Let be a unit vector perpendicular to the vectors and , and makes an angle of with the vector . If makes an angle of with the vector , then the value of is : [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let $\hat{a}$ be a unit vector perpendicular to the vectors $\vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\vec{c} = 2 \hat{i} + 3 \hat{j} - \hat{k}$ , and makes an angle of $\cos^{- 1} (- \frac{1}{3})$ with the vector $\hat{i} + \hat{j} + \hat{k}$ . If $\hat{a}$ makes an angle of $\frac{π}{3}$ with the vector $\hat{i} + α \hat{j} + \hat{k}$ , then the value of $α$ is : [2025]

1 $\sqrt{3}$

2 $\sqrt{6}$

3 $- \sqrt{6}$

4 $- \sqrt{3}$

Ans.
(3)

Let $\vec{m} = \hat{i} + \hat{j} + \hat{k}$

Since, $\vec{b} \times \vec{c} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 3 \\ 2 & 3 & - 1 \end{matrix} |$ $= - 7 \hat{i} + 7 \hat{j} + 7 \hat{k} = - 7 (\hat{i} - \hat{j} - \hat{k})$

$∵ \hat{a} parallel to \hat{b} \times \hat{c}$

$∴ \hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}} or \frac{- \hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

Now, $\cos θ = \frac{\hat{a} \cdot \vec{m}}{| \vec{m} |} = \frac{1 - 1 - 1}{\sqrt{3} \sqrt{3}} = \frac{- 1}{3}$ ${∵ \cos^{- 1} (\frac{- 1}{3}) = θ}$

or $\cos θ = \frac{- 1 + 1 + 1}{\sqrt{3} \sqrt{3}} = \frac{1}{3}$ (rejected)

Hence, $\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$

Now, $\cos (\frac{π}{3}) = \frac{\hat{a} \cdot (\hat{i} + α \hat{j} + \hat{k})}{\sqrt{1 + α^{2} + 1}} \Rightarrow \frac{1}{2} = \frac{1 - α - 1}{\sqrt{3} \sqrt{α^{2} + 2}}$

$\Rightarrow \sqrt{3} \sqrt{α^{2} + 2} = - 2 α \Rightarrow α = \frac{- \sqrt{3 (α^{2} + 2)}}{2}, i . e ., α < 0$

$\Rightarrow 3 α^{2} + 6 = 4 α^{2} \Rightarrow α = - \sqrt{6}$ .

Want Us to Email you About Special Offers & Updates?