Question

The distance of line 3y – 2z – 1 = 0 = 3x – z + 4 from the point (2, – 1, 6) is :

(1) √26

(2) 2√5

(3) 2√6

(4) 4√2

Answer 1

Correct option is (3) 2√6

Direction ratio's of given line

\(\begin{vmatrix} \hat i&\hat j&\hat k\\0&3&-2\\3&0&-1\end{vmatrix} = \hat i (-3) - \hat j(6) + \hat k (-9)\)

\(= -3\hat i - 6\hat j - 9\hat k\)

Let z = 0

⇒ y = \(\frac 13\) and x = \(-\frac 43\)

∴ Line in Cartesian form is

\(\cfrac{x + \frac 43}{-3} = \cfrac{y - \frac 13}{-6} = \cfrac z{-9}\)

Let point of shortest distance be P(λ) i.e.

\(P(-\lambda - \frac 43, -2\lambda + \frac 13, -3\lambda)\) and \(Q(2, -1,6)\)

For shortest distance 

\(\vec {PQ} . (\hat i + 2\hat j + 3\hat k) = 0\)

\(\left((\frac {10}3 + \lambda)\hat i + (2\lambda - \frac 43) \hat j + (6 + 3\lambda)\hat k\right).(\hat i + 2\hat j + 3\hat k) = 0\)

⇒ \(\lambda = -\frac 43\)

\(\therefore P = (0,3,4)\)

\(\therefore |PQ| = 2\sqrt 6\)

Answer 2

Option : (3). 2√6

3y – 2z – 1 = 0 = 3x – z + 4

3y – 2z – 1 = 0   D.R's ⇒ (0,3,–2)

3x – z + 4 = 0     D.R's ⇒ (3,-1,0)

Let DR's of given line are a, b, c

Now,

3b – 2c = 0 & 3a – c = 0

∴6a = 3b = 2c

a : b : c = 3 : 6 : 9

Any pt on line

3K – 1, 6K + 1, 9K + 1

Now,

3(3K – 1) + 6(6K + 1)1 + 9(9K + 1) = 0

⇒ K = \(\frac{1}{3}\)

Point on line ⇒ (0,3,4)

Given point (2,-1,6)

⇒ Distance = \(\sqrt{4+16+4}\) 

= 2√6