Question

Vector(s) perpendicular to both the vectors
$\overrightarrow{A}=3\hat{i}+5\hat{j}+2\hat{k}$ and
$\overrightarrow{B}=2\hat{i}+4\hat{j}+6\hat{k}$ is/are

• A. $22\hat{i}-14\hat{j}+2\hat{k}$
• B. $6\hat{i}+20\hat{j}+12\hat{k}$
• C. $-(22\hat{i}-14\hat{j}+2\hat{k})$
• D. $-(6\hat{i}+20\hat{j}+12\hat{k})$

Answer 1

Answer: (A), (C)

$-(22\hat{i}-14\hat{j}+2\hat{k})$

The vector perpendicular to both the vectors can be derived from the cross product of the two vectors as,

$\lambda \left(\overrightarrow{A}\times \overrightarrow{B}\right)=\lambda \left(\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 5& 2\\ 2& 4& 6\end{array}\right|$

$\Rightarrow \lambda \left\{\hat{i}\right[(5\times 6)-(2\times 4)]-\hat{j}[(3\times 6)-(2\times 2)]+\hat{k}[(3\times 4)-(5\times 2)\left]\right\}$

$=\lambda (22\hat{i}-14\hat{j}+2\hat{k})$

Therefore, the vectors perpendicular to both the vectors are $\pm \left(22\hat{i}-14\hat{j}+2\hat{k}\right)$
corresponding to $\lambda =\pm 1.$
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