Question

If $\overrightarrow{A}=2\hat{i}+3\hat{j}+6\hat{k}$ and $\overrightarrow{B}=3\hat{i}-6\hat{j}+2\hat{k}$ then vector perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{B}$ has magnitude $K$ times that of $6\hat{i}+2\hat{j}-3\hat{k}$. Then $K=$

• A. $1$
• B. $3$
• C. $7$
• D. $9$

Answer 1

Answer: (C)

$7$

It is given that , $A=2\hat{i}+3\hat{j}+6\hat{k}$ and $B=3\hat{i}-6\hat{j}+2\hat{k}.$

Let a vector R which is perpendicular to both A and B. it means, R is the cross product of A and B.

i.e., $R=\pm (A\times B)$

or, $|R|=|(A\times B)|$

so, magnitude of cross product of A and B, $|(A\times B)|=|A||B|sin\alpha$ , where $\alpha$ is angle between A and B.

|A| = $\sqrt{(2²+3²+6²)}=\sqrt{(4+9+36)}=7$

|B| = $\sqrt{3²+(-6)²+2²}=7$

and angle between them, $\alpha ={\cos }^{-1}$

$={\cos }^{-1}\frac{(2i+3j+6k)\cdot (3i-6j+2k)}{7\cdot 7}$

$={\cos }^{-1}\frac{(6-18+12)}{7.7}$

$={\cos }^{-1}\left(0\right)=\pi /2$

so, angle between them, $\alpha =90°$

now, $|R|=|(A\times B)|=|A||B|sin\alpha$

$=7\cdot 7\sin 90°=49unit.$

a/c to question,

$|R|=k|(6i+2j-3k)|$

or, $49=k\sqrt{(-6)²+\left(2\right)²+(-3)²}$

or, $49=k\times 7$

or, $k=7$

hence, value of k = 7