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(C) Let $\vec A = 2\hat i + 2\hat j - \hat k$ and $\vec B = 6\hat i - 3\hat j + 2\hat k$.The vector perpendicular to both $\vec A$ and $\vec B$ is giv

Vedclass · Accepted Answer

$\frac{\hat i - 10\hat j - 18\hat k}{5\sqrt{17}}$

Vedclass · Answer

(C) Let $\vec A = 2\hat i + 2\hat j - \hat k$ and $\vec B = 6\hat i - 3\hat j + 2\hat k$.The vector perpendicular to both $\vec A$ and $\vec B$ is given by their cross product $\vec C = \vec A 	imes \vec B$.$\vec C = \begin{vmatrix} \hat i & \hat j & \hat k \ 2 & 2 & -1 \ 6 & -3 & 2 \end{vmatrix} = \hat i(4 - 3) - \hat j(4 + 6) + \hat k(-6 - 12) = \hat i - 10\hat j - 18\hat k$.The unit vector $\hat n$ is given by $\frac{\vec C}{|\vec C|}$.$|\vec C| = \sqrt{1^2 + (-10)^2 + (-18)^2} = \sqrt{1 + 100 + 324} = \sqrt{425} = \sqrt{25 	imes 17} = 5\sqrt{17}$.Therefore,$\hat n = \frac{\hat i - 10\hat j - 18\hat k}{5\sqrt{17}}$.

Column $I$	Column $II$
$(A) (A+B)$	$(p)$ North-east
$(B) (A-B)$	$(q)$ Vertically upwards
$(C) (A \times B)$	$(r)$ Vertically downwards
$(D) (A \times B) \times (A \times B)$	$(s)$ None

What is the unit vector perpendicular to the vectors $2\hat i + 2\hat j - \hat k$ and $6\hat i - 3\hat j + 2\hat k$?

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