Find the co-ordinates of a point lying on the line which is at a distance 10 units from (2,

Engineering

Mathematics

Equation of Straight Line in 3D

Distance from a Plane

Question

Find the co-ordinates of a point lying on the line $\frac{x - 2}{3} = \frac{y + 3}{4} = \frac{z - 1}{7}$ which is at a distance 10 units from (2,–3,1).

(−32,−37,−71)

None of these

(32,37,71)

(−28,−43,−69)

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Given that the required point lies on the line

\frac{x - 2}{3} = \frac{y + 3}{4} = \frac{z - 1}{7}

The required point is at a distance of 10 units from (2,–3,1)

By option verification,

(A) Substituting the point (32,37,71)

⟹ \frac{32 - 2}{3} = \frac{37 + 3}{4} = \frac{71 - 1}{7}

⇒ 10 = 10 = 10

Therefore, distance is

\sqrt{(32 - 2)^{2} + (37 + 3)^{2} + (71 - 1)^{2}} = 86.02

units

Hence, (32,37,71) is not the required point.

(B) Substituting the point (–28,–43,–69)

⟹ \frac{- 28 - 2}{3} = \frac{- 43 + 3}{4} = \frac{- 69 - 1}{7}

⇒ –10=–10 =–10

Therefore, distance is

\sqrt{(- 28 - 2)^{2} + (- 43 + 3)^{2} + (- 69 - 1)^{2}} = 86.02

units

Hence, (–28,–43,–69) is not the required point.

⟹ \frac{- 32 - 2}{3} = \frac{- 37 + 3}{4} = \frac{- 71 - 1}{7}

⇒ 11.3 = 11.3 =10.28

Therefore, distance is

\sqrt{(- 32 - 2)^{2} + (- 37 + 3)^{2} + (- 71 - 1)^{2}} = 86.57

units

Hence, (–32,–37,–71) is not the required point.