If P(3, 2, 6) is a point in space and Q is a | vedclass

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(A) Given point $P = (3, 2, 6)$.Point $Q$ on the line is $(1 - 3\mu, -1 + \mu, 2 + 5\mu)$.The vector $\vec{PQ} = (1 - 3\mu - 3)\hat{i} + (-1 + \mu - 2

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$\frac{1}{4}$

Vedclass · Answer

(A) Given point $P = (3, 2, 6)$.Point $Q$ on the line is $(1 - 3\mu, -1 + \mu, 2 + 5\mu)$.The vector $\vec{PQ} = (1 - 3\mu - 3)\hat{i} + (-1 + \mu - 2)\hat{j} + (2 + 5\mu - 6)\hat{k} = (-2 - 3\mu)\hat{i} + (-3 + \mu)\hat{j} + (-4 + 5\mu)\hat{k}$.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,$\vec{PQ} \cdot \vec{n} = 0$.$(-2 - 3\mu)(1) + (-3 + \mu)(-4) + (-4 + 5\mu)(3) = 0$.$-2 - 3\mu + 12 - 4\mu - 12 + 15\mu = 0$.$8\mu - 2 = 0$.$8\mu = 2$.$\mu = \frac{2}{8} = \frac{1}{4}$.

If $P(3, 2, 6)$ is a point in space and $Q$ is 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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