Question

Show that vectors A= 2 i - 3 j - k and B= -6i+9j+3k are parallel.

Answer 1

Given ,

A = 2i - 3j - k

B = -6i + 9j + 3k

We know that , if two vectors P and Q are parallel to each other , then their cross product is zero i.e

P × Q = 0

Thus ,

$\tt \implies \vec{A }× \vec{B} = \{2i - 3j - k \} \times \{ - 6i + 9j + 3k\}$

$\tt \implies \vec{A }× \vec{B} = \{ - 3 \times 3 - 9 ( - 1)\} - j \{2 \times 3 - ( - 6) ( - 1) \} + k \{ 2 \times 9 - ( - 6) ( - 3)\}$

$\tt \implies \vec{A }× \vec{B} = i\{ - 9 + 9\} - j \{6 - 6 \} + k \{ 18 - 18\}$

$\tt \implies \vec{A }× \vec{B} = \{ 0\} - j \{0 \} + k \{ 0\}$

$\tt \implies \vec{A }× \vec{B} =0$

Therefore , the given two vectors A and B are parallel

_________________ Keep Smiling ☺