Let P(3, 2, 6) be a point in space and Q be a | vedclass

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(A) The position vector of point $Q$ on the line is given by $\vec{q} = (1 - 3\mu)\hat{i} + (-1 + \mu)\hat{j} + (2 + 5\mu)\hat{k}$.Given point $P$ is

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$\frac{1}{4}$

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(A) The position vector of point $Q$ on the line is given by $\vec{q} = (1 - 3\mu)\hat{i} + (-1 + \mu)\hat{j} + (2 + 5\mu)\hat{k}$.Given point $P$ is $(3, 2, 6)$,so its position vector is $\vec{p} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.The vector $\vec{PQ} = \vec{q} - \vec{p} = (-3\mu - 2)\hat{i} + (\mu - 3)\hat{j} + (5\mu - 4)\hat{k}$.The normal vector to the plane $x - 4y + 3z = 1$ is $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$.Since $\vec{PQ}$ is parallel to the plane,it must be perpendicular to the normal vector $\vec{n}$,so $\vec{PQ} \cdot \vec{n} = 0$.$(-3\mu - 2)(1) + (\mu - 3)(-4) + (5\mu - 4)(3) = 0$.$-3\mu - 2 - 4\mu + 12 + 15\mu - 12 = 0$.$8\mu - 2 = 0$.$8\mu = 2 \Rightarrow \mu = \frac{2}{8} = \frac{1}{4}$.

Let $P(3, 2, 6)$ be a point in space and $Q$ be a point on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$. Then the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is

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