Reason: One of the two vectors is a scalar multiple of the other vector. If $\mathbf{A} \times \mathbf{B} = \mathbf{B} \times \mathbf{A}$ , then $\mathbf{A}$ and $\mathbf{B}$ are parallel or one of them is zero.

Step-by-Step Solution:

Step 1: Given $\mathbf{A} \times \mathbf{B} = \mathbf{B} \times \mathbf{A}$ .

Step 2: We know that $\mathbf{B} \times \mathbf{A} = -(\mathbf{A} \times \mathbf{B})$ .

Step 3: Therefore, $\mathbf{A} \times \mathbf{B} = -(\mathbf{A} \times \mathbf{B})$ , which implies $2(\mathbf{A} \times \mathbf{B}) = 0$ , so $\mathbf{A} \times \mathbf{B} = 0$ .

Step 4: The cross product of two vectors is zero if and only if the vectors are parallel or one of them is the zero vector.

Step 5: If the vectors are parallel, one is a scalar multiple of the other.

Why other options are wrong:
- Option A: Atleast one of the two vectors must be the zero vector. While this is a possibility, it's not the only condition. The vectors could also be parallel.
- Option B: $\mathbf{A} \times \mathbf{B}=\mathbf{B} \times \mathbf{A}$ is true for any two vectors. This is incorrect because $\mathbf{A} \times \mathbf{B} = -(\mathbf{B} \times \mathbf{A})$ .
- Option D: The two vectors must be perpendicular to each other. This is incorrect because if they are perpendicular, their cross product is non-zero.

Key Concepts:
- Cross product properties
- Parallel vectors
- Zero vector

Memory Aids:
- Remember that $\mathbf{A} \times \mathbf{B} = |A||B|\sin(\theta)\mathbf{n}$ , where $\theta$ is the angle between the vectors. If $\sin(\theta) = 0$ , then $\theta = 0$ or $\theta = \pi$ , meaning the vectors are parallel.

Quick Tips:
- When dealing with cross products, remember the anti-commutative property: $\mathbf{A} \times \mathbf{B} = -(\mathbf{B} \times \mathbf{A})$ .

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Application +4 | -1

If

a = 2\widehat i + 5\widehat j

and

b = 2\widehat i - \widehat j,

then the unit vector along

a + b

will be

View answer and explanation

Correct option: D

\frac{{\widehat i + \widehat j}}{{\sqrt 2 }}

Option D (

\frac{{\widehat i + \widehat j}}{{\sqrt 2 }}

) is correct because it represents the unit vector in the direction of the sum of the given vectors.

Solution approach

Correct Answer + Core Reason: $\frac{{\widehat i + \widehat j}}{{\sqrt 2 }}$ . This is the unit vector along the sum of the given vectors.

Step-by-Step Solution:

Step 1: Given vectors $a = 2\widehat i + 5\widehat j$ and $b = 2\widehat i - \widehat j$ .

Step 2: Find the sum of the vectors: $a + b = (2 + 2)\widehat i + (5 - 1)\widehat j = 4\widehat i + 4\widehat j$ .

Step 3: Calculate the magnitude of the sum: $|a + b| = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ .

Step 4: Find the unit vector along $a + b$ : $\frac{a + b}{|a + b|} = \frac{4\widehat i + 4\widehat j}{4\sqrt{2}} = \frac{\widehat i + \widehat j}{\sqrt{2}}$ .

Why other options are wrong:
- **Option A ( $\frac{{\widehat i - \widehat j}}{{\sqrt 2 }}$ ): This is incorrect because it has the wrong sign for the j component.
- Option B ( $\widehat i + \widehat j$ ): This is incorrect because it is not a unit vector.
- Option C ( $\sqrt 2 (\widehat i + \widehat j)$ ): This is incorrect because it is not a unit vector.

Key Concepts:
- Vector addition
- Magnitude of a vector
- Unit vector

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

Quick Tips:**
- Find the sum of the vectors first, then divide by its magnitude to get the unit vector.

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Analyzing +4 | -1

Let a, b and c be three non-zero vectors which are pairwise non-collinear. If

a + 3b

collinear with c and

b + 2c

is collinear with a, then

a + 3b + 6c

is

View answer and explanation

Correct option: D

0

Option D (0) is correct because the collinearity conditions lead to a linear combination of a, b, and c equaling the zero vector.

Solution approach

Correct Answer + Core Reason: 0. The given collinearity conditions imply that $a + 3b + 6c = 0$ .

Step-by-Step Solution:

Step 1: Since a + 3b is collinear with c, we can write $a + 3b = \lambda c$ for some scalar $\lambda$ .

Step 2: Since b + 2c is collinear with a, we can write $b + 2c = \mu a$ for some scalar $\mu$ .

Step 3: From the first equation, $a + 3b + 6c = \lambda c + 6c = (\lambda + 6)c$ .

Step 4: From the second equation, $a + 3b + 6c = a + 3(\mu a - 2c) + 6c = (1 + 3\mu)a$ .

Step 5: Equating the two expressions, we have $(\lambda + 6)c = (1 + 3\mu)a$ .

Step 6: Since a and c are non-collinear, we must have $\lambda + 6 = 0$ and $1 + 3\mu = 0$ .

Step 7: Thus, $\lambda = -6$ and $\mu = -\frac{1}{3}$ .

Step 8: Substituting $\lambda = -6$ into $a + 3b = \lambda c$ , we get $a + 3b = -6c$ , so $a + 3b + 6c = 0$ .

Why other options are wrong:
- Option A: $a + c$ : Incorrect, the correct linear combination equals the zero vector.
- Option B: $a$ : Incorrect, the correct linear combination equals the zero vector.
- Option C: $c$ : Incorrect, the correct linear combination equals the zero vector.

Key Concepts:
- Collinearity of vectors
- Linear combination of vectors

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

Quick Tips:
- Use the collinearity conditions to express the vectors in terms of scalar multiples.

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Application +4 | -1

Let

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

are three non-coplanar vectors such that

\overrightarrow {{r_1}} = \vec a - \vec b + \vec c,\;\overrightarrow {{r_2}} = \vec b + \vec c - \vec a,\;\overrightarrow {{r_3}} = \vec c + \vec a + \vec b,\;\vec r = 2\vec a - 3\vec b + 4\vec c

If

\vec r = {\lambda _1}\overrightarrow {{r_1}} + {\lambda _2}\overrightarrow {{r_2}} + {\lambda _3}\overrightarrow {{r_3}}

, then

View answer and explanation

Correct option: B

{\lambda _1} + {\lambda _3} = 3

The correct answer is B. By expressing

\vec r

as a linear combination of

\overrightarrow {{r_1}}

,

\overrightarrow {{r_2}}

, and

\overrightarrow {{r_3}}

and solving the resulting system of equations, we find that

{\lambda _1} + {\lambda _3} = 3

.

Solution approach

Correct Answer + Core Reason: ${\lambda _1} + {\lambda _3} = 3$ . This is found by solving the system of equations resulting from the linear combination.

Step-by-Step Solution:

Step 1: Express $\vec r$ as a linear combination of $\overrightarrow {{r_1}}$ , $\overrightarrow {{r_2}}$ , and $\overrightarrow {{r_3}}$ : $\vec r = {\lambda _1}\overrightarrow {{r_1}} + {\lambda _2}\overrightarrow {{r_2}} + {\lambda _3}\overrightarrow {{r_3}}$ .

Step 2: Substitute the given vectors: $2\vec a - 3\vec b + 4\vec c = {\lambda _1}(\vec a - \vec b + \vec c) + {\lambda _2}(\vec b + \vec c - \vec a) + {\lambda _3}(\vec c + \vec a + \vec b)$ .

Step 3: Group the terms with $\vec a$ , $\vec b$ , and $\vec c$ : $2\vec a - 3\vec b + 4\vec c = (\lambda_1 - \lambda_2 + \lambda_3)\vec a + (-\lambda_1 + \lambda_2 + \lambda_3)\vec b + (\lambda_1 + \lambda_2 + \lambda_3)\vec c$ .

Step 4: Equate the coefficients of $\vec a$ , $\vec b$ , and $\vec c$ :
$\lambda_1 - \lambda_2 + \lambda_3 = 2$
$-\lambda_1 + \lambda_2 + \lambda_3 = -3$
$\lambda_1 + \lambda_2 + \lambda_3 = 4$

Step 5: Solve the system of equations. Adding the first two equations gives $2\lambda_3 = -1$ , so $\lambda_3 = -\frac{1}{2}$ .

Step 6: Adding the first and third equations gives $2\lambda_1 + 2\lambda_3 = 6$ , so $\lambda_1 + \lambda_3 = 3$ .

Why other options are wrong:
- Option A: ${\lambda _1} = 7$ is incorrect based on the system of equations.
- Option C: ${\lambda _1} + {\lambda _2} + {\lambda _3} = 3$ is incorrect; it should be 4.
- Option D: ${\lambda _3} + {\lambda _2} = 2$ is incorrect based on the solved values.

Key Concepts:
- Linear combination of vectors
- System of equations
- Non-coplanar vectors

Memory Aids:
- Remember to equate coefficients of corresponding vectors.

Quick Tips:
- When dealing with linear combinations, always equate coefficients to form a system of equations.

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Application +4 | -1

If the vectors

\vec a = 2\hat i + 3\hat j + 6\hat k

are collinear and

\left| {\vec b} \right| = 21

, then

\vec b =

View answer and explanation

Correct option: A

\pm 3\left( {2\hat i + 3\hat j + 6\hat k} \right)

Option A (

\pm 3\left( {2\hat i + 3\hat j + 6\hat k} \right)

) is correct because it represents a vector collinear with

\vec a

and having a magnitude of 21.

Solution approach

Correct Answer + Core Reason: $\pm 3\left( {2\hat i + 3\hat j + 6\hat k} \right)$ . The vector is found using the collinearity condition and the given magnitude.

Step-by-Step Solution:

Step 1: Since $\vec a$ and $\vec b$ are collinear, $\vec b = \lambda \vec a = \lambda (2\hat i + 3\hat j + 6\hat k)$ .

Step 2: Given $\left| {\vec b} \right| = 21$ , we have $\left| {\lambda (2\hat i + 3\hat j + 6\hat k)} \right| = 21$ .

Step 3: Simplify: $\left| \lambda \right| \sqrt {2^2 + 3^2 + 6^2} = 21 \Rightarrow \left| \lambda \right| \sqrt {4 + 9 + 36} = 21 \Rightarrow \left| \lambda \right| \sqrt {49} = 21 \Rightarrow \left| \lambda \right| (7) = 21 \Rightarrow \left| \lambda \right| = 3 \Rightarrow \lambda = \pm 3$ .

Step 4: Therefore, $\vec b = \pm 3(2\hat i + 3\hat j + 6\hat k) = \pm (6\hat i + 9\hat j + 18\hat k)$ .

Why other options are wrong:
- Option B: $\pm \left( {2\hat i + 3\hat j - 6\hat k} \right)$ is incorrect because its magnitude is not 21.
- Option C: $\pm 21\left( {2\hat i + 3\hat j + 6\hat k} \right)$ is incorrect because $\vec b$ would not be collinear with $\vec a$ after normalization.
- Option D: $\pm 21\left( {\hat i + \hat j + \hat k} \right)$ is incorrect because it is not collinear with $\vec a$ and its magnitude is not 21.

Key Concepts:
- Collinear vectors
- Magnitude of a vector

Memory Aids:
- Two vectors are collinear if one is a scalar multiple of the other.
- The magnitude of a vector $\vec a = x\hat i + y\hat j + z\hat k$ is $\sqrt {x^2 + y^2 + z^2}$ .

Quick Tips:
- Always check if the final vector satisfies both the collinearity and magnitude conditions.
- Remember that if two vectors are collinear, they are parallel.

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Application +4 | -1

If the vector b is collinear with the vector

a = (2\sqrt 2 , - 1,4)

and

\left| b \right| = 0

,then

View answer and explanation

Correct option: C

2a \pm b = 0

Option C (

2a \pm b = 0

) is correct because it represents the relationship between vectors a and b when b is collinear with a and |b| = 10.

Solution approach

Correct Answer + Core Reason: $2a \pm b = 0$ . This relationship holds when b is collinear with a and |b| = 10.

Step-by-Step Solution:

Step 1: Since b is collinear with a, we can write $b = \lambda a$ , where $\lambda$ is a scalar.
$b = \lambda (2\sqrt{2}\hat i - \hat j + 4\hat k) = 2\sqrt{2}\lambda \hat i - \lambda \hat j + 4\lambda \hat k$

Step 2: Given that $|b| = 10$ , we can find the value of $\lambda$ .
$|b| = \sqrt{(2\sqrt{2}\lambda)^2 + (-\lambda)^2 + (4\lambda)^2} = 10$
$\sqrt{8\lambda^2 + \lambda^2 + 16\lambda^2} = 10$
$\sqrt{25\lambda^2} = 10$
$5|\lambda| = 10$
$|\lambda| = 2$
$\lambda = \pm 2$

Step 3: Substitute the value of $\lambda$ back into the equation $b = \lambda a$ .
$b = \pm 2a$
$2a \pm b = 0$

Why other options are wrong:
- Option A: $a \pm b = 0$ : This is incorrect because the magnitudes of a and b are different.
- Option B: $a \pm 2b = 0$ : This is incorrect because the scalar multiple is on the wrong vector.
- Option D: None of these: This is incorrect because option C is the correct answer.

Key Concepts:
- Collinear vectors
- Magnitude of a vector

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

Quick Tips:
- Remember to consider both positive and negative values for the scalar multiple when dealing with collinear vectors.

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Analyzing +4 | -1

The lines

x=a y-1=z-2

and

x=3 y-2=b z-2,(a b \neq 0)

are coplanar, if:

View answer and explanation

Correct option: B

b=1, a \in R-\{0\}

Option B (b=1, a ∈ R-{0}) is correct because it satisfies the coplanarity condition derived from setting the scalar triple product of the direction vectors and the vector connecting the lines to zero.

Solution approach

Correct Answer + Core Reason: $b=1, a \in R - \{0\}$ . This condition ensures the lines are coplanar.

Step-by-Step Solution:

Step 1: Rewrite the equations of the lines in parametric form.\nLine 1: $x = ay - 1 = z - 2 \Rightarrow \frac{{x}}{{1}} = \frac{{y - \frac{{1}}{{a}}}}{{\frac{{1}}{{a}}}} = \frac{{z - 2}}{{1}}$ \nLine 2: $x = 3y - 2 = bz - 2 \Rightarrow \frac{{x}}{{1}} = \frac{{y - \frac{{2}}{{3}}}}{{\frac{{1}}{{3}}}} = \frac{{z - \frac{{2}}{{b}}}}{{\frac{{1}}{{b}}}}$

Step 2: Find the direction vectors and a point on each line.\nLine 1: Direction vector $\vec{{d_1}} = <1, \frac{{1}}{{a}}, 1>$ , Point $P_1 = (0, \frac{{1}}{{a}}, 2)$ \nLine 2: Direction vector $\vec{{d_2}} = <1, \frac{{1}}{{3}}, \frac{{1}}{{b}}>$ , Point $P_2 = (0, \frac{{2}}{{3}}, \frac{{2}}{{b}})$

Step 3: Find the vector connecting the two points.
$\vec{{P_1P_2}} = <0, \frac{{2}}{{3}} - \frac{{1}}{{a}}, \frac{{2}}{{b}} - 2> = <0, \frac{{2a - 3}}{{3a}}, \frac{{2 - 2b}}{{b}}>$

Step 4: Apply the coplanarity condition (scalar triple product equals zero).
$\vec{{P_1P_2}} \cdot (\vec{{d_1}} \times \vec{{d_2}}) = 0$
$\begin{{vmatrix}} 0 & \frac{{2a - 3}}{{3a}} & \frac{{2 - 2b}}{{b}} \\ 1 & \frac{{1}}{{a}} & 1 \\ 1 & \frac{{1}}{{3}} & \frac{{1}}{{b}} \end{{vmatrix}} = 0$
$0 - \frac{{2a - 3}}{{3a}} (\frac{{1}}{{b}} - 1) + \frac{{2 - 2b}}{{b}} (\frac{{1}}{{3}} - \frac{{1}}{{a}}) = 0$
$-\frac{{2a - 3}}{{3a}} (\frac{{1 - b}}{{b}}) + \frac{{2 - 2b}}{{b}} (\frac{{a - 3}}{{3a}}) = 0$
$\frac{{(2a - 3)(b - 1)}}{{3ab}} + \frac{{2(b - 1)(a - 3)}}{{3ab}} = 0$
$(b - 1) [(2a - 3) + 2(a - 3)] = 0$
$(b - 1) [2a - 3 + 2a - 6] = 0$
$(b - 1) (4a - 9) = 0$ \nEither $b = 1$ or $a = \frac{{9}}{{4}}$ .

Step 5: Analyze the solutions.\nIf $b = 1$ , then $a$ can be any real number except 0. If $a = \frac{{9}}{{4}}$ , then $b$ can be any real number except 0.\nHowever, the options suggest $b=1$ and $a \in R - \{0\}$ .

Why other options are wrong:
- Option A (a=2, b=2): This doesn't satisfy the coplanarity condition.
- Option C (a=1, b ∈ R-{0}): This doesn't satisfy the coplanarity condition.
- Option D (a=2, b=3): This doesn't satisfy the coplanarity condition.

Key Concepts:
- Coplanar lines
- Direction vectors
- Scalar triple product

Memory Aids:
- The scalar triple product of three vectors is zero if they are coplanar.

Quick Tips:
- Rewrite the equations in parametric form to find direction vectors and points.
- Apply the coplanarity condition using the scalar triple product.

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Understanding +4 | -1

If

ABCD

be a parallelogram and

M

be the point of intersection of the diagonals. If

O

is any point, then

\overrightarrow {\mathbf{OA}} + \overrightarrow {\mathbf{OB}} + \overrightarrow {\mathbf{OC}} + \overrightarrow {\mathbf{OD}}

is

View answer and explanation

Correct option: B

4\overrightarrow {\mathbf{OM}}

Option B (

4\overrightarrow {\mathbf{OM}}

) is correct because the sum of the position vectors of the vertices of a parallelogram with respect to any point O is four times the position vector of the intersection point M of the diagonals.

Solution approach

Correct Answer + Core Reason: $4\overrightarrow {\mathbf{OM}}$ . The sum of the position vectors of the vertices of a parallelogram with respect to any point O is four times the position vector of the intersection point M of the diagonals.

Step-by-Step Solution:

Step 1: Given: $ABCD$ is a parallelogram, $M$ is the intersection of the diagonals, and $O$ is any point.

Step 2: Recall that the diagonals of a parallelogram bisect each other. Therefore, $M$ is the midpoint of both $AC$ and $BD$ .

Step 3: Apply the midpoint formula for vectors:
- $\overrightarrow {OM} = \frac{1}{2} (\overrightarrow {OA} + \overrightarrow {OC})$ => $2\overrightarrow {OM} = \overrightarrow {OA} + \overrightarrow {OC}$
- $\overrightarrow {OM} = \frac{1}{2} (\overrightarrow {OB} + \overrightarrow {OD})$ => $2\overrightarrow {OM} = \overrightarrow {OB} + \overrightarrow {OD}$

Step 4: Add the two equations:
$2\overrightarrow {OM} + 2\overrightarrow {OM} = (\overrightarrow {OA} + \overrightarrow {OC}) + (\overrightarrow {OB} + \overrightarrow {OD})$
$4\overrightarrow {OM} = \overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} + \overrightarrow {OD}$

Why other options are wrong:
- Option A: $3\overrightarrow {\mathbf{OM}}$ : This is incorrect because the correct relationship is $\overrightarrow {\mathbf{OA}} + \overrightarrow {\mathbf{OB}} + \overrightarrow {\mathbf{OC}} + \overrightarrow {\mathbf{OD}} = 4\overrightarrow {\mathbf{OM}}$ .
- Option C: $\overrightarrow {\mathbf{OM}}$ : This is incorrect because the correct relationship is $\overrightarrow {\mathbf{OA}} + \overrightarrow {\mathbf{OB}} + \overrightarrow {\mathbf{OC}} + \overrightarrow {\mathbf{OD}} = 4\overrightarrow {\mathbf{OM}}$ .
- Option D: $2\overrightarrow {\mathbf{OM}}$ : This is incorrect because the correct relationship is $\overrightarrow {\mathbf{OA}} + \overrightarrow {\mathbf{OB}} + \overrightarrow {\mathbf{OC}} + \overrightarrow {\mathbf{OD}} = 4\overrightarrow {\mathbf{OM}}$ .

Key Concepts:
- Parallelogram properties
- Midpoint formula for vectors
- Vector addition

Memory Aids:
- Diagonals of a parallelogram bisect each other.

Quick Tips:
- When dealing with parallelograms, remember the properties of their diagonals and sides.

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Application +4 | -1

If three points A, B and C are collinear, whose position vectors are

\hat i - 2\hat j - 8\hat k,\,\,\,5\hat i - 2\hat k{\text{ }}\,\,{\text{and}}\,\,\,{\text{ }}11\hat i + 3\hat j + 7\hat k

respectively, then the ratio in which B divides AC is

View answer and explanation

Correct option: B

02:03

Option B (02:03) is correct because it represents the ratio in which point B divides the line segment AC, calculated using the section formula.

Solution approach

Correct Answer + Core Reason: 02:03. The ratio is found using the section formula for collinear points.

Step-by-Step Solution:

Step 1: Let B divide AC in the ratio $\lambda :1$ . Then, the position vector of B is given by: $\vec{b} = \frac{{\lambda \vec{c} + \vec{a}}}{{\lambda + 1}}$

Step 2: Substitute the given position vectors: $5\hat i - 2\hat k = \frac{{\lambda (11\hat i + 3\hat j + 7\hat k) + (\hat i - 2\hat j - 8\hat k)}}{{\lambda + 1}}$

Step 3: Equate the coefficients of $\hat i$ , $\hat j$ , and $\hat k$ :
- For $\hat i$ : $5(\lambda + 1) = 11\lambda + 1 \Rightarrow 5\lambda + 5 = 11\lambda + 1 \Rightarrow 6\lambda = 4 \Rightarrow \lambda = \frac{2}{3}$
- For $\hat j$ : $0 = 3\lambda - 2 \Rightarrow \lambda = \frac{2}{3}$
- For $\hat k$ : $-2(\lambda + 1) = 7\lambda - 8 \Rightarrow -2\lambda - 2 = 7\lambda - 8 \Rightarrow 9\lambda = 6 \Rightarrow \lambda = \frac{2}{3}$

Step 4: The ratio is $\lambda : 1 = \frac{2}{3} : 1 = 2:3$

Why other options are wrong:
- Option A: 01:02: Incorrect ratio. The calculation using the section formula does not result in this ratio.
- Option C: 02:01: Incorrect ratio. The calculation using the section formula does not result in this ratio.
- Option D: 01:01: Incorrect ratio. The calculation using the section formula does not result in this ratio.

Key Concepts:
- Section formula for collinear points
- Position vectors

Memory Aids:
- Remember the section formula: $\vec{r} = \frac{{m\vec{b} + n\vec{a}}}{{m + n}}$

Quick Tips:
- Equate coefficients of the unit vectors to solve for the ratio.
- Verify the ratio using all three components ( $\hat i$ , $\hat j$ , $\hat k$ ).

72 Easy

72 of 257

Application +4 | -1

If the position vectors of the points

A, B, C, D

be

2 \mathbf{i}+3 \mathbf{j}+5 \mathbf{k}, \quad \mathbf{i}+2 \mathbf{j}+3 \mathbf{k},-5 \mathbf{i}+4 \mathbf{j}-2 \mathbf{k}

and

\mathbf{i}+10 \mathbf{j}+10 \mathbf{k}

respectively, then

View answer and explanation

Correct option: B

\overrightarrow{A B} \| \overrightarrow{C D}

Option B (

\overrightarrow{A B} || \overrightarrow{C D}

) is correct because the vectors

\overrightarrow{A B}

and

\overrightarrow{C D}

are scalar multiples of each other, indicating they are parallel.

Solution approach

Correct Answer + Core Reason: $\overrightarrow{A B} || \overrightarrow{C D}$ . The vectors are parallel because one is a scalar multiple of the other.

Step-by-Step Solution:

Step 1: Find the vector $\overrightarrow{A B}$ .
$\overrightarrow{A B} = (\mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}) - (2 \mathbf{i} + 3 \mathbf{j} + 5 \mathbf{k}) = - \mathbf{i} - \mathbf{j} - 2 \mathbf{k}$

Step 2: Find the vector $\overrightarrow{C D}$ .
$\overrightarrow{C D} = (\mathbf{i} + 10 \mathbf{j} + 10 \mathbf{k}) - (-5 \mathbf{i} + 4 \mathbf{j} - 2 \mathbf{k}) = 6 \mathbf{i} + 6 \mathbf{j} + 12 \mathbf{k}$

Step 3: Check if $\overrightarrow{A B}$ and $\overrightarrow{C D}$ are parallel.
$\overrightarrow{C D} = -6 \overrightarrow{A B} = -6(-\mathbf{i} - \mathbf{j} - 2 \mathbf{k}) = 6 \mathbf{i} + 6 \mathbf{j} + 12 \mathbf{k}$ \nSince $\overrightarrow{C D}$ is a scalar multiple of $\overrightarrow{A B}$ , they are parallel.

Final Answer: $\overrightarrow{A B} || \overrightarrow{C D}$

Why other options are wrong:
- **Option A ( $\overrightarrow{A B}=\overrightarrow{C D}$ ): This is incorrect because the vectors are not equal, only parallel.
- Option C ( $\overrightarrow{A B} \perp \overrightarrow{C D}$ ): This is incorrect because the dot product of the vectors is not zero, so they are not perpendicular.
- Option D (None of these): This is incorrect because option B is the correct answer.

Key Concepts:
- Position vectors
- Vector subtraction
- Parallel vectors

Memory Aids:
- Two vectors are parallel if one is a scalar multiple of the other.

Quick Tips:**
- Check if the vectors are scalar multiples of each other to determine if they are parallel.

73 Easy

73 of 257

Application +4 | -1

The value of λ, for which the four points

2\widehat \mathbf{i} + 3\widehat \mathbf{j} - \mathbf{\hat k},\mathrm{\;\;}\widehat \mathbf{i} - 2\widehat \mathbf{j} + 3\mathbf{\hat k},3\widehat \mathbf{i} + 4\widehat \mathbf{j} - 2\mathbf{\hat k}\mathrm{\;},\widehat \mathbf{i} - 6\widehat \mathbf{j} + \mathrm{\lambda }\mathbf{\hat k}\mathrm{\;}

are coplanar, is

View answer and explanation

Correct option: C

6

Option C (6) is correct because it is the value of λ that makes the four points coplanar, which is determined by setting the scalar triple product of the vectors formed by these points equal to zero.

Solution approach

Correct Answer + Core Reason: 6. The value of λ is found by setting the scalar triple product of the vectors formed by the points to zero.

Step-by-Step Solution:

Step 1: Define the points as vectors: $\mathbf{\vec A} = 2\widehat \mathbf{i} + 3\widehat \mathbf{j} - \mathbf{\hat k}$ , $\mathbf{\vec B} = \widehat \mathbf{i} - 2\widehat \mathbf{j} + 3\mathbf{\hat k}$ , $\mathbf{\vec C} = 3\widehat \mathbf{i} + 4\widehat \mathbf{j} - 2\mathbf{\hat k}$ and $\mathbf{\vec D} = \widehat \mathbf{i} - 6\widehat \mathbf{j} + \mathrm{\lambda }\mathbf{\hat k}$ .

Step 2: Find the vectors $\overrightarrow \mathbf{AB}$ , $\overrightarrow \mathbf{AC}$ and $\overrightarrow \mathbf{AD}$ :
$\overrightarrow \mathbf{AB} = \mathbf{\vec B} - \mathbf{\vec A} = (1-2)\widehat \mathbf{i} + (-2-3)\widehat \mathbf{j} + (3-(-1))\mathbf{\hat k} = -\widehat \mathbf{i} - 5\widehat \mathbf{j} + 4\mathbf{\hat k}$
$\overrightarrow \mathbf{AC} = \mathbf{\vec C} - \mathbf{\vec A} = (3-2)\widehat \mathbf{i} + (4-3)\widehat \mathbf{j} + (-2-(-1))\mathbf{\hat k} = \widehat \mathbf{i} + \widehat \mathbf{j} - \mathbf{\hat k}$
$\overrightarrow \mathbf{AD} = \mathbf{\vec D} - \mathbf{\vec A} = (1-2)\widehat \mathbf{i} + (-6-3)\widehat \mathbf{j} + (\lambda-(-1))\mathbf{\hat k} = -\widehat \mathbf{i} - 9\widehat \mathbf{j} + (\lambda + 1)\mathbf{\hat k}$

Step 3: For the points to be coplanar, the scalar triple product of these vectors must be zero:
$\left[ {\overrightarrow \mathbf{AB} \;\overrightarrow \mathbf{AC} \;\overrightarrow \mathbf{AD} } \right]$

74 Easy

74 of 257

Application +4 | -1

If the vectors

2 \mathbf{i}-3 \mathbf{j}+4 \mathbf{k}, \mathbf{i}+2 \mathbf{j}-\mathbf{k}

and

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

are coplanar, then

x=

View answer and explanation

Correct option: A

\frac{8}{5}

Option A (8/5) is correct because the determinant of the matrix formed by the components of the coplanar vectors must be zero, which leads to x = 8/5.

Solution approach

Correct Answer + Core Reason: $\frac{8}{5}$ . Coplanar vectors have a determinant of zero.

Step-by-Step Solution:

Step 1: The vectors $2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ , $\mathbf{i} + 2\mathbf{j} - \mathbf{k}$ , and $x\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ are coplanar if the determinant of the matrix formed by their components is zero.

Step 2: Set up the determinant:
$\begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ x & -1 & 2 \end{vmatrix} = 0$

Step 3: Expand the determinant:
$2(4 - 1) - (-3)(2 + x) + 4(-1 - 2x) = 0$
$2(3) + 3(2 + x) + 4(-1 - 2x) = 0$
$6 + 6 + 3x - 4 - 8x = 0$

Step 4: Simplify the equation:
$8 - 5x = 0$

Step 5: Solve for $x$ :
$5x = 8$
$x = \frac{8}{5}$

Why other options are wrong:
- Option B (5/8): This value is incorrect. It might arise from an error in solving for x after expanding the determinant.
- Option C (0): This value is incorrect. It might arise from an error in expanding the determinant or assuming x = 0 without proper calculation.
- Option D (1): This value is incorrect. It might arise from an error in expanding the determinant or solving for x.

Key Concepts:
- Coplanar vectors
- Determinants

Memory Aids:
- Coplanar vectors: The volume of the parallelepiped formed by them is zero, hence the determinant is zero.

Quick Tips:
- Remember the determinant representation for coplanar vectors.
- Double-check the signs when expanding the determinant.

75 Easy

75 of 257

Analyzing +4 | -1

Let

\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}+\hat{\mathbf{k}}

and

\mathbf{c}=\alpha \mathbf{a}+\beta \mathbf{b}

. If the vectors,

\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, 3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}

and

\mathbf{c}

are coplanar, then

\frac{\alpha}{\beta}

is equal to

View answer and explanation

Correct option: D

-3

Option D (-3) is correct because the coplanarity condition leads to a determinant equation, which simplifies to

\frac{\alpha}{\beta} = -3

.

Solution approach

Correct Answer + Core Reason: -3. The condition for coplanarity of three vectors is that their scalar triple product is zero.

Step-by-Step Solution:

Step 1: Define the vectors $\mathbf{v_1} = \hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}}$ , $\mathbf{v_2} = 3\hat{\mathbf{i}} + 2\hat{\mathbf{j}} - \hat{\mathbf{k}}$ , and $\mathbf{c} = \alpha \mathbf{a} + \beta \mathbf{b} = \alpha(\hat{\mathbf{i}} + \hat{\mathbf{j}}) + \beta(\hat{\mathbf{j}} + \hat{\mathbf{k}}) = \alpha \hat{\mathbf{i}} + (\alpha + \beta)\hat{\mathbf{j}} + \beta \hat{\mathbf{k}}$ .

Step 2: For the vectors to be coplanar, their scalar triple product must be zero: $[\mathbf{v_1} \mathbf{v_2} \mathbf{c}] = 0$ .

Step 3: Write the scalar triple product as a determinant:
$\begin{vmatrix} 1 & -2 & 1 \\ 3 & 2 & -1 \\ \alpha & \alpha + \beta & \beta \end{vmatrix} = 0$ .

Step 4: Evaluate the determinant. Using the operation $C_2 \rightarrow C_2 - (C_1 + C_3)$ , we get:
$\begin{vmatrix} 1 & -4 & 1 \\ 3 & 0 & -1 \\ \alpha & 0 & \beta \end{vmatrix} = 0$ .

Step 5: Expanding along the second column, we have: $-(-4)(3\beta + \alpha) = 0$ , which simplifies to $4(3\beta + \alpha) = 0$ .

Step 6: Solve for the ratio $\frac{\alpha}{\beta}$ : $3\beta + \alpha = 0 \Rightarrow \alpha = -3\beta \Rightarrow \frac{\alpha}{\beta} = -3$ .

Why other options are wrong:
- Option A (1): This is incorrect because it doesn't satisfy the coplanarity condition.
- Option B (2): This is incorrect because it doesn't satisfy the coplanarity condition.
- Option C (3): This is incorrect because it doesn't satisfy the coplanarity condition.

Key Concepts:
- Coplanarity of vectors
- Scalar triple product
- Determinants

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

Quick Tips:
- Remember the determinant representation of the scalar triple product.
- Use column or row operations to simplify the determinant before evaluating.

76 Easy

76 of 257

Application +4 | -1

ABCD

is a quadrilateral ,

P,Q

are the mid points of

\overrightarrow {\mathbf{BC}}

and

\overrightarrow {\mathbf{AD}}

, then

\overrightarrow {\mathbf{AB}} + \overrightarrow {\mathbf{DC}}

is equal to

View answer and explanation

Correct option: D

2\overrightarrow {\mathbf{QP}}

Option D (

2\overrightarrow {\mathbf{QP}}

) is correct because it correctly relates the given vectors using midpoint properties.

Solution approach

Correct Answer + Core Reason: $2\overrightarrow {\mathbf{QP}}$ . This is derived from the properties of midpoints in a quadrilateral.

Step-by-Step Solution:

Step 1: Understand the problem: $ABCD$ is a quadrilateral, $P$ and $Q$ are midpoints of $BC$ and $AD$ respectively. We need to find the relationship between $\overrightarrow {\mathbf{AB}} + \overrightarrow {\mathbf{DC}}$ and $\overrightarrow {\mathbf{QP}}$ .

Step 2: Express $\overrightarrow {\mathbf{QP}}$ in terms of other vectors: $\overrightarrow {\mathbf{QP}} = \overrightarrow {\mathbf{QA}} + \overrightarrow {\mathbf{AB}} + \overrightarrow {\mathbf{BP}}$ . Also, $\overrightarrow {\mathbf{QP}} = \overrightarrow {\mathbf{QD}} + \overrightarrow {\mathbf{DC}} + \overrightarrow {\mathbf{CP}}$ .

Step 3: Add the two expressions for $\overrightarrow {\mathbf{QP}}$ : $2\overrightarrow {\mathbf{QP}} = (\overrightarrow {\mathbf{QA}} + \overrightarrow {\mathbf{QD}} ) + (\overrightarrow {\mathbf{AB}} + \overrightarrow {\mathbf{DC}} ) + (\overrightarrow {\mathbf{BP}} + \overrightarrow {\mathbf{CP}} )$ .

Step 4: Simplify using midpoint properties: Since $Q$ is the midpoint of $AD$ , $\overrightarrow {\mathbf{QA}} + \overrightarrow {\mathbf{QD}} = \vec 0$ . Since $P$ is the midpoint of $BC$ , $\overrightarrow {\mathbf{BP}} + \overrightarrow {\mathbf{CP}} = \vec 0$ .

Step 5: Therefore, $2\overrightarrow {\mathbf{QP}} = \overrightarrow {\mathbf{AB}} + \overrightarrow {\mathbf{DC}}$ .

Step 6: Final Answer: $2\overrightarrow {\mathbf{QP}}$

Why other options are wrong:
- Option A: Incorrect. It should be 2, not 3.
- Option B: Incorrect. It should be 2, not 1.
- Option C: Incorrect. It should be 2, not 4.

Key Concepts:
- Vector addition
- Midpoint properties

Memory Aids:
- Remember that the sum of vectors from a midpoint to the endpoints is zero.

Quick Tips:
- Express the required vector in terms of other vectors using known points.
- Use midpoint properties to simplify the expression.

77 Easy

77 of 257

Application +4 | -1

In a

\mathrm{\Delta }ABC

, if

\vec AB = \hat i - 7\hat j + \hat k

and

\vec BC = 3\hat i + \hat j + 2\hat k

, then

\left| {\vec CA} \right| =

View answer and explanation

Correct option: A

\sqrt {61}

Option A (

\sqrt {61}

) is correct because it is the magnitude of the vector

\vec{CA}

, which is the negative of

\vec{AC}

, and

\vec{AC}

is the sum of

\vec{AB}

and

\vec{BC}

.

Solution approach

Correct Answer + Core Reason: $\sqrt {61}$ . The magnitude of $\vec{CA}$ is calculated by finding $\vec{AC}$ first.

Step-by-Step Solution:

Step 1: Find $\vec{AC}$ using the vector addition rule: $\vec{AC} = \vec{AB} + \vec{BC}$ .

Step 2: Substitute the given vectors: $\vec{AC} = (\hat i - 7\hat j + \hat k) + (3\hat i + \hat j + 2\hat k) = 4\hat i - 6\hat j + 3\hat k$ .

Step 3: Find $\vec{CA}$ : $\vec{CA} = -\vec{AC} = -4\hat i + 6\hat j - 3\hat k$ .

Step 4: Calculate the magnitude of $\vec{CA}$ : $\left| {\vec{{CA}}} \right| = \sqrt{{(-4)^2 + (6)^2 + (-3)^2}} = \sqrt{{16 + 36 + 9}} = \sqrt{{61}}$ .

Why other options are wrong:
- Option B: $\sqrt {52}$ : This is incorrect because it is likely a calculation error in the magnitude.
- Option C: $\sqrt {51}$ : This is incorrect because it is likely a calculation error in the magnitude.
- Option D: $\sqrt {41}$ : This is incorrect because it is likely a calculation error in the magnitude.

Key Concepts:
- Vector addition
- Magnitude of a vector

Memory Aids:
- Remember that $\vec{CA} = -\vec{AC}$ .

Quick Tips:
- Always pay attention to the direction of the vectors.

78 Easy

78 of 257

Application +4 | -1

If

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

and

\mathbf{c}=3 \mathbf{i}+p \mathbf{j}+5 \mathbf{k}

are coplanar then the value of

p

will be [RPET

\left.\mathbf{1 9 8} \mathbf{8}_5, \mathbf{8 6}, \mathbf{8 8}, \mathbf{9 1}\right]

View answer and explanation

Correct option: A

-6

Option A (-6) is correct because the scalar triple product of coplanar vectors is zero, leading to the determinant equation that solves for p = -6.

Solution approach

Correct Answer + Core Reason: -6. Coplanar vectors have a scalar triple product of zero.

Step-by-Step Solution:

Step 1: The vectors $\mathbf{a}$ , $\mathbf{b}$ , and $\mathbf{c}$ are coplanar if their scalar triple product is zero, i.e., $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0$ .

Step 2: The scalar triple product can be represented as a determinant:
$\begin{vmatrix} 1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & p & 5 \end{vmatrix} = 0$

Step 3: Expand the determinant:
$1(10 + p) - (-1)(5 + 3) + 1(p - 6) = 0$

Step 4: Simplify the equation:
$10 + p + 8 + p - 6 = 0$

Step 5: Combine like terms:
$2p + 12 = 0$

Step 6: Solve for $p$ :
$2p = -12$
$p = -6$

Why other options are wrong:
- Option B (-2): This value is incorrect. It might arise from an error in expanding or simplifying the determinant.
- Option C (6): This value is incorrect. It might arise from an error in sign when solving for p.
- Option D (2): This value is incorrect. It might arise from an error in expanding or simplifying the determinant, or incorrect sign.

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

Memory Aids:
- Coplanar vectors: The volume of the parallelepiped formed by them is zero.

Quick Tips:
- Remember the determinant representation of the scalar triple product.
- Double-check the signs when expanding the determinant.

79 Easy

79 of 257

Analyzing +4 | -1

p = 2a – 3b, q = a – 2b + c and r = -3a + b + 2c, where a, b, c being non – zero non – coplanar vectors, then the vector -2a + 3b – c is equal to

View answer and explanation

Correct option: B

\frac{{ - 7q + r}}{5}

Option B (

\frac{{ - 7q + r}}{5}

) is the correct answer because it represents the linear combination of vectors p, q, and r that equals -2a + 3b - c.

Solution approach

Correct Answer + Core Reason: $\frac{{ - 7q + r}}{5}$ . This is found by expressing -2a + 3b - c as a linear combination of p, q, and r and solving for the coefficients.

Step-by-Step Solution:

Step 1: Let -2a + 3b – c = xp + yq + zr

Step 2: Substitute the given expressions for p, q, and r: -2a + 3b – c = x(2a – 3b) + y(a – 2b + c) + z(-3a + b + 2c)

Step 3: Group the coefficients of a, b, and c: -2a + 3b – c = (2x + y – 3z)a + (-3x – 2y + z)b + (y + 2z)c

Step 4: Equate the coefficients of a, b, and c on both sides:
2x + y – 3z = -2
-3x – 2y + z = 3\ny + 2z = -1

Step 5: Solve the system of equations. From the third equation, y = -1 - 2z. Substitute this into the first two equations:
2x + (-1 - 2z) – 3z = -2 => 2x - 5z = -1
-3x – 2(-1 - 2z) + z = 3 => -3x + 5z = 1

Step 6: Add the two equations to eliminate z: -x = 0 => x = 0

Step 7: Substitute x = 0 into 2x - 5z = -1: -5z = -1 => z = 1/5

Step 8: Substitute z = 1/5 into y = -1 - 2z: y = -1 - 2(1/5) = -1 - 2/5 = -7/5

Step 9: Therefore, -2a + 3b - c = 0p + (-7/5)q + (1/5)r = $\frac{{ - 7q + r}}{5}$

Why other options are wrong:
- Option A: p – 4q = (2a - 3b) - 4(a - 2b + c) = -2a + 5b - 4c, which is not equal to -2a + 3b - c.
- Option C: 2p – 3q + r = 2(2a - 3b) - 3(a - 2b + c) + (-3a + b + 2c) = (4a - 6b) - (3a - 6b + 3c) + (-3a + b + 2c) = -2a + b - c, which is not equal to -2a + 3b - c.
- Option D: 4p – 2r = 4(2a - 3b) - 2(-3a + b + 2c) = 8a - 12b + 6a - 2b - 4c = 14a - 14b - 4c, which is not equal to -2a + 3b - c.

Key Concepts:
- Vector algebra
- Linear combination of vectors
- Solving systems of linear equations

Memory Aids:
- Remember to carefully distribute and combine like terms when working with vector equations.

Quick Tips:
- When solving for coefficients in a linear combination, set up a system of equations by equating corresponding components.

80 Easy

80 of 257

Analyzing +4 | -1

Let a, b and c be three non – zero vectors such that no two of these are collinear. If the vector a + 2b is collinear with c and b + 3c is collinear with

(\lambda

being some non – zero scalar), then a + 2b + 6c is equal to

View answer and explanation

Correct option: A

0

The correct answer is A (0) because, given the collinearity conditions, the vector a + 2b + 6c simplifies to zero.

Solution approach

Correct Answer + Core Reason: 0. The vector sum is zero due to the collinearity conditions.

Step-by-Step Solution:

Step 1: Given that a + 2b is collinear with c, we can write $a + 2b = \lambda c$ ...(i).

Step 2: Also, b + 3c is collinear with a, so $b + 3c = \mu a$ ...(ii).

Step 3: We want to find a + 2b + 6c. Using (i), $a + 2b + 6c = (a + 2b) + 6c = \lambda c + 6c = (\lambda + 6)c$ ...(iii).

Step 4: Using (ii), $a + 2b + 6c = a + 2(b + 3c) = a + 2(\mu a) = (1 + 2\mu)a$ ...(iv).

Step 5: From (iii) and (iv), $(\lambda + 6)c = (1 + 2\mu)a$ . Since a and c are non-zero and non-collinear, we must have $\lambda + 6 = 0$ and $1 + 2\mu = 0$ .

Step 6: This implies $\lambda = -6$ and $\mu = -\frac{1}{2}$ .

Step 7: Substituting $\lambda = -6$ into (iii), we get $a + 2b + 6c = (\lambda + 6)c = (-6 + 6)c = 0$ .

Why other options are wrong:
- Option B: $\lambda b$ is incorrect because the vector sum simplifies to zero, not a scalar multiple of b.
- Option C: $\lambda c$ is incorrect because the vector sum simplifies to zero, not a scalar multiple of c.
- Option D: $\lambda a$ is incorrect because the vector sum simplifies to zero, not a scalar multiple of a.

Key Concepts:
- Collinearity of vectors
- Linear combination of vectors

Memory Aids:
- If two vectors are collinear, one is a scalar multiple of the other.

Quick Tips:
- Use the given collinearity conditions to express the vectors in terms of each other.
- Look for ways to simplify the expression using the given conditions.

81 Easy

81 of 257

Analyzing +4 | -1

Let us define the length of a vector

a\hat i + b\hat j + c\hat k

as

|a| + |b| + |c|.

This definition coincides with the usual definition of length of a vector

a\hat i + b\hat j + c\hat k

if and only if

View answer and explanation

Correct option: B

Any two of a, b and c are zero

Option B (Any two of a, b and c are zero) is correct because the given length definition coincides with the usual definition only when at least two of the components are zero.

Solution approach

Correct Answer + Core Reason: Any two of a, b and c are zero. This condition ensures that the given length definition coincides with the usual Euclidean length.

Step-by-Step Solution:

Step 1: Set up the equation equating the two length definitions:
$|a| + |b| + |c| = \sqrt{{a^2 + b^2 + c^2}}$

Step 2: Square both sides:
$(|a| + |b| + |c|)^2 = a^2 + b^2 + c^2$
$a^2 + b^2 + c^2 + 2|ab| + 2|bc| + 2|ca| = a^2 + b^2 + c^2$

Step 3: Simplify the equation:
$2|ab| + 2|bc| + 2|ca| = 0$
$|ab| + |bc| + |ca| = 0$

Step 4: Since absolute values are non-negative, the only way for the sum to be zero is if each term is zero:
$|ab| = 0$ , $|bc| = 0$ , $|ca| = 0$ \nThis implies that ab = 0, bc = 0, and ca = 0.

Step 5: The only way for all these products to be zero is if at least two of a, b, and c are zero.

Step 6: Final Answer: Any two of a, b and c are zero

Why other options are wrong:
- Option A (a = b = c = 0): Incorrect. While this satisfies the condition, it's not the only solution. Any two being zero is sufficient.
- Option C (Any one of a, b and c is zero): Incorrect. If only one is zero, the equation $|ab| + |bc| + |ca| = 0$ will not hold.
- Option D (a + b + c = 0): Incorrect. This condition doesn't guarantee that $|a| + |b| + |c| = \sqrt{{a^2 + b^2 + c^2}}$ .

Key Concepts:
- Euclidean length of a vector
- Absolute values

Memory Aids:
- The sum of absolute values is zero only if each term is zero.

Quick Tips:
- Squaring equations involving square roots can help simplify them.

82 Easy

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Application +4 | -1

Let

\mathbf{\vec a}\;,\mathbf{\vec b},,\overrightarrow {\mathbf{c}\;} \;\;

be three non-zero vectors such that no two of these are collinear. If the vector

\mathbf{\vec a} + 2\;\mathbf{\vec b}

is collinear with

\overrightarrow {\mathbf{c}\;} \;,

then

\overrightarrow {\mathbf{a}\;} + 2\overrightarrow {\mathbf{b}\;} + 6\;\overrightarrow {\mathbf{c}\;} \;\;\mathrm{equals}\;\;\;

View answer and explanation

Correct option: C

\lambda \;\mathbf{\vec c}\;\left( {\lambda \ne 0,\;\mathrm{a\;scalar}} \right)

Option C (

\lambda \;\mathbf{\vec c}\;(\lambda \ne 0,\;\mathrm{a\;scalar}))

is correct because

\mathbf{\vec a} + 2\mathbf{\vec b}

is a scalar multiple of

\mathbf{\vec c}

, and adding

6\mathbf{\vec c}

results in another scalar multiple of

\mathbf{\vec c}

.

Solution approach

Correct Answer + Core Reason: $\lambda \;\mathbf{\vec c}\;(\lambda \ne 0,\;\mathrm{a\;scalar})$ . The result is a scalar multiple of vector $\mathbf{\vec c}$ .

Step-by-Step Solution:

Step 1: Given that $\mathbf{\vec a} + 2\mathbf{\vec b}$ is collinear with $\mathbf{\vec c}$ , we can write $\mathbf{\vec a} + 2\mathbf{\vec b} = k\mathbf{\vec c}$ for some scalar k.

Step 2: We want to find an expression for $\mathbf{\vec a} + 2\mathbf{\vec b} + 6\mathbf{\vec c}$ .

Step 3: Substitute $k\mathbf{\vec c}$ for $\mathbf{\vec a} + 2\mathbf{\vec b}$ in the expression: $\mathbf{\vec a} + 2\mathbf{\vec b} + 6\mathbf{\vec c} = k\mathbf{\vec c} + 6\mathbf{\vec c}$ .

Step 4: Factor out $\mathbf{\vec c}$ : $k\mathbf{\vec c} + 6\mathbf{\vec c} = (k + 6)\mathbf{\vec c}$ .

Step 5: Let $\lambda = k + 6$ . Then, the expression becomes $\lambda\mathbf{\vec c}$ .

Final Answer: $\lambda\mathbf{\vec c}$

Why other options are wrong:
- Option A: This is incorrect because the resulting vector is a scalar multiple of $\mathbf{\vec c}$ , not $\mathbf{\vec a}$ .
- Option B: This is incorrect because the resulting vector is a scalar multiple of $\mathbf{\vec c}$ , not $\mathbf{\vec b}$ .
- Option D: This is incorrect because the resulting vector is a non-zero scalar multiple of $\mathbf{\vec c}$ , not the zero vector.

Key Concepts:
- Collinear vectors
- Scalar multiplication of vectors
- Vector addition

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

Quick Tips:
- When vectors are collinear, express one as a scalar multiple of the other.

83 Easy

83 of 257

Application +4 | -1

In a triangle

A B C

, if

2 \overrightarrow{A C}=3 \overrightarrow{C B}

, then

2 \overrightarrow{O A}+3 \overrightarrow{O B}

equals

View answer and explanation

Correct option: A

5 \overrightarrow{O C}

Option A (

5 \overrightarrow{O C}

) is correct because it follows from the section formula applied to the given vector equation

2 \overrightarrow{A C}=3 \overrightarrow{C B}

.

Solution approach

Correct Answer + Core Reason: $5 \overrightarrow{O C}$ . This result comes from applying the section formula to the given vector equation.

Step-by-Step Solution:

Step 1: Express $\overrightarrow{A C}$ and $\overrightarrow{C B}$ in terms of position vectors with respect to origin $O$ .
$\overrightarrow{A C} = \overrightarrow{O C} - \overrightarrow{O A}$
$\overrightarrow{C B} = \overrightarrow{O B} - \overrightarrow{O C}$

Step 2: Substitute these expressions into the given equation $2 \overrightarrow{A C} = 3 \overrightarrow{C B}$ .
$2(\overrightarrow{O C} - \overrightarrow{O A}) = 3(\overrightarrow{O B} - \overrightarrow{O C})$

Step 3: Expand the equation.
$2 \overrightarrow{O C} - 2 \overrightarrow{O A} = 3 \overrightarrow{O B} - 3 \overrightarrow{O C}$

Step 4: Rearrange the equation to isolate $2 \overrightarrow{O A} + 3 \overrightarrow{O B}$ .
$2 \overrightarrow{O A} + 3 \overrightarrow{O B} = 2 \overrightarrow{O C} + 3 \overrightarrow{O C}$

Step 5: Simplify the equation.
$2 \overrightarrow{O A} + 3 \overrightarrow{O B} = 5 \overrightarrow{O C}$

Final Answer: $5 \overrightarrow{O C}$

Why other options are wrong:
- **Option B ( $-\overrightarrow{O C}$ ): This is incorrect because it has the wrong sign and magnitude.
- Option C ( $\overrightarrow{O C}$ ): This is incorrect because it has the wrong magnitude.
- Option D (None of these): This is incorrect because option A is the correct answer.

Key Concepts:
- Position vectors
- Section formula

Memory Aids:**
- Remember that $\overrightarrow{A B} = \mathbf{b} - \mathbf{a}$ , where $\mathbf{a}$ and $\mathbf{b}$ are the position vectors of points $A$ and $B$ , respectively.

Quick Tips:
- Express all vectors in terms of position vectors with respect to the origin to simplify the equation.

84 Easy

84 of 257

Application +4 | -1

If

\mathbf{a}=2 \mathbf{i}+\mathbf{j}-8 \mathbf{k}

and

\mathbf{b}=\mathbf{i}+3 \mathbf{j}-4 \mathbf{k}

, then the magnitude of

\mathbf{a}+\mathbf{b}=

View answer and explanation

Correct option: A

13

Option A (13) is correct because the magnitude of the sum of the vectors is calculated by adding the vectors and then finding the magnitude of the resultant vector.

Solution approach

Correct Answer + Core Reason: 13. The magnitude of the sum of the vectors is calculated by adding the vectors and then finding the magnitude of the resultant vector.

Step-by-Step Solution:

Step 1: Add the vectors $\mathbf{a}$ and $\mathbf{b}$ .
$\mathbf{a} + \mathbf{b} = (2 \mathbf{i} + \mathbf{j} - 8 \mathbf{k}) + (\mathbf{i} + 3 \mathbf{j} - 4 \mathbf{k})$

Step 2: Combine like terms.
$\mathbf{a} + \mathbf{b} = (2+1) \mathbf{i} + (1+3) \mathbf{j} + (-8-4) \mathbf{k} = 3 \mathbf{i} + 4 \mathbf{j} - 12 \mathbf{k}$

Step 3: Calculate the magnitude of the resultant vector.
$|\mathbf{a} + \mathbf{b}| = \sqrt{3^2 + 4^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$

Final Answer: 13

Why other options are wrong:
- **Option B ( $\frac{{13}}{{3}}$ ): This is incorrect because it seems to be dividing the magnitude by 3 without any logical reason.
- Option C ( $\frac{{3}}{{13}}$ ): This is incorrect as it is the reciprocal of a possible intermediate calculation and has no relation to the correct magnitude.
- Option D ( $\frac{{4}}{{13}}$ ): This is incorrect as it is also a random fraction with no relation to the correct magnitude.

Key Concepts:
- Vector addition
- Magnitude of a vector

Memory Aids:**
- Remember that the magnitude of a vector $\mathbf{v} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}$ is given by $|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$

Quick Tips:
- Always add the vectors first before finding the magnitude.
- Be careful with the signs when adding vectors.

85 Easy

85 of 257

Application +4 | -1

If

{\vec a}

is a vector of magnitude 50 collinear with the vector

\vec b = 6\hat i - 8\hat j - \frac{{15}}{2}\hat k

and makes an acute angle with the positive direction of

z

-axis, then

\vec a =

View answer and explanation

Correct option: B

- 24\hat i + 32\hat j + 30\hat k

Option B (

- 24\hat i + 32\hat j + 30\hat k

) is correct because it is collinear with the given vector, has a magnitude of 50, and makes an acute angle with the positive z-axis.

Solution approach

Correct Answer + Core Reason: $- 24\hat i + 32\hat j + 30\hat k$ . This vector satisfies the collinearity, magnitude, and angle conditions.

Step-by-Step Solution:

Step 1: Since ${\vec a}$ is collinear to $\vec b = 6\hat i - 8\hat j - \frac{{15}}{2}\hat k$ , we can write ${\vec a} = m\vec b$ for some scalar $m$ .

Step 2: So, ${\vec a} = m(6\hat i - 8\hat j - \frac{{15}}{2}\hat k)$ .

Step 3: Find the magnitude of ${\vec a}$ : $|{\vec a}| = |m|\sqrt{6^2 + (-8)^2 + (-\frac{15}{2})^2} = |m|\sqrt{36 + 64 + \frac{225}{4}} = |m|\sqrt{\frac{400 + 225}{4}} = |m|\sqrt{\frac{625}{4}} = |m|\frac{25}{2}$ .

Step 4: Given $|{\vec a}| = 50$ , so $50 = |m|\frac{25}{2} \Rightarrow |m| = 4 \Rightarrow m = \pm 4$ .

Step 5: Since ${\vec a}$ makes an acute angle with the positive z-axis, its z-component must be positive. Therefore, $m$ must be negative, so $m = -4$ .

Step 6: Substitute $m = -4$ into ${\vec a} = m(6\hat i - 8\hat j - \frac{{15}}{2}\hat k)$ : ${\vec a} = -4(6\hat i - 8\hat j - \frac{{15}}{2}\hat k) = -24\hat i + 32\hat j + 30\hat k$ .

Step 7: Final Answer: ${\vec a} = -24\hat i + 32\hat j + 30\hat k$ .

Why other options are wrong:
- Option A: This option has a negative z-component, so it makes an obtuse angle with the positive z-axis.
- Option C: This option has a magnitude that is not equal to 50.
- Option D: None of these is incorrect because option B is correct.

Key Concepts:
- Collinear vectors
- Magnitude of a vector
- Angle between vectors

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

Quick Tips:
- Always check the magnitude and direction of the resulting vector.

86 Easy

86 of 257

Analyzing +4 | -1

Let A and B be points with position vectors a and b with respect to the origin O. If the point C on OA is such that 2AC = CO, CD is parallel to OB and

\left| {CD} \right| = 3\left| {OB} \right|,

then AD is equal to

View answer and explanation

Correct option: C

3b - \frac{a}{3}

Option C (

3b - \frac{a}{3}

) is correct because, using the section formula and given conditions, the vector AD can be expressed as

3b - \frac{a}{3}

.

Solution approach

Correct Answer + Core Reason: $3b - \frac{a}{3}$ . This is the vector AD calculated using the section formula and given conditions.

Step-by-Step Solution:

Step 1: Given that OA = a, OB = b, and 2AC = CO, we can find OC using the section formula.

Step 2: Since 2AC = CO, we have OC = 2/3 * OA = (2/3)a.

Step 3: Also, CD is parallel to OB and |CD| = 3|OB|, so CD = 3b.

Step 4: Now, OD = OC + CD = (2/3)a + 3b.

Step 5: We want to find AD, which is OD - OA = ((2/3)a + 3b) - a = 3b - (1/3)a.

Why other options are wrong:
- Option A: $3b - \frac{a}{2}$ is incorrect because the correct calculation gives $3b - \frac{a}{3}$ .
- Option B: $3b + \frac{a}{2}$ is incorrect because the correct calculation gives $3b - \frac{a}{3}$ .
- Option D: $3b + \frac{a}{3}$ is incorrect because the correct calculation gives $3b - \frac{a}{3}$ .

Key Concepts:
- Section formula
- Vector addition
- Parallel vectors

Memory Aids:
- Section formula: If a point C divides AB in the ratio m:n, then OC = (n*OA + m*OB)/(m+n).

Quick Tips:
- Use the section formula to find the position vector of a point dividing a line segment in a given ratio.
- Remember that parallel vectors are scalar multiples of each other.

87 Easy

87 of 257

Application +4 | -1

If the point whose position vectors are

2\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}

,

6\widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k}

and

14\widehat \mathbf{i} - 5\widehat \mathbf{j} + p\mathbf{\hat k}

are collinear, then the value of

p

is

View answer and explanation

Correct option: B

4

The value of p is 4, which makes the points collinear, meaning they lie on the same line. This is verified by setting the determinant of the matrix formed by the position vectors to zero.

Solution approach

Correct Answer + Core Reason: 4. The points are collinear if the determinant of the matrix formed by their position vectors is zero.

Step-by-Step Solution:

Step 1: Write the position vectors of the points: $\mathbf{\vec a} = 2\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}$ , $\mathbf{\vec b} = 6\widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k}$ , and $\mathbf{\vec c} = 14\widehat \mathbf{i} - 5\widehat \mathbf{j} + p\mathbf{\hat k}$ .

Step 2: For the points to be collinear, the vectors $\vec{ab}$ and $\vec{ac}$ must be parallel. This means that $\vec{ab} = k \vec{ac}$ for some scalar k. Alternatively, the scalar triple product of $\vec{a}$ , $\vec{b}$ , and $\vec{c}$ must be zero if we consider the origin as a reference point. However, a simpler approach is to consider the area of the triangle formed by the points to be zero, which means the determinant formed by the coordinates must be zero.

Step 3: Set up the determinant equation: $\left| {\begin{array}{cccccccccccccccccccc} 2&1&1\\ 6&{ - 1}&2\\ {14}&{ - 5}&\mathrm{p} \end{array}} \right| = 0$ .

Step 4: Expand the determinant: $2[(-1)p - (2)(-5)] - 1[(6)p - (2)(14)] + 1[(6)(-5) - (-1)(14)] = 0$ .

Step 5: Simplify the equation: $2[-p + 10] - [6p - 28] + [-30 + 14] = 0 \Rightarrow -2p + 20 - 6p + 28 - 16 = 0 \Rightarrow -8p + 32 = 0$ .

Step 6: Solve for p: $-8p = -32 \Rightarrow p = 4$ .

Step 7: Final Answer: The value of p is 4.

Why other options are wrong:
- Option A: 2: If p=2, the determinant is not zero, and the points are not collinear.
- Option B: 4: This is the correct answer.
- Option C: 6: If p=6, the determinant is not zero, and the points are not collinear.
- Option D: 8: If p=8, the determinant is not zero, and the points are not collinear.

Key Concepts:
- Collinearity of points.
- Determinant of a matrix.

Memory Aids:
- Points are collinear if the area of the triangle formed by them is zero.

Quick Tips:
- Remember the determinant expansion rules.
- Check your calculations carefully to avoid errors.

88 Easy

88 of 257

Application +4 | -1

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

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

and

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

respectively. If A is chosen as the origin, then the position vectors of B and C are

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it represents the position vectors of B and C relative to A as the origin.

Solution approach

Correct Answer + Core Reason: $- \widehat j + 2\widehat k,\widehat i + \widehat j + 3\widehat k$ . These are the position vectors of B and C with respect to A.

Step-by-Step Solution:

Step 1: Given position vectors $OA = \widehat i + 2\widehat j - \widehat k$ , $OB = \widehat i + \widehat j + \widehat k$ , and $OC = 2\widehat i + 3\widehat j + 2\widehat k$ .

Step 2: Find the position vector of B with respect to A: $AB = OB - OA = (\widehat i + \widehat j + \widehat k) - (\widehat i + 2\widehat j - \widehat k) = (1 - 1)\widehat i + (1 - 2)\widehat j + (1 - (-1))\widehat k = 0\widehat i - \widehat j + 2\widehat k = -\widehat j + 2\widehat k$ .

Step 3: Find the position vector of C with respect to A: $AC = OC - OA = (2\widehat i + 3\widehat j + 2\widehat k) - (\widehat i + 2\widehat j - \widehat k) = (2 - 1)\widehat i + (3 - 2)\widehat j + (2 - (-1))\widehat k = \widehat i + \widehat j + 3\widehat k$ .

Why other options are wrong:
- **Option A ( $\widehat i + 2\widehat k,\widehat i + \widehat j + 3\widehat k$ ): This is incorrect because the position vector of B is calculated incorrectly.
- Option B ( $\widehat j + 2\widehat k,\widehat i + \widehat j + 3\widehat k$ ): This is incorrect because the position vector of B is calculated incorrectly.
- Option C ( $- \widehat j + 2\widehat k,\widehat i - \widehat j + 3\widehat k$ ): This is incorrect because the position vector of C is calculated incorrectly.

Key Concepts:
- Position vector
- Vector subtraction

Memory Aids:**
- Remember that the position vector of B with respect to A is $AB = OB - OA$ .

Quick Tips:
- Subtract the position vector of the reference point (A) from the position vector of the point of interest (B or C).

89 Easy

89 of 257

Application +4 | -1

If

\hat i - 3\hat j + 5 \hat k

bisects the angle between

a

and

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

where

a

is a unit vector, then

View answer and explanation

Correct option: D

a = \frac{1}{{105}}(41\hat i - 88\hat j - 40\hat k)

Option D (

a = \frac{1}{{105}}(41\hat i - 88\hat j - 40\hat k)

) is correct because it is derived from the angle bisector property and the condition that 'a' is a unit vector.

Solution approach

Correct Answer + Core Reason: $a = \frac{1}{{105}}(41\hat i - 88\hat j - 40\hat k)$ . This is found using the angle bisector property and the unit vector condition.

Step-by-Step Solution:

Step 1: Let $\vec{b} = -\hat i + 2\hat j + 2\hat k$ . Then $\hat{b} = \frac{{\vec{b}}}{|\vec{b}|} = \frac{{ - \hat i + 2\hat j + 2\hat k}}{{\sqrt{(-1)^2 + 2^2 + 2^2}}} = \frac{{ - \hat i + 2\hat j + 2\hat k}}{3}$ .

Step 2: Since $\hat i - 3\hat j + 5\hat k$ bisects the angle between $\vec{a}$ and $\vec{b}$ , we have $\lambda (\hat i - 3\hat j + 5\hat k) = \hat{a} + \hat{b}$ , where $\lambda$ is a scalar.

Step 3: Therefore, $\hat{a} = \lambda (\hat i - 3\hat j + 5\hat k) - \frac{{ - \hat i + 2\hat j + 2\hat k}}{3} = (\lambda + \frac{1}{3})\hat i + (-3\lambda - \frac{2}{3})\hat j + (5\lambda - \frac{2}{3})\hat k$ .

Step 4: Since $\hat{a}$ is a unit vector, its magnitude is 1: $|\hat{a}|^2 = 1$ .\nSo, $(\lambda + \frac{1}{3})^2 + (-3\lambda - \frac{2}{3})^2 + (5\lambda - \frac{2}{3})^2 = 1$ .\nThis simplifies to $35\lambda^2 - 2\lambda - \frac{8}{9} = 0$ . Multiplying by 9, we get $315\lambda^2 - 18\lambda - 8 = 0$ .\nSolving for $\lambda$ : $\lambda = \frac{{18 \pm \sqrt{18^2 - 4(315)(-8)}}}{{2(315)}} = \frac{{18 \pm \sqrt{32724}}}{{630}} = \frac{{18 \pm 181. }}{{630}}$ . We take $\lambda = \frac{2}{35}$ .

Step 5: Substitute $\lambda = \frac{2}{35}$ into the expression for $\hat{a}$ : $\hat{a} = (\frac{2}{35} + \frac{1}{3})\hat i + (-3(\frac{2}{35}) - \frac{2}{3})\hat j + (5(\frac{2}{35}) - \frac{2}{3})\hat k = \frac{{41}}{{105}}\hat i - \frac{{88}}{{105}}\hat j - \frac{{40}}{{105}}\hat k$ .

Why other options are wrong:
- Option A: $a = \frac{1}{{105}}(41\hat i + 88\hat j - 40\hat k)$ : Incorrect signs for the j and k components.
- Option B: $a = \frac{1}{{105}}(41\hat i + 88\hat j + 40\hat k)$ : Incorrect signs for the j and k components.
- Option C: $a = \frac{1}{{105}}( - 41\hat i + 88\hat j - 40\hat k)$ : Incorrect sign for the i component.

Key Concepts:
- Angle bisector property
- Unit vectors
- Vector addition

Memory Aids:
- The angle bisector divides the angle into two equal parts.

Quick Tips:
- Remember to normalize vectors to find unit vectors.
- Pay attention to signs when performing vector operations.

90 Easy

90 of 257

Understanding +4 | -1

If forces

\overrightarrow{{P}}, \overrightarrow{{Q}}, \overrightarrow{{R}}

acting at a point can be represented by the sides of a triangle taken in order, then

View answer and explanation

Correct option: A

\overrightarrow{{P}}+\overrightarrow{{Q}}+\overrightarrow{{R}}=\overrightarrow{{0}}

Option A (

\overrightarrow{{P}}+\overrightarrow{{Q}}+\overrightarrow{{R}}=\overrightarrow{{0}}

) is correct because if the forces can be represented by the sides of a triangle taken in order, they form a closed triangle and their vector sum is zero.

Solution approach

Correct Answer + Core Reason: $\overrightarrow{{P}}+\overrightarrow{{Q}}+\overrightarrow{{R}}=\overrightarrow{{0}}$ . This represents the condition for equilibrium when forces are represented by the sides of a triangle.

Step-by-Step Solution:

Step 1: According to the triangle law of vector addition, if three vectors can be represented by the three sides of a triangle taken in order, then their sum is zero.

Step 2: In this case, the forces $\overrightarrow{{P}}, \overrightarrow{{Q}},$ and $\overrightarrow{{R}}$ can be represented by the sides of a triangle taken in order.

Step 3: Therefore, the vector sum of these forces is zero: $\overrightarrow{{P}}+\overrightarrow{{Q}}+\overrightarrow{{R}}=\overrightarrow{{0}}$ .

Why other options are wrong:
- Option B: This is incorrect because it implies that $\overrightarrow{{P}} + \overrightarrow{{R}} = \overrightarrow{{Q}}$ , which is not generally true for a triangle formed by these vectors.
- Option C: This is incorrect because it implies that $\overrightarrow{{P}} + \overrightarrow{{Q}} = \overrightarrow{{R}}$ , which is not generally true for a triangle formed by these vectors.
- Option D: This is incorrect because it implies that $\overrightarrow{{P}} = \overrightarrow{{Q}} + \overrightarrow{{R}}$ , which is not generally true for a triangle formed by these vectors.

Key Concepts:
- Triangle law of vector addition
- Equilibrium of forces

Memory Aids:
- Remember that if vectors form a closed loop (like a triangle), their sum is zero.

Quick Tips:
- Visualize the vectors forming a closed triangle to understand the concept.

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