Let a be a unit vector perpendicular to the vectors b = i -2 j +3 k and c =2 i +3 j - k , and makes an angle of ^ -1 (-(1)/(3)) with the vector i + j + k . If a makes an angle of (π)/(3) with the vector i +α j + k

Correct answer: B. - 6 aligned & | array ccc i & j & k 1 & -2 & 3 2 & 3 & -1 array |= i (-7)+7 j +7 k & a = ± (-7 i +7 j +7 k ) 7^2+7^2+7^2 = ±( - i + j + k 3 ) aligned aligned & Now, θ= ± (-1+1+1) 3 · 3 = ± (1)/(3) & ^ -1 ((-1)/(3)) a = -(- i + j + k ) 3 & a = i - j - k 3 aligned aligned & (π)/(3)= 1-α-1 3 · α^2+2 & (1)/(2)= -α 3 · α^2+2 arrow α < 0 & 3(α^2+2)=4 α^2 & 6=α^2 & α= ± 6 aligned Clearly, α=- 6

JEE Main 202529 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Vector Algebra

Let a be a unit vector perpendicular to the vectors b = i -2 j +3 k...

Let

\hat{a} be a unit vector perpendicular to the vectors

\overrightarrow{\mathrm{b}}=\hat{i}-2 \hat{j}+3 \hat{k}

and

\overrightarrow{\mathrm{c}}=2 \hat{i}+3 \hat{j}-\hat{k}

, and makes an angle of

\cos ^{-1}

\left(-\frac{1}{3}\right)

with the vector

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

. If

\hat{\mathrm{a}}

makes an angle of

\frac{\pi}{3}

with the vector

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

, then the value of

\alpha

is :

Options

A $\sqrt{6}$
B $-\sqrt{6}$
C $-\sqrt{3}$
D $\sqrt{3}$

Correct answer

B. $-\sqrt{6}$

Step-by-step solution

\begin{aligned} & \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{array}\right|=\hat{i}(-7)+7 \hat{j}+7 \hat{k} \\ & \hat{a}= \pm \frac{(-7 \hat{i}+7 \hat{j}+7 \hat{k})}{\sqrt{7^2+7^2+7^2}}= \pm\left(\frac{-\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\right) \end{aligned}

\begin{aligned} & \text{ Now, } \cos \theta= \pm \frac{(-1+1+1)}{\sqrt{3} \cdot \sqrt{3}}= \pm \frac{1}{3} \\ & \Rightarrow \cos ^{-1}\left(\frac{-1}{3}\right) \Rightarrow \hat{a}=\frac{-(-\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}} \\ & \hat{a}=\frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}} \end{aligned}

\begin{aligned} & \cos \frac{\pi}{3}=\frac{1-\alpha-1}{\sqrt{3} \cdot \sqrt{\alpha^2+2}} \\ & \frac{1}{2}=\frac{-\alpha}{\sqrt{3} \cdot \sqrt{\alpha^2+2}} \rightarrow \alpha < 0 \\ & 3\left(\alpha^2+2\right)=4 \alpha^2 \\ & 6=\alpha^2 \\ & \alpha= \pm \sqrt{6} \end{aligned}

Clearly,

\alpha=-\sqrt{6}$

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