The point P is the intersection of the straight line joining the 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 the point T (2, 1, 4) to QR, then the length of the line segment PS is

Engineering

Mathematics

Equation of Straight Line in 3D

Question

$2 \sqrt{2}$

$\sqrt{2}$

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

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Line QR $\frac{x - 1}{1} = \frac{y + 1}{4} = \frac{z - 4}{1} = r$

Let coordinates of point P are (r + 1, 4r – 1, r + 4) which lies on the plane

5x – 4y – z = 1

$\Rightarrow$ 5r + 5 – 4 (4r – 1) – (r + 4) = 1

$\Rightarrow$ – 12r + 5 = 1 $\Rightarrow$ 12r = 4 $\Rightarrow$ r = $\frac{1}{3}$

$∴ P \equiv (\begin{matrix} \frac{4}{3}, & \frac{1}{3}, & \frac{13}{3} \end{matrix})$

Let coordinate of S are ( $λ$ + 1, 4 $λ$ – 1, $λ$ + 4) ; ST $⊥$ QR

$∴$ ( $λ$ + 1 – 2) 1 + (4 $λ$ – 1 – 1) 4 + ( $λ$ + 4 – 4) 1 = 0

$λ$ – 1 + 16 $λ$ + $λ$ = 0 $\Rightarrow λ = \frac{1}{2}$

Coordinate of S are $(\begin{matrix} \frac{3}{2}, & 1, & \frac{9}{2} \end{matrix})$

Now, length PS = $\sqrt{{(\frac{4}{3} - \frac{3}{2})}^{2} + {(\frac{1}{3} - 1)}^{2} + {(\frac{13}{3} - \frac{9}{2})}^{2}} = \sqrt{\frac{1}{36} + \frac{4}{9} + \frac{1}{36}} = \frac{1}{\sqrt{2}}$