Question

The cartesian equations of a line are $6 x - 2 = 3 y + 1 = 2 z - 2$ . The vector equation of the line is

• A. $r = \left( \frac { 1 } { 3 } i - \frac { 1 } { 3 } j + k \right) + \lambda ( i + 2 j + 3 k )$
• B.
• C.
• D. None of these

Answer 1

Answer: (A)

$r = \left( \frac { 1 } { 3 } i - \frac { 1 } { 3 } j + k \right) + \lambda ( i + 2 j + 3 k )$

Answer 2

The given line is $6 x - 2 = 3 y + 1 = 2 z - 2$

⇒ $\frac { x - 1 / 3 } { 1 } = \frac { y + 1 / 3 } { 2 } = \frac { z - 1 } { 3 }$

This show that the given line passes through (1/3, –1/3) and has direction ratio 1, 2, 3.

Position vector and is parallel to vector . Hence, $r = \left( \frac { 1 } { 3 } i - \frac { 1 } { 3 } j + k \right) + \lambda ( i + 2 j + 3 k )$