6. ⇒ (MHT CET 2023 10th May Evening Shift )

If $\overset{―}{a} = m \overset{―}{b} + nc$ , where $\overset{―}{a} = 4 \hat{i} + 13 \hat{j} - 18 \hat{k}, \overset{―}{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \overset{―}{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ , then $m + n =$

A. 1

B. 2

C. 3

D. -1

Correct answer option is (A)

Given:

$\begin{aligned} \bar{a} = 4 \hat{i} + 13 \hat{j} - 18 \hat{k} \\ \bar{b} = \hat{i} - 2 \hat{j} + 3 \hat{k} \\ \bar{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k} \end{aligned}$

Also, $\overset{―}{a} = m \overset{―}{b} + n \overset{―}{c}$

$\begin{aligned} \Rightarrow 4 \hat{i} + 13 \hat{j} - 18 \hat{k} = m (\hat{i} - 2 \hat{j} + 3 \hat{k}) + n (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) \\ \Rightarrow 4 \hat{i} + 13 \hat{j} - 18 \hat{k} \\ = (m + 2 n) \hat{i} + (- 2 m + 3 n) \hat{j} + (3 m - 4 n) \hat{k} \end{aligned}$

Comparing, we get

$m + 2 n = 4 and - 2 m + 3 n = 13$

Solving above equations, we get $m = - 2 $ a n d $ n = 3$

$∴ m + n = - 2 + 3 = 1$

7. ⇒ (MHT CET 2023 9th May Evening Shift )

Two adjacent sides of a parallelogram are $2 \hat{i} - 4 \hat{j} + 5 \hat{k}$ and $\hat{i} - 2 \hat{j} - 3 \hat{k}$ , then the unit vector parallel to its diagonal is

A. $\frac{3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k}$

B. $\frac{2}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{3}{7} \hat{k}$

C. $\frac{6}{7} \hat{i} - \frac{2}{7} \hat{j} + \frac{3}{7} \hat{k}$

D. $\frac{1}{7} \hat{i} + \frac{1}{7} \hat{j} - \frac{3}{7} \hat{k}$

Correct answer option is (A)

Let $\vec{a}$ and $\vec{b}$ be the adjacent sides of a parallelogram, where

$\begin{aligned} \vec{a} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} \\ \vec{b} = \hat{i} - 2 \hat{j} - 3 \hat{k} \end{aligned}$

Let diagonal be $\vec{c}$

$\begin{aligned} \vec{c} & = \vec{a} + \vec{b} \\ \vec{c} & = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} + \hat{i} - 2 \hat{j} - 3 \hat{k} \\ = 3 \hat{i} - 6 \hat{j} + 2 \hat{k} \end{aligned}$

$Magnitude of \begin{aligned} \vec{c} & = \sqrt{3^{2} + (- 6)^{2} + (2)^{2}} \\ = \sqrt{49} = 7 \end{aligned}$

$∴$ Unit vector in direction of diagonal $\vec{c}$ is

$\begin{aligned} = \frac{\vec{c}}{| \vec{c} |} \\ = \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) \\ = \frac{3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k} \end{aligned}$

8. ⇒ (MHT CET 2023 9th May Evening Shift )

If $D, E$ and $F$ are the mid-points of the sides $BC$ , $CA$ and $AB$ of triangle $ABC$ respectively, then $\overset{―}{AD} + \frac{2}{3} \overset{―}{BE} + \frac{1}{3} \overset{―}{CF} =$

A. $\frac{1}{2} \overset{―}{AB}$

B. $\frac{1}{2} \overset{―}{AC}$

C. $\frac{1}{2} \overset{―}{BC}$

D. $\frac{2}{3} \overset{―}{AC}$

Correct answer option is (B)

Let the position vector of A, B, C, D, E, F be $\overset{―}{a}, \overset{―}{b}, \overset{―}{c}, \overset{―}{d}, \overset{―}{e}, \overset{―}{f}$ respectively.

$∴ \overset{―}{d} = \frac{\overset{―}{b} + \overset{―}{c}}{2}, \overset{―}{e} = \frac{\overset{―}{c} + \overset{―}{a}}{2}, \overset{―}{f} = \frac{\overset{―}{a} + \overset{―}{b}}{2}$

Now, $\overset{―}{AD} + \frac{2}{3} \overset{―}{BE} + \frac{1}{3} \overset{―}{CF}$

$\begin{aligned} = \overset{―}{d} - \overset{―}{a} + \frac{2}{3} (\overset{―}{e} - \overset{―}{b}) + \frac{1}{3} (\overset{―}{f} - \overset{―}{c}) \\ = \frac{\overset{―}{b} + \overset{―}{c}}{2} - \overset{―}{a} + \frac{2}{3} (\frac{\overset{―}{c} + \overset{―}{a}}{2} - \overset{―}{b}) + \frac{1}{3} (\frac{\overset{―}{a} + \overset{―}{b}}{2} - \overset{―}{c}) \\ = \frac{\overset{―}{b} + \overset{―}{c} - 2 \overset{―}{a}}{2} + \frac{\overset{―}{c} + \overset{―}{a} - 2 \overset{―}{b}}{3} + \frac{\overset{―}{a} + \overset{―}{b} - 2 \overset{―}{c}}{6} \\ = \frac{3 \overset{―}{c} - 3 \overset{―}{a}}{6} \\ = \frac{3}{6} (\overset{―}{c} - \overset{―}{a}) \\ = \frac{1}{2} \overset{―}{AC} \end{aligned}$

9. ⇒ (MHT CET 2023 9th May Morning Shift )

If two vertices of a triangle are $A (3, 1, 4)$ and $B (- 4, 5, - 3)$ and the centroid of the triangle is $G (- 1, 2, 1)$ , then the third vertex $C$ of the triangle is

A. $(2, 0, 2)$

B. $(- 2, 0, 2)$

C. $(0, - 2, 2)$

D. $(2, - 2, 0)$

Correct answer option is (B)

Let $\overset{―}{a}, \overset{―}{b}, \overset{―}{c}$ and $\overset{―}{g}$ be the position vectors of $A, B, C$ and $G$ respectively.

$\begin{aligned} \bar{a} = 3 \hat{i} + 1 \hat{j} + 4 \hat{k}, \\ \bar{b} = - 4 \hat{i} + 5 \hat{j} - 3 \hat{k}, \\ \bar{g} = - \hat{i} + 2 \hat{j} + \hat{k}, \end{aligned}$

$G$ is centroid of $△ ABC$ .

$\begin{aligned} ∴ \bar{g} & = \frac{\bar{a} + \bar{b} + \bar{c}}{3} \\ 3 \bar{g} & = \bar{a} + \bar{b} + \bar{c} \\ 3 (- \hat{i} + 2 \hat{j} + \hat{k}) = 3 \hat{i} + \hat{j} + 4 \hat{k} - 4 \hat{i} + 5 \hat{j} - 3 \hat{k} + \bar{c} \\ ∴ \bar{c} & = - 3 \hat{i} + 6 \hat{j} + 3 \hat{k} - 3 \hat{i} - \hat{j} - 4 \hat{k} + 4 \hat{i} - 5 \hat{j} + 3 \hat{k} \\ = - 2 \hat{i} + 0 \hat{j} + 2 \hat{k} \end{aligned}$

$∴$ Third vertex $C \equiv (- 2, 0, 2)$

10. ⇒ (MHT CET 2023 9th May Morning Shift )

Let two non-collinear vectors $\hat{a}$ and $\hat{b}$ form an acute angle. A point $P$ moves, so that at any time $t$ the position vector $\overset{―}{OP}$ , where $O$ is origin, is given by $\hat{a} \sin t + \hat{b} \cos t$ , when $P$ is farthest from origin $O$ , let $M$ be the length of $\overset{―}{OP}$ and $\hat{u}$ be the unit vector along $\overset{―}{OP}$ , then

A. $\hat{u} = \frac{\hat{a} + \hat{b}}{| \hat{a} + \hat{b} |}$ and $M = (1 + \hat{a} \cdot \hat{b})^{\frac{1}{2}}$

B. $\hat{u} = \frac{\hat{a} - \hat{b}}{| \hat{a} - \hat{b} |}$ and $M = (1 + \hat{a} \cdot \hat{b})^{\frac{1}{2}}$

C. $\hat{u} = \frac{\hat{a} + \hat{b}}{| \hat{a} + \hat{b} |}$ and $M = (1 + 2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}$

D. $\hat{u} = \frac{\hat{a} - \hat{b}}{| \hat{a} - \hat{b} |}$ and $M = (1 - 2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}$

Correct answer option is (A)

$\begin{aligned} M = | \vec{OP} | \\ M = \sqrt{(\hat{a} \sin t + \hat{b} \cos t)^{2}} \\ = \sqrt{(\hat{a} \sin t)^{2} + (\hat{b} \cos t)^{2} + 2 (\hat{a} \sin t) \cdot (\hat{b} \cos t)} \\ = \sqrt{\sin^{2} t + \cos^{2} t + \hat{a} \cdot \hat{b} (2 \sin t \cos t)} \\ = \sqrt{1 + \hat{a} \cdot \hat{b} (\sin 2 t)} \\ Maximum value of \sin 2 t = 1 \\ ∴ 2 t = \sin^{- 1} (1) \\ ∴ t = \frac{π}{4} \\ ∴ M = \sqrt{1 + \hat{a} \cdot \hat{b} (l)} \\ = (1 + \hat{a} \cdot \hat{b})^{\frac{1}{2}} \\ Now, \hat{u} = \frac{\overset{―}{OP}}{| \overset{―}{OP} |} \\ = \frac{\hat{a} \sin t + \hat{b} \cos t}{| \hat{a} \sin t + \hat{b} \cos t |} \\ = \frac{\hat{a} (\frac{1}{\sqrt{2}}) + \hat{b} (\frac{1}{\sqrt{2}})}{| \hat{a} (\frac{1}{\sqrt{2}}) + \hat{b} (\frac{1}{\sqrt{2}}) |} \end{aligned}$

Unit vector of OP is

$\hat{u} = \frac{\hat{a} + \hat{b}}{| (\hat{a} + \hat{b}) |}$

⇒Vectors