Find a vector x→ which is perpendicular to both A→ and B→ but has magnitude equal to that ofB→ . Rule: Inter change coeff. Of i^ and j^ and change sign of one of the vectors, A→= 3i^−2j^+k^,B→=4i^+3j^−2k^

1 10 ( i ^ − 10 j ^ + 17 k ^ )

Find a vector x→ which is perpendicular to both A→ and B→ but has magnitude equal to that ofB→ . Rule: Inter change coeff. Of i^ and j^ and change sign of one of the vectors, A→= 3i^−2j^+k^,B→=4i^+3j^−2k^

1 10 ( i ^ − 10 j ^ + 17 k ^ )

29 390 ( i ^ + 10 j ^ + 17 k ^ )

29 390 ( i ^ − 10 j ^ + 17 k ^ )

1 10 ( i ^ − 10 j ^ + 17 k ^ )

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Find a vector $\vec{x}$ which is perpendicular to both $\vec{A}$ and $\vec{B}$ but has magnitude equal to that of $\vec{B}$ .
Rule: Inter change coeff. Of $\hat{i}$ and $\hat{j}$ and change sign of one of the vectors,

$\vec{A} = 3 \hat{i} - 2 \hat{j} + \hat{k}, \vec{B} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\sqrt{\frac{29}{390}} (\hat{i} + 10 \hat{j} + 17 \hat{k})$

$\frac{1}{\sqrt{10}} (\hat{i} + 10 \hat{j} + 17 \hat{k})$

$\sqrt{\frac{29}{390}} (\hat{i} - 10 \hat{j} + 17 \hat{k})$

$\frac{1}{\sqrt{10}}$ $(\hat{i} - 10 \hat{j} + 17 \hat{k})$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$|\vec{B}| = \sqrt{4^{2} + 3^{2} + {(- 2)}^{2}} = \sqrt{29}$

$\vec{A} \times \vec{B} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 2 & 1 \\ 4 & 3 & - 2 \end{matrix} |$ $= \hat{i} [(4 - 3) - \hat{j} (- 6 - 4) + \hat{k} (9 + 8)$

$|\vec{A} \times \vec{B}| = \sqrt{1^{2} + 10^{2} + 17^{2}} = \sqrt{390}$

$\vec{n} = \frac{\vec{A} \times \vec{B}}{| \vec{A} \times \vec{B} |} = \frac{\hat{i} + 10 \hat{j} + 17 \hat{k}}{\sqrt{390}}$

$\vec{x} = | \vec{B} | \hat{n} = \frac{\sqrt{29} (\hat{i} + 10 \hat{j} + 17 \hat{k})}{\sqrt{390}}$

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th