If the line intersects the plane 2x + 3y – z + 13 = 0 at a point P and the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to

Engineering

Mathematics

Distance from a Plane

Equation of Straight Line in 3D and Distance Of a Point From Line

Question

If the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{- 1}$ intersects the plane 2x + 3y – z + 13 = 0 at a point P and the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to

$2 \sqrt{7}$

$\sqrt{14}$

$2 \sqrt{14}$

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$\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{- 1} = λ$

x = 3λ + 2, y = 2λ = 1, z = –λ + 1

Intersection with plane 2x + 3y – z + 13 = 0

2(3λ + 2) + 3(2λ –1) (-λ + 1) + 13 = 0

13λ + 13 = 0

λ = –1

$∴$ P(–1, – 3, 2)

Intersection with plane

3x + y + 4z = 16

3(3λ + 2) + (2λ – 1) +4(–λ + 1) = 16

λ = 1

Q(5, 1,0)

$PQ = \sqrt{6^{2} + 4^{2} + 2^{2}} = \sqrt{56} = 2 \sqrt{14}$