The length of the perpendicular drawn from th | vedclass

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(B) The plane contains the point $(1, 1, 0)$ and is parallel to the vectors $\vec{b_1} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b_2} = -\hat{i} + \ha

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$\sqrt{3}$

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(B) The plane contains the point $(1, 1, 0)$ and is parallel to the vectors $\vec{b_1} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b_2} = -\hat{i} + \hat{j} - 2\hat{k}$.The normal vector to the plane is $\vec{n} = \vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -1 \ -1 & 1 & -2 \end{vmatrix} = \hat{i}(-4+1) - \hat{j}(-2-1) + \hat{k}(1+2) = -3\hat{i} + 3\hat{j} + 3\hat{k}$.Dividing by $-3$,we can take the normal vector as $\vec{n} = \hat{i} - \hat{j} - \hat{k}$.The equation of the plane is $1(x-1) - 1(y-1) - 1(z-0) = 0$,which simplifies to $x - y - z = 0$.The perpendicular distance from the point $(2, 1, 4)$ to the plane $x - y - z = 0$ is given by $d = \frac{|2 - 1 - 4|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{|-3|}{\sqrt{3}} = \sqrt{3}$.

The length of the perpendicular drawn from the point $(2, 1, 4)$ to the plane containing the lines $\vec r = (\hat i + \hat j) + \lambda (\hat i + 2\hat j - \hat k)$ and $\vec r = (\hat i + \hat j) + \mu (-\hat i + \hat j - 2\hat k)$ is

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