Question

The distance of the point (-3, 2, 3) from the line passing through (4, 6, -2) and having direction ratios -1, 2, 3 is ______units.

Options

• 2√17

• 4√17

• 2√19

• 4√19

Answer 1

The distance of the point (-3, 2, 3) from the line passing through (4, 6, -2) and having direction ratios -1, 2, 3 is 2√19  units.

Explanation:

Let P ≡ (–3, 2, 3) 

Equation of a line passing through (4, 6, –2) and having d.r.’s -1, 2, 3 are

\[\frac{x-4}{-1}=\frac{y-6}{2}=\frac{z+2}{3}\]

Let \[\frac{x-4}{-1}=\frac{y-6}{2}=\frac{z+2}{3}=\lambda\]

⇒ x = 4 − λ, y = 2λ + 6, z = 3λ − 2

Co-ordinations of point

Q = (4 − λ, 2λ + 6, 3λ − 2)

∴ the d.r’s of PQ are

(4 − λ − (−3), 6 + 2λ − 2, −2 + 3λ − 3)

= 7 − λ, 4 + 2λ, −5 + 3λ

Since PQ is perpendicular to the given lines
∴ −1(7 − λ) + (2)(4 + 2λ) + 3(−5 + 3λ) = 0
⇒ −7 + λ + 8 + 4λ − 15 + 9λ = 0
⇒ λ = 1
∴ Co-ordinates of Q are (3, 8, 1)

\[\therefore\quad\mathrm{PQ}=\sqrt{\left(3-\left(-3\right)\right)^2+\left(8-2\right)^2+\left(1-3\right)^2}\]

= √76
= 2 √19