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(A) The equation of the line $QR$ passing through $Q(2, 3, 5)$ and $R(1, -1, 4)$ is given by $\frac{x-2}{1-2} = \frac{y-3}{-1-3} = \frac{z-5}{4-5} \Ri

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$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) The equation of the line $QR$ passing through $Q(2, 3, 5)$ and $R(1, -1, 4)$ is given by $\frac{x-2}{1-2} = \frac{y-3}{-1-3} = \frac{z-5}{4-5} \Rightarrow \frac{x-2}{-1} = \frac{y-3}{-4} = \frac{z-5}{-1} = \lambda$.Any point on this line is $P(\lambda+2, 4\lambda+3, \lambda+5)$.Since $P$ lies on the plane $5x - 4y - z = 1$,we have $5(\lambda+2) - 4(4\lambda+3) - (\lambda+5) = 1$.$5\lambda + 10 - 16\lambda - 12 - \lambda - 5 = 1 \Rightarrow -12\lambda - 7 = 1 \Rightarrow -12\lambda = 8 \Rightarrow \lambda = -\frac{2}{3}$.Thus,$P = (-\frac{2}{3}+2, 4(-\frac{2}{3})+3, -\frac{2}{3}+5) = (\frac{4}{3}, \frac{1}{3}, \frac{13}{3})$.Now,let $S$ be the foot of the perpendicular from $T(2, 1, 4)$ to the line $QR$. Let $S = (\mu+2, 4\mu+3, \mu+5)$.The direction ratios of $TS$ are $(\mu+2-2, 4\mu+3-1, \mu+5-4) = (\mu, 4\mu+2, \mu+1)$.Since $TS \perp QR$,the dot product of the direction ratios of $TS$ and $QR(1, 4, 1)$ is zero: $1(\mu) + 4(4\mu+2) + 1(\mu+1) = 0$.$\mu + 16\mu + 8 + \mu + 1 = 0 \Rightarrow 18\mu = -9 \Rightarrow \mu = -\frac{1}{2}$.Thus,$S = (-\frac{1}{2}+2, 4(-\frac{1}{2})+3, -\frac{1}{2}+5) = (\frac{3}{2}, 1, \frac{9}{2})$.The length $PS = \sqrt{(\frac{3}{2}-\frac{4}{3})^2 + (1-\frac{1}{3})^2 + (\frac{9}{2}-\frac{13}{3})^2} = \sqrt{(\frac{1}{6})^2 + (\frac{2}{3})^2 + (\frac{1}{6})^2} = \sqrt{\frac{1}{36} + \frac{16}{36} + \frac{1}{36}} = \sqrt{\frac{18}{36}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

Point $P$ is the intersection of the line joining points $Q(2, 3, 5)$ and $R(1, -1, 4)$ with the plane $5x - 4y - z = 1$. If $S$ is the foot of the perpendicular drawn from point $T(2, 1, 4)$ to the line $QR$,find the length of the line segment $PS$.

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