The equation of the line passing through the points ${a_1}{\bf{\hat i}} + {a_2}{\bf{\hat j}} + {a_3}{\bf{\hat k}}\;{\rm{and}}\;{b_1}{\bf{\hat i}} + {b_2}{\bf{\hat j}} + {b_3}{\bf{\hat k}}$ is

${a_1}\left( {1 - t} \right){\bf{\hat i}} + {a_2}\left( {1 - t} \right){\bf{\hat j}} + {a_3}\left( {1 - t} \right){\bf{\hat k}} + \left( {\;{b_1}{\bf{\hat i}} + {b_2}{\bf{\hat j}} + {b_3}{\bf{\hat k}}} \right)t$

$\left( {{a_1}{\bf{\hat i}} + {a_2}{\bf{\hat j}} + {a_3}{\bf{\hat k}}} \right) + t\left( {\;{b_1}{\bf{\hat i}} + {b_2}{\bf{\hat j}} + {b_3}{\bf{\hat k}}} \right)$

$\left( {\;{a_1}{\bf{\hat i}} + {a_2}{\bf{\hat j}} + {a_3}{\bf{\hat k}}\;} \right) - t\left( {{b_1}{\bf{\hat i}} + {b_2}{\bf{\hat j}} + {b_3}{\bf{\hat k}}} \right)$

If $D,\;E$ and $F$ are respectively the mid points of $AB,\;AC\;{\rm{and\;}}BC\;{\rm{in\;}}\Delta ABC,\;{\rm{then\;}}$ $\overrightarrow {{\bf{BE}}} + \overrightarrow {{\bf{AF}}} $ is equal to

$\frac{1}{2}\overrightarrow {{\bf{BF}}} $

$2\;\overrightarrow {{\bf{BF}}} $

$\frac{3}{2}\overrightarrow {{\bf{BF}}} $

ADDITION OF VECTORS MCQs for JEE MATHEMATICS — Easy, Page 9

241 Easy

241 of 257

Application +4 | -1

If

\vec{a}=(1,-1,2), \vec{b}=(-2,3,5)

,

\vec{c}=(2,-2,4)

and

\hat{i}

is the unit vector in the

x

-direction, then

(\vec{a}-2 \vec{b}+3 \vec{c}) \hat{i}=

View answer and explanation

Correct option: A

11

Option A (11) is correct because after substituting the vectors and performing the dot product with

\hat{i}

, the result is 11.

Solution approach

Correct Answer + Core Reason: 11. The dot product of the resulting vector with $\hat{i}$ is 11.

Step-by-Step Solution:

Step 1: Given $\vec{a} = (1, -1, 2), \vec{b} = (-2, 3, 5), \vec{c} = (2, -2, 4)$ .

Step 2: Calculate $2\vec{b} = 2(-2, 3, 5) = (-4, 6, 10)$ and $3\vec{c} = 3(2, -2, 4) = (6, -6, 12)$ .

Step 3: Calculate $\vec{a} - 2\vec{b} + 3\vec{c} = (1, -1, 2) - (-4, 6, 10) + (6, -6, 12) = (1 + 4 + 6, -1 - 6 - 6, 2 - 10 + 12) = (11, -13, 4)$ .

Step 4: The unit vector in the x-direction is $\hat{i} = (1, 0, 0)$ .

Step 5: Calculate $(\vec{a} - 2\vec{b} + 3\vec{c}) \cdot \hat{i} = (11, -13, 4) \cdot (1, 0, 0) = 11(1) + (-13)(0) + 4(0) = 11$ .

Why other options are wrong:
- Option B (15): This is incorrect as it is not the correct dot product.
- Option C (18): This is incorrect as it is not the correct dot product.
- Option D (36): This is incorrect as it is not the correct dot product.

Key Concepts:
- Vector operations (addition, subtraction, scalar multiplication)
- Dot product of vectors

Memory Aids:
- Remember the dot product formula: $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$

Quick Tips:
- Pay attention to the signs when performing vector operations.

242 Easy

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Application +4 | -1

ABCDE is a pentagon. Forces AB, AE, DC and ED act at a point. Which force should be added to this system to make the resultant 2AC?

View answer and explanation

Correct option: C

BC

Option C (BC) is correct because adding BC to the given forces results in a vector sum of 2AC, as derived from vector addition properties.

Solution approach

Correct Answer + Core Reason: BC. Adding BC to the system results in 2AC due to vector addition.

Step-by-Step Solution:

Step 1: Analyze the given forces: AB, AE, DC, and ED.

Step 2: Use vector addition: AE + ED = AD. So, the forces become AB, AD, and DC.

Step 3: Use vector addition again: AD + DC = AC. So, the forces become AB and AC.

Step 4: The resultant of the given forces is AB + AC. We want the total resultant to be 2AC.

Step 5: Let the force to be added be X. Then, AB + AC + X = 2AC.

Step 6: Solve for X: X = 2AC - AB - AC = AC - AB = AC + BA = BC.

Why other options are wrong:
- Option A: AC: Adding AC would result in AB + 2AC, which is not equal to 2AC.
- Option B: AD: Adding AD would result in AB + AC + AD, which is not equal to 2AC.
- Option D: BD: Adding BD would result in AB + AC + BD, which is not equal to 2AC.

Key Concepts:
- Vector addition
- Resultant force

Memory Aids:
- Remember that AB + BC = AC.

Quick Tips:
- Break down complex vector expressions into simpler ones using vector addition rules.
- Visualize the vectors to understand their relationships.

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Application +4 | -1

The value of

\lambda

for which the four points

2\hat i + 3\hat j - \hat k,\,\,\hat i + 2\hat j + 3\hat k,\,\,3\hat i + 4\hat j - 2\hat k\,\,\,{\text{and}}\,\,\,\hat i - \lambda \hat j + 6\hat k\,

are coplanar

View answer and explanation

Correct option: C

-2

Option C (-2) is correct because it satisfies the condition for coplanar points, which can be determined using the determinant of the matrix formed by the position vectors.

Solution approach

Correct Answer + Core Reason: -2. The value of $\lambda$ is found by setting the determinant of the matrix formed by the position vectors to zero, indicating coplanarity.

Step-by-Step Solution:

Step 1: Define the position vectors.
$A = 2\hat i + 3\hat j - \hat k$
$B = \hat i + 2\hat j + 3\hat k$
$C = 3\hat i + 4\hat j - 2\hat k$
$D = \hat i - \lambda \hat j + 6\hat k$

Step 2: Find the vectors AB, AC, and AD.
$AB = B - A = (\hat i + 2\hat j + 3\hat k) - (2\hat i + 3\hat j - \hat k) = -\hat i - \hat j + 4\hat k$
$AC = C - A = (3\hat i + 4\hat j - 2\hat k) - (2\hat i + 3\hat j - \hat k) = \hat i + \hat j - \hat k$
$AD = D - A = (\hat i - \lambda \hat j + 6\hat k) - (2\hat i + 3\hat j - \hat k) = -\hat i - (\lambda + 3)\hat j + 7\hat k$

Step 3: Set up the determinant equation for coplanar points.\nFor points to be coplanar, the scalar triple product of the vectors AB, AC, and AD must be zero. This can be expressed as:
$\left| {\begin{array}{cccccccccccccccccccc} -1&-1&4 \\ 1&1&-1 \\ -1&{-(\lambda + 3)}&7 \end{array}} \right| = 0$

Step 4: Evaluate the determinant.
$-1(1(7) - (-1)(-\lambda - 3)) - (-1)(1(7) - (-1)(-1)) + 4(1(-\lambda - 3) - 1(-1)) = 0$
$-1(7 - (\lambda + 3)) + 1(7 - 1) + 4(-\lambda - 3 + 1) = 0$
$-1(4 - \lambda) + 6 + 4(-\lambda - 2) = 0$
$-4 + \lambda + 6 - 4\lambda - 8 = 0$
$-3\lambda - 6 = 0$

Step 5: Solve for $\lambda$ .
$-3\lambda = 6$
$\lambda = -2$

Why other options are wrong:
- Option A: Using $\lambda = 8$ would result in a non-zero determinant, indicating the points are not coplanar.
- Option B: Using $\lambda = 0$ would result in a non-zero determinant, indicating the points are not coplanar.
- Option D: Using $\lambda = 6$ would result in a non-zero determinant, indicating the points are not coplanar.

Key Concepts:
- Coplanar points
- Scalar triple product
- Determinant of a matrix
- Vector subtraction

Memory Aids:
- Four points are coplanar if the scalar triple product of the vectors formed by them is zero.

Quick Tips:
- Remember the determinant formula for coplanarity.
- Carefully perform vector subtraction and determinant evaluation.

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Application +4 | -1

The vector equation of the line joining the points

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

and

-2 \mathbf{j}+3 \mathbf{k}

is

View answer and explanation

Correct option: C

\mathbf{r}=(\mathbf{i}-2 \mathbf{j}+\mathbf{k})+t(2 \mathbf{k}-\mathbf{i})

Option C (

\mathbf{r}=(\mathbf{i}-2 \mathbf{j}+\mathbf{k})+t(2 \mathbf{k}-\mathbf{i})

) is correct because it represents the line passing through the given points.

Solution approach

Correct Answer + Core Reason: $\mathbf{r}=(\mathbf{i}-2 \mathbf{j}+\mathbf{k})+t(2 \mathbf{k}-\mathbf{i})$ . The vector equation of the line is $\mathbf{r}=(\mathbf{i}-2 \mathbf{j}+\mathbf{k})+t(2 \mathbf{k}-\mathbf{i})$ .

Step-by-Step Solution:

Step 1: State the general form of the vector equation of a line passing through two points.\nVector equation of a straight line passing through two points $\mathbf{a}$ and $\mathbf{b}$ is, $\mathbf{r}=\mathbf{a}+t(\mathbf{b}-\mathbf{a})$

Step 2: Substitute the given points into the equation.
$\Rightarrow \mathbf{r}=(\mathbf{i}-2 \mathbf{j}+\mathbf{k})+t(2 \mathbf{k}-\mathbf{i})$

Why other options are wrong:
- **Option A ( $\mathbf{r}=t(\mathbf{i}+\mathbf{j}+\mathbf{k})$ ): Incorrect because it doesn't pass through the given points.
- Option B ( $\mathbf{r}=t_1(\mathbf{i}-2 \mathbf{j}+\mathbf{k})+t_2(3 \mathbf{k}-2 \mathbf{j})$ ): Incorrect because it is not in the correct form of a line equation.
- Option D ( $\mathbf{r}=t(2 \mathbf{k}-\mathbf{i})$ ): Incorrect because it doesn't pass through the given points.

Key Concepts:
- Vector equation of a line

Memory Aids:
- Remember the formula for the vector equation of a line passing through two points.

Quick Tips:**
- Use the formula $\mathbf{r}=\mathbf{a}+t(\mathbf{b}-\mathbf{a})$ to find the vector equation of a line.

245 Easy

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Application +4 | -1

For a non – zero vector

a

the set of real numbers satisfying

\left| {(5 - x)a} \right| < \left| {2a} \right|

consists of all

x

such that

View answer and explanation

Correct option: B

3 < x < 7

Option B (

3 < x < 7

) is correct because solving the inequality

\left| {(5 - x)a} \right| < \left| {2a} \right|

leads to the solution

3 < x < 7

.

Solution approach

Correct Answer + Core Reason: $3 < x < 7$ . The inequality is solved by using the properties of absolute values and vector magnitudes.

Step-by-Step Solution:

Step 1: Given the inequality $\left| {(5 - x)a} \right| < \left| {2a} \right|$ .

Step 2: Using the property of magnitudes, $\left| {(5 - x)a} \right| = \left| {5 - x} \right|\left| a \right|$ .

Step 3: So, $\left| {5 - x} \right|\left| a \right| < \left| {2a} \right| = 2\left| a \right|$ .

Step 4: Since a is a non-zero vector, $\left| a \right| > 0$ . Divide both sides by $\left| a \right|$ : $\left| {5 - x} \right| < 2$ .

Step 5: This means $-2 < 5 - x < 2$ .

Step 6: Subtract 5 from all parts of the inequality: $-7 < -x < -3$ .

Step 7: Multiply by -1 and reverse the inequality signs: $3 < x < 7$ .

Why other options are wrong:
- **Option A ( $0 < x < 3$ ): This is incorrect due to errors in solving the inequality.
- Option C ( $- 7 < x < - 3$ ): This is incorrect due to errors in solving the inequality.
- Option D ( $- 7 < x < 3$ ): This is incorrect due to errors in solving the inequality.

Key Concepts:
- Properties of absolute values.
- Properties of vector magnitudes.
- Solving inequalities.

Memory Aids:**
- Remember the properties of absolute values: $\left| x \right| < a$ means $-a < x < a$ .

Quick Tips:
- Be careful with signs when solving inequalities.
- Remember to reverse the inequality signs when multiplying by a negative number.

246 Easy

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Analyzing +4 | -1

If the vectors

6\hat i - 2\hat j + 3\hat k,\,\,2\hat i + 3\hat j - 6\hat k

and

3\hat i + 6\hat j - 2\hat k

form a triangle, then it is

View answer and explanation

Correct option: B

obtuse angled

Option B (obtuse angled) is correct because the sides of the triangle formed by the given vectors satisfy the condition

AC^2 < AB^2 + BC^2

, indicating an obtuse angle.

Solution approach

Correct Answer + Core Reason: obtuse angled. The triangle formed has sides that satisfy the obtuse angle condition.

Step-by-Step Solution:

Step 1: Let A, B, and C be the points with position vectors $6\hat i - 2\hat j + 3\hat k$ , $2\hat i + 3\hat j - 6\hat k$ , and $3\hat i + 6\hat j - 2\hat k$ , respectively.

Step 2: Find the vectors representing the sides of the triangle:
$AB = (2\hat i + 3\hat j - 6\hat k) - (6\hat i - 2\hat j + 3\hat k) = -4\hat i + 5\hat j - 9\hat k$
$BC = (3\hat i + 6\hat j - 2\hat k) - (2\hat i + 3\hat j - 6\hat k) = \hat i + 3\hat j + 4\hat k$
$AC = (3\hat i + 6\hat j - 2\hat k) - (6\hat i - 2\hat j + 3\hat k) = -3\hat i + 8\hat j - 5\hat k$

Step 3: Calculate the magnitudes of the sides:
$|AB| = \sqrt{(-4)^2 + 5^2 + (-9)^2} = \sqrt{16 + 25 + 81} = \sqrt{122}$
$|BC| = \sqrt{1^2 + 3^2 + 4^2} = \sqrt{1 + 9 + 16} = \sqrt{26}$
$|AC| = \sqrt{(-3)^2 + 8^2 + (-5)^2} = \sqrt{9 + 64 + 25} = \sqrt{98}$

Step 4: Check the triangle inequality and the Pythagorean theorem:
$AB^2 = 122, BC^2 = 26, AC^2 = 98$ \nSince $AC^2 < AB^2 + BC^2$ (98 < 122 + 26 = 148), the triangle is obtuse angled.

Why other options are wrong:
- Option A: right angled is incorrect because $AC^2 \neq AB^2 + BC^2$ .
- Option C: equilateral is incorrect because the side lengths are not equal.
- Option D: isosceles is incorrect because only one side length is calculated to be equal to another.

Key Concepts:
- Vector subtraction
- Magnitude of a vector
- Triangle inequality
- Pythagorean theorem

Memory Aids:
- Obtuse angled triangle: $c^2 < a^2 + b^2$
- Right angled triangle: $c^2 = a^2 + b^2$
- Acute angled triangle: $c^2 > a^2 + b^2$

Quick Tips:
- Calculate the side lengths using vector subtraction and magnitudes.
- Check the Pythagorean theorem to determine the type of triangle.

247 Easy

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Understanding +4 | -1

The equation of the line passing through the points

{a_1}\mathbf{\hat i} + {a_2}\mathbf{\hat j} + {a_3}\mathbf{\hat k}\;\mathrm{and}\;{b_1}\mathbf{\hat i} + {b_2}\mathbf{\hat j} + {b_3}\mathbf{\hat k}

is

View answer and explanation

Correct option: C

{a_1}\left( {1 - t} \right)\mathbf{\hat i} + {a_2}\left( {1 - t} \right)\mathbf{\hat j} + {a_3}\left( {1 - t} \right)\mathbf{\hat k} + \left( {\;{b_1}\mathbf{\hat i} + {b_2}\mathbf{\hat j} + {b_3}\mathbf{\hat k}} \right)t

Option C is correct because it represents the parametric equation of a line passing through two points in vector form.

Solution approach

Correct Answer + Core Reason: ${a_1}\left( {1 - t} \right)\mathbf{\hat i} + {a_2}\left( {1 - t} \right)\mathbf{\hat j} + {a_3}\left( {1 - t} \right)\mathbf{\hat k} + \left( {\;{b_1}\mathbf{\hat i} + {b_2}\mathbf{\hat j} + {b_3}\mathbf{\hat k}} \right)t$ is correct because it represents the equation of a line passing through two points.

Step-by-Step Solution:

Step 1: The equation of a line passing through points $\vec{a}$ and $\vec{b}$ can be written as $\vec{r} = (1-t)\vec{a} + t\vec{b}$ .

Step 2: Given points are ${a_1}\mathbf{\hat i} + {a_2}\mathbf{\hat j} + {a_3}\mathbf{\hat k}$ and ${b_1}\mathbf{\hat i} + {b_2}\mathbf{\hat j} + {b_3}\mathbf{\hat k}$ .

Step 3: Substitute the given points into the equation of the line: $\vec{r} = (1-t)({a_1}\mathbf{\hat i} + {a_2}\mathbf{\hat j} + {a_3}\mathbf{\hat k}) + t({b_1}\mathbf{\hat i} + {b_2}\mathbf{\hat j} + {b_3}\mathbf{\hat k})$ .

Step 4: Simplify the equation: $\vec{r} = {a_1}(1-t)\mathbf{\hat i} + {a_2}(1-t)\mathbf{\hat j} + {a_3}(1-t)\mathbf{\hat k} + {b_1}t\mathbf{\hat i} + {b_2}t\mathbf{\hat j} + {b_3}t\mathbf{\hat k}$ .

Step 5: Final Answer: ${a_1}\left( {1 - t} \right)\mathbf{\hat i} + {a_2}\left( {1 - t} \right)\mathbf{\hat j} + {a_3}\left( {1 - t} \right)\mathbf{\hat k} + \left( {\;{b_1}\mathbf{\hat i} + {b_2}\mathbf{\hat j} + {b_3}\mathbf{\hat k}} \right)t$

Why other options are wrong:
- Option A: This option only represents a linear combination of the two position vectors, without ensuring it passes through both points.
- Option B: Similar to option A, this option does not guarantee the line passes through both given points.
- Option D: This is incorrect as the correct equation is given in option C.

Key Concepts:
- Equation of a line in vector form
- Parametric equation of a line

Memory Aids:
- Remember the formula: $\vec{r} = (1-t)\vec{a} + t\vec{b}$

Quick Tips:
- Always remember the standard form of the equation of a line in vector form.

248 Easy

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Application +4 | -1

The vector c, directed along the internal bisector of the angle between the vectors

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

and

b = - 2\widehat i - \widehat j + 2\widehat k

with

\left| c \right| = 5\sqrt 6 ,

View answer and explanation

Correct option: A

\frac{5}{3}(\widehat i - 7\widehat j + 2\widehat k)

Option A (

\frac{5}{3}(\widehat i - 7\widehat j + 2\widehat k)

) is the correct answer because it represents the vector along the internal bisector with the given magnitude.

Solution approach

Correct Answer + Core Reason: $\frac{5}{3}(\widehat i - 7\widehat j + 2\widehat k)$ . The vector is calculated by finding the unit vectors of a and b, adding them, and scaling to the correct magnitude.

Step-by-Step Solution:

Step 1: Find the magnitudes of vectors a and b.
$|a| = \sqrt{7^2 + (-4)^2 + (-4)^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$
$|b| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

Step 2: Find the unit vectors of a and b.
$\hat{a} = \frac{a}{|a|} = \frac{7\hat i - 4\hat j - 4\hat k}{9}$
$\hat{b} = \frac{b}{|b|} = \frac{-2\hat i - \hat j + 2\hat k}{3}$

Step 3: Find the vector along the internal 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 4: Scale the vector to the correct magnitude.\nLet $c = \lambda(\frac{\hat i - 7\hat j + 2\hat k}{9})$ .
$|c| = 5\sqrt{6} = \frac{\lambda}{9}\sqrt{1^2 + (-7)^2 + 2^2} = \frac{\lambda}{9}\sqrt{1 + 49 + 4} = \frac{\lambda}{9}\sqrt{54} = \frac{\lambda}{9}3\sqrt{6} = \frac{\lambda}{3}\sqrt{6}$
$\lambda = 15$
$c = \frac{15}{9}(\hat i - 7\hat j + 2\hat k) = \frac{5}{3}(\hat i - 7\hat j + 2\hat k)$

Why other options are wrong:
- Option B: Incorrect calculation of the vector components and/or magnitude.
- Option C: Incorrect sign in the j component.
- Option D: Incorrect signs in the i and j components.

Key Concepts:
- Vector addition
- Unit vectors
- Internal bisector of an angle between two vectors
- Magnitude of a vector

Memory Aids:
- The internal bisector is along the direction of the sum of the unit vectors.

Quick Tips:
- Always find the unit vectors first before adding them.
- Remember to scale the resulting vector to the desired magnitude.

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Analyzing +4 | -1

If

\mathbf{\vec a},\mathbf{\vec b},\mathbf{\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

\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c} = \vec 0

The correct answer is A (

\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c} = \vec 0

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

Solution approach

Correct Answer + Core Reason: $\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c} = \vec 0$ . The orthocenter and centroid coincide in an equilateral triangle. 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 $\frac{{\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c}}}{3}$ .

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 $\frac{{\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c}}}{3} = \vec 0$ .

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

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.

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Application +4 | -1

On the xy plane where O is the origin, given points,

A(1,0),\,\,B(0,1){\text{ }}\,\,{\text{and}}\,\,{\text{ }}C(1,1).

Let P, Q and R be moving point on the line

OA,\,\,OB,\,\,OC

respectively such that

OP = 45t(OA),\,\,\,\,OQ = 60t(OB),\,\,\,\,OR = (1 - t)(OC)

with

t > 0.

If the three points P, Q and R are collinear, then the value of t is equal to

View answer and explanation

Correct option: B

7/187

Option B (7/187) is correct because it is the value of t that makes the points P, Q, and R collinear, derived by equating the slopes of PQ and QR.

Solution approach

Correct Answer + Core Reason: 7/187. This value of t makes the points P, Q, and R collinear.

Step-by-Step Solution:

Step 1: Find the position vectors of P, Q, and R:
$\vec{OP} = 45t \vec{OA} = 45t(\hat i) = 45t\hat i$
$\vec{OQ} = 60t \vec{OB} = 60t(\hat j) = 60t\hat j$
$\vec{OR} = (1 - t) \vec{OC} = (1 - t)(\hat i + \hat j) = (1 - t)\hat i + (1 - t)\hat j$

Step 2: Find the vectors $\vec{PQ}$ and $\vec{QR}$ :
$\vec{PQ} = \vec{OQ} - \vec{OP} = 60t\hat j - 45t\hat i = -45t\hat i + 60t\hat j$
$\vec{QR} = \vec{OR} - \vec{OQ} = (1 - t)\hat i + (1 - t)\hat j - 60t\hat j = (1 - t)\hat i + (1 - 61t)\hat j$

Step 3: Since P, Q, and R are collinear, $\vec{PQ}$ and $\vec{QR}$ are parallel. Therefore, $\vec{PQ} = \lambda \vec{QR}$ for some scalar $\lambda$ .
$-45t\hat i + 60t\hat j = \lambda [(1 - t)\hat i + (1 - 61t)\hat j]$

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Application +4 | -1

If a = (2, 5) and b = (1, 4), then the vector parallel to (a + b) is

View answer and explanation

Correct option: C

(1, 3)

Option C ((1, 3)) is correct because the sum of the vectors a and b is parallel to (1, 3).

Solution approach

Correct Answer + Core Reason: (1, 3). The sum of the vectors is parallel to this vector.

Step-by-Step Solution:

Step 1: Given vectors $a = (2, 5)$ and $b = (1, 4)$ .

Step 2: Find the sum of the vectors: $a + b = (2 + 1, 5 + 4) = (3, 9)$ .

Step 3: A vector parallel to $(3, 9)$ is any scalar multiple of it. We can write $(3, 9) = 3(1, 3)$ .

Step 4: Therefore, the vector $(1, 3)$ is parallel to $(a + b)$ .

Why other options are wrong:
- Option A ((3, 5)): This is incorrect because it is not parallel to (3, 9).
- Option B ((1, 1)): This is incorrect because it is not parallel to (3, 9).
- Option D ((8, 5)): This is incorrect because it is not parallel to (3, 9).

Key Concepts:
- Vector addition
- Parallel vectors

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

Quick Tips:
- Find the sum of the vectors, then look for a scalar multiple among the options.

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Application +4 | -1

If

D,\;E

and

F

are respectively the mid points of

AB,\;AC\;\mathrm{and\;}BC\;\mathrm{in\;}\Delta ABC,\;\mathrm{then\;}

\overrightarrow {\mathbf{BE}} + \overrightarrow {\mathbf{AF}}

is equal to

View answer and explanation

Correct option: A

\overrightarrow {\mathbf{DC}}

Option A (

\overrightarrow{\mathbf{DC}}

) is correct because by expressing the vectors in terms of position vectors and simplifying,

\overrightarrow{\mathbf{BE}} + \overrightarrow{\mathbf{AF}}

equals

\overrightarrow{\mathbf{DC}}

.

Solution approach

Correct Answer + Core Reason: $\overrightarrow{\mathbf{DC}}$ . $\overrightarrow{\mathbf{BE}} + \overrightarrow{\mathbf{AF}} = \overrightarrow{\mathbf{DC}}$ after expressing the vectors in terms of position vectors and simplifying.

Step-by-Step Solution:

Step 1: Let $\overrightarrow{\mathbf{OA}} = \mathbf{\vec a}$ , $\overrightarrow{\mathbf{OB}} = \mathbf{\vec b}$ , and $\overrightarrow{\mathbf{OC}} = \mathbf{\vec c}$ .

Step 2: Express the position vectors of the midpoints in terms of $\vec a$ , $\vec b$ , and $\vec c$ .
$\overrightarrow{\mathbf{OD}} = \frac{{\mathbf{\vec a} + \mathbf{\vec b}}}{2}$ , $\overrightarrow{\mathbf{OE}} = \frac{{\mathbf{\vec a} + \mathbf{\vec c}}}{2}$ , $\overrightarrow{\mathbf{OF}} = \frac{{\mathbf{\vec b} + \mathbf{\vec c}}}{2}$

Step 3: Express $\overrightarrow{\mathbf{AF}}$ and $\overrightarrow{\mathbf{BE}}$ in terms of position vectors.
$\overrightarrow{\mathbf{AF}} = \overrightarrow{\mathbf{OF}} - \overrightarrow{\mathbf{OA}} = \frac{1}{2}(\mathbf{\vec b} + \mathbf{\vec c}) - \mathbf{\vec a}$
$\overrightarrow{\mathbf{BE}} = \overrightarrow{\mathbf{OE}} - \overrightarrow{\mathbf{OB}} = \frac{1}{2}(\mathbf{\vec a} + \mathbf{\vec c}) - \mathbf{\vec b}$

Step 4: Express $\overrightarrow{\mathbf{DC}}$ in terms of position vectors.
$\overrightarrow{\mathbf{DC}} = \overrightarrow{\mathbf{OC}} - \overrightarrow{\mathbf{OD}} = \mathbf{\vec c} - \frac{1}{2}(\mathbf{\vec a} + \mathbf{\vec b})$

Step 5: Calculate $\overrightarrow{\mathbf{AF}} + \overrightarrow{\mathbf{BE}}$ .
$\overrightarrow{\mathbf{AF}} + \overrightarrow{\mathbf{BE}} = \frac{1}{2}(\mathbf{\vec b} + \mathbf{\vec c}) - \mathbf{\vec a} + \frac{1}{2}(\mathbf{\vec a} + \mathbf{\vec c}) - \mathbf{\vec b} = \mathbf{\vec c} - \frac{1}{2}(\mathbf{\vec a} + \mathbf{\vec b})$

Step 6: Compare $\overrightarrow{\mathbf{AF}} + \overrightarrow{\mathbf{BE}}$ with $\overrightarrow{\mathbf{DC}}$ .
Since $\overrightarrow{\mathbf{AF}} + \overrightarrow{\mathbf{BE}} = \mathbf{\vec c} - \frac{1}{2}(\mathbf{\vec a} + \mathbf{\vec b}) = \overrightarrow{\mathbf{DC}}$ , the answer is $\overrightarrow{\mathbf{DC}}$ .

Why other options are wrong:
- Option B: Incorrect. $\frac{1}{2}\overrightarrow{\mathbf{BF}}$ is not equal to $\overrightarrow{\mathbf{DC}}$ .
- Option C: Incorrect. $2\overrightarrow{\mathbf{BF}}$ is not equal to $\overrightarrow{\mathbf{DC}}$ .
- Option D: Incorrect. $\frac{3}{2}\overrightarrow{\mathbf{BF}}$ is not equal to $\overrightarrow{\mathbf{DC}}$ .

Key Concepts:
- Position vectors
- Midpoint formula
- Vector addition and subtraction

Memory Aids:
- Remember how to express vectors in terms of position vectors.

Quick Tips:
- Express all vectors in terms of position vectors to simplify the problem.

253 Easy

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Application +4 | -1

The vectors

\widehat i + 2\widehat j + 3\widehat k,\lambda \widehat i + 4\widehat j + 7\widehat k, - 3\widehat i - 2\widehat j - 5\widehat k

are collinear, if

\lambda

is equal to

View answer and explanation

Correct option: A

3

Option A (3) is correct because it represents the value of

\lambda

for which the given vectors are collinear, calculated using the determinant condition for collinearity.

Solution approach

Correct Answer + Core Reason: 3. The value of $\lambda$ is calculated using the determinant condition for collinearity of vectors.

Step-by-Step Solution:

Step 1: For the vectors $\widehat i + 2\widehat j + 3\widehat k$ , $\lambda \widehat i + 4\widehat j + 7\widehat k$ , and $-3\widehat i - 2\widehat j - 5\widehat k$ to be coplanar (or collinear in 2D), the scalar triple product must be zero. This can be expressed as a determinant:
$\left| {\begin{array}{cccccccccccccccccccc} 1&2&3 \\ \lambda &4&7 \\ { - 3}&{ - 2}&{ - 5} \end{array}} \right| = 0$

Step 2: Evaluate the determinant.
$1(4(-5) - 7(-2)) - 2(\lambda(-5) - 7(-3)) + 3(\lambda(-2) - 4(-3)) = 0$
$1(-20 + 14) - 2(-5\lambda + 21) + 3(-2\lambda + 12) = 0$
$-6 + 10\lambda - 42 - 6\lambda + 36 = 0$
$4\lambda - 12 = 0$
$4\lambda = 12$
$\lambda = 3$

Why other options are wrong:
- Option B: 4: This is incorrect because it does not satisfy the determinant condition for collinearity.
- Option C: 5: This is incorrect because it does not satisfy the determinant condition for collinearity.
- Option D: 6: This is incorrect because it does not satisfy the determinant condition for collinearity.

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

Memory Aids:
- The scalar triple product is zero for coplanar vectors.

Quick Tips:
- Remember the determinant condition for coplanarity.

254 Easy

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Application +4 | -1

The position vectors of P and Q are

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

and

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

respectively. If the distance between them is 7, then the value of a will be

View answer and explanation

Correct option: A

– 5, 1

Option A (– 5, 1) is correct because these are the values of 'a' that satisfy the distance condition between the two points.

Solution approach

Correct Answer + Core Reason: – 5, 1. These values satisfy the distance condition between the points.

Step-by-Step Solution:

Step 1: Given position vectors $P = 5\widehat i + 4\widehat j + a\widehat k$ and $Q = -\widehat i + 2\widehat j - 2\widehat k$ , and the distance between them is 7.

Step 2: Find the vector $PQ = Q - P = (-1 - 5)\widehat i + (2 - 4)\widehat j + (-2 - a)\widehat k = -6\widehat i - 2\widehat j - (a + 2)\widehat k$ .

Step 3: Calculate the magnitude of $PQ$ : $|PQ| = \sqrt{(-6)^2 + (-2)^2 + (-a - 2)^2} = \sqrt{36 + 4 + (a + 2)^2} = \sqrt{40 + (a + 2)^2}$ .

Step 4: We are given that $|PQ| = 7$ , so $7 = \sqrt{40 + (a + 2)^2}$ .

Step 5: Square both sides: $49 = 40 + (a + 2)^2$ , which gives $(a + 2)^2 = 9$ .

Step 6: Take the square root: $a + 2 = \pm 3$ .

Step 7: Solve for a: $a = -2 \pm 3$ , so $a = -2 + 3 = 1$ or $a = -2 - 3 = -5$ .

Why other options are wrong:
- Option B (5, 1): This is incorrect because a = 5 does not satisfy the distance condition.
- Option C (0, 5): This is incorrect because neither a = 0 nor a = 5 satisfy the distance condition.
- Option D (1, 0): This is incorrect because a = 0 does not satisfy the distance condition.

Key Concepts:
- Position vector
- Vector subtraction
- Magnitude of a vector
- Distance between two points

Memory Aids:
- Remember that the distance between two points is the magnitude of the vector connecting them.

Quick Tips:
- Find the vector connecting the two points, then use the distance formula to solve for the unknown variable.

255 Easy

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Analyzing +4 | -1

If the position vectors of three vertices A, B and C of a

\Delta ABC

are

7\hat j + 10\hat k,\,\, - \hat i + 6\hat j + 6\hat k\,\,\,\,{\text{and}}\,\,\, - 4\hat i + 9\hat j + 6\hat 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 satisfies both the conditions for being right-angled and isosceles based on the side lengths.

Solution approach

Correct Answer + Core Reason: right angled and isosceles also. The triangle is right-angled and isosceles because two sides are equal and the Pythagorean theorem holds.

Step-by-Step Solution:

Step 1: Given the position vectors of A, B, and C are $A = 7\hat j + 10\hat k$ , $B = -\hat i + 6\hat j + 6\hat k$ , and $C = -4\hat i + 9\hat j + 6\hat k$ , respectively.

Step 2: Find the vectors $\vec{AB}$ , $\vec{BC}$ , and $\vec{AC}$ .
$\vec{AB} = B - A = (-1 - 0)\hat i + (6 - 7)\hat j + (6 - 10)\hat k = -\hat i - \hat j - 4\hat k$ .
$\vec{BC} = C - B = (-4 - (-1))\hat i + (9 - 6)\hat j + (6 - 6)\hat k = -3\hat i + 3\hat j + 0\hat k$ .
$\vec{AC} = C - A = (-4 - 0)\hat i + (9 - 7)\hat j + (6 - 10)\hat k = -4\hat i + 2\hat j - 4\hat k$ .

Step 3: Calculate the magnitudes of the vectors.
$|AB| = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18}$ .
$|BC| = \sqrt{(-3)^2 + (3)^2 + (0)^2} = \sqrt{9 + 9 + 0} = \sqrt{18}$ .
$|AC| = \sqrt{(-4)^2 + (2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$ .

Step 4: Check if the triangle is isosceles. Since $|AB| = |BC| = \sqrt{18}$ , the triangle is isosceles.

Step 5: Check if the triangle is right-angled. Check if the Pythagorean theorem holds: $|AC|^2 = |AB|^2 + |BC|^2$ .
$6^2 = (\sqrt{18})^2 + (\sqrt{18})^2 \Rightarrow 36 = 18 + 18 \Rightarrow 36 = 36$ . The Pythagorean theorem holds, so the triangle is right-angled.

Step 6: Final Answer: The triangle is right-angled and isosceles.

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

Key Concepts:
- Position vectors
- Vector subtraction
- 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 find the vectors representing the sides of the triangle first.
- Calculate the magnitudes of the vectors to determine the side lengths.
- Check if the triangle is isosceles or equilateral by comparing the side lengths.
- Check if the triangle is right-angled by verifying the Pythagorean theorem.

256 Easy

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Application +4 | -1

If vectors

{a_1} = x\widehat i - \widehat j + \widehat k

and

{a_2} = \widehat i + y\widehat j + z\widehat k

are collinear, then a possible unit vector parallel to the vector

x\widehat i + y\widehat j + z\widehat k

is

View answer and explanation

Correct option: D

\frac{1}{{\sqrt 3 }}\left( {\widehat i - \widehat j + \widehat k} \right)

Option D (

\frac{1}{{\sqrt 3 }}(\widehat i - \widehat j + \widehat k)

) is correct because it represents a possible unit vector parallel to the given vector when the collinearity condition is satisfied.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{\sqrt 3 }}\left( {\widehat i - \widehat j + \widehat k} \right)$ . The unit vector is found by using the collinearity condition and normalizing the resulting vector.

Step-by-Step Solution:

Step 1: Given ${a_1} = x\widehat i - \widehat j + \widehat k$ and ${a_2} = \widehat i + y\widehat j + z\widehat k$ are collinear, then $\frac{x}{1} = \frac{{ - 1}}{y} = \frac{1}{z} = \lambda$ .

Step 2: Express $x$ , $y$ , and $z$ in terms of $\lambda$ : $x = \lambda ,\,\,y = - \frac{1}{\lambda },\,\,z = \frac{1}{\lambda }$ .

Step 3: Find the vector $x\widehat i + y\widehat j + z\widehat k = \lambda \widehat i - \frac{1}{\lambda }\widehat j + \frac{1}{\lambda }\widehat k$ .

Step 4: Find the unit vector parallel to this vector:
$\frac{{ \pm \left( {\left( \lambda \right)\widehat i - \left( {\frac{1}{\lambda } \right)\widehat j + \left( {\frac{1}{\lambda } \right)\widehat k} \right)}}{{\sqrt {{{(\lambda )}^2} + {{\left( {\frac{{ - 1}}{\lambda } \right)}^2} + {{\left( {\frac{1}{\lambda } \right)}^2}} }} = \pm \frac{{({\lambda ^2}\widehat i - \widehat j + \widehat k)}}{{\sqrt {{\lambda ^4} + 2} }}$

Step 5: Take $\lambda = 1$ : $\frac{{ \pm (\widehat i - \widehat j + \widehat k)}}{{\sqrt 3 }}$

Why other options are wrong:
- Option A: $\frac{1}{{\sqrt 2 }}\left( { - \widehat j + \widehat k} \right)$ : This is incorrect because it does not satisfy the collinearity condition.
- Option B: $\frac{1}{{\sqrt 2 }}\left( {\widehat i - \widehat j} \right)$ : This is incorrect because it does not satisfy the collinearity condition.
- Option C: $\frac{1}{{\sqrt 3 }}\left( {\widehat i + \widehat j - \widehat k} \right)$ : This is incorrect because the signs of the components do not match the collinearity condition.

Key Concepts:
- Collinear vectors
- Unit vectors

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

Quick Tips:
- When finding unit vectors, always normalize the vector by dividing by its magnitude.
- Pay attention to the signs of the components when dealing with collinear vectors.

257 Easy

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Application +4 | -1

If the position vectors of P and Q are

(\hat i + 3\hat j - 7\hat k)\,\,{\text{and}}\,\,(5\hat i - 2\hat j + 4\hat k)

then

\left| {PQ} \right|

is

View answer and explanation

Correct option: D

\sqrt {162}

Option D (

\sqrt {162}

) is correct because the magnitude of the vector PQ is calculated to be

\sqrt {162}

.

Solution approach

Correct Answer + Core Reason: $\sqrt {162}$ . This is the magnitude of the vector PQ.

Step-by-Step Solution:

Step 1: Given position vectors $P = \widehat i + 3\widehat j - 7\widehat k$ and $Q = 5\widehat i - 2\widehat j + 4\widehat k$ .

Step 2: Find the vector $PQ = Q - P = (5 - 1)\widehat i + (-2 - 3)\widehat j + (4 - (-7))\widehat k = 4\widehat i - 5\widehat j + 11\widehat k$ .

Step 3: Calculate the magnitude of $PQ$ : $|PQ| = \sqrt{4^2 + (-5)^2 + 11^2} = \sqrt{16 + 25 + 121} = \sqrt{162}$ .

Why other options are wrong:
- **Option A ( $\sqrt {158}$ ): This is incorrect because it is not the correct magnitude.
- Option B ( $\sqrt {160}$ ): This is incorrect because it is not the correct magnitude.
- Option C ( $\sqrt {161}$ ): This is incorrect because it is not the correct magnitude.

Key Concepts:
- Position vector
- Vector subtraction
- Magnitude of a vector

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

Quick Tips:
- Find the vector PQ first, then calculate its magnitude.

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