The point on the line at a distance of 6 from the point (2,–3,

Engineering

Mathematics

Equation of Straight Line in 3D

Distance from a Plane

Question

The point on the line $\frac{x - 2}{1} = \frac{y + 3}{- 2} = \frac{z + 5}{- 2}$ at a distance of 6 from the point (2,–3,–5) is

(3,–5,–3)

(4,–7,–9)

(0,2,–1)

(–3,5,3)

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Given,

A= (2,–3,–5)

Let $\frac{x - 2}{1} = \frac{y + 3}{- 2} = \frac{z + 5}{- 2} = λ$

∴ x = λ + 2, y = –2–3, z = –2λ – 5

$PA = \sqrt{(λ + 2 - 2)^{2} + (- 2 λ - 3 + 3)^{2} + (- 2 λ - 5 + 5)^{2}}$

$6 = \sqrt{λ^{2} + 4 λ^{2} + 4 λ^{2}} = \sqrt{9 λ^{2}} = \sqrt{{(3 λ)}^{2}}$

6 = 3λ ⇒ λ = 2

Now,

x = λ + 2y = –2(2) – 3 z = –2 – 5

= 2 + 2 = –4 – 3 = –2(2) – 5

x = 4 y = –7 = –9

∴ P(x,y,z) = (4,–7,–9)

Hence Option B is correct.