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Question

(D) The equation of the line $PM$ passing through $P(2, -1, 3)$ and perpendicular to the plane $x + 2y - z = 0$ is $\frac{x - 2}{1} = \frac{y + 1}{2}

Vedclass · Accepted Answer

$3 \sqrt{14}$

Vedclass · Answer

(D) The equation of the line $PM$ passing through $P(2, -1, 3)$ and perpendicular to the plane $x + 2y - z = 0$ is $\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 3}{-1} = \lambda$.Any point on this line is $(\lambda + 2, 2\lambda - 1, -\lambda + 3)$.For the foot of the perpendicular $M$,this point must satisfy the plane equation: $(\lambda + 2) + 2(2\lambda - 1) - (-\lambda + 3) = 0$.$\lambda + 2 + 4\lambda - 2 + \lambda - 3 = 0 \implies 6\lambda = 3 \implies \lambda = \frac{1}{2}$.The coordinates of $M$ are $(\frac{1}{2} + 2, 2(\frac{1}{2}) - 1, -\frac{1}{2} + 3) = (\frac{5}{2}, 0, \frac{5}{2})$.Let $Q(\alpha, \beta, \gamma)$ be the image of $P$. Since $M$ is the midpoint of $PQ$,we have $\frac{\alpha + 2}{2} = \frac{5}{2}$,$\frac{\beta - 1}{2} = 0$,and $\frac{\gamma + 3}{2} = \frac{5}{2}$.Solving these,we get $\alpha = 3, \beta = 1, \gamma = 2$. Thus,$Q = (3, 1, 2)$.The distance of $Q(3, 1, 2)$ from the plane $3x + 2y + z + 29 = 0$ is given by $d = \frac{|3(3) + 2(1) + 1(2) + 29|}{\sqrt{3^2 + 2^2 + 1^2}}$.$d = \frac{|9 + 2 + 2 + 29|}{\sqrt{9 + 4 + 1}} = \frac{42}{\sqrt{14}} = \frac{42 \sqrt{14}}{14} = 3 \sqrt{14}$.

Let the image of the point $P(2, -1, 3)$ in the plane $x + 2y - z = 0$ be $Q$. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point $Q$ is $.........$.

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