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(C) The unit vector $\hat{n}$ perpendicular to the plane containing vectors $\vec{a}$ and $\vec{b}$ is given by $\hat{n} = \pm \frac{\vec{a} 	imes \v

Vedclass · Accepted Answer

$\frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7}$

Vedclass · Answer

(C) The unit vector $\hat{n}$ perpendicular to the plane containing vectors $\vec{a}$ and $\vec{b}$ is given by $\hat{n} = \pm \frac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|}$.First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -6 & -3 \ 4 & 3 & -1 \end{vmatrix}$$= \hat{i}((-6)(-1) - (-3)(3)) - \hat{j}((2)(-1) - (-3)(4)) + \hat{k}((2)(3) - (-6)(4))$$= \hat{i}(6 + 9) - \hat{j}(-2 + 12) + \hat{k}(6 + 24)$$= 15\hat{i} - 10\hat{j} + 30\hat{k}$.Now,find the magnitude $|\vec{a} 	imes \vec{b}| = \sqrt{15^2 + (-10)^2 + 30^2} = \sqrt{225 + 100 + 900} = \sqrt{1225} = 35$.The unit vector is $\pm \frac{15\hat{i} - 10\hat{j} + 30\hat{k}}{35} = \pm \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7}$.Comparing this with the given options,option $C$ is $\frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7}$.

Find the unit vector perpendicular to the plane containing the vectors $\vec{a} = 2\hat{i} - 6\hat{j} - 3\hat{k}$ and $\vec{b} = 4\hat{i} + 3\hat{j} - \hat{k}$.

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