If the line vec{r} = (hat{i} - 2hat{j} - hat{ | vedclass

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(A) line is parallel to a plane if the direction vector of the line is perpendicular to the normal vector of the plane.Let the direction vector of the

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$2$

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(A) line is parallel to a plane if the direction vector of the line is perpendicular to the normal vector of the plane.Let the direction vector of the line be $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$.Let the normal vector of the plane be $\vec{n} = 3\hat{i} - 2\hat{j} - m\hat{k}$.Since the line is parallel to the plane,the dot product of $\vec{b}$ and $\vec{n}$ must be zero,i.e.,$\vec{b} \cdot \vec{n} = 0$.$(2\hat{i} + \hat{j} + 2\hat{k}) \cdot (3\hat{i} - 2\hat{j} - m\hat{k}) = 0$.$(2)(3) + (1)(-2) + (2)(-m) = 0$.$6 - 2 - 2m = 0$.$4 - 2m = 0$.$2m = 4$.$m = 2$.

If the line $\vec{r} = (\hat{i} - 2\hat{j} - \hat{k}) + \lambda (2\hat{i} + \hat{j} + 2\hat{k})$ is parallel to the plane $\vec{r} \cdot (3\hat{i} - 2\hat{j} - m\hat{k}) = 14$,then the value of $m$ is:

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