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 \left(\right. 2 ,

Engineering

Mathematics

Line of Greatest Slope

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)

(- 28, - 43, - 69)

(- 32, - 37, - 71)

None of these

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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.

(- 32, - 37, - 71)

⟹ \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.

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