Question

Q1.35. Show that vectors A = 2i - 3j- k and B = 6i+ 9j + 3k are parallel.​

Answer 1

Step-by-step explanation:

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

Explanation:

Given,

A = 2i - 3j - k

B = -6j + 9j + 3k

Two vectors are said to be parallel if there cross product is null vector.

Thus we have to show that,

$\vec{A}\times\vec{B}=\vec{0}$

Now consider,

$\vec{A}\times\vec{B}=(2\hat{i}-3\hat{j}-\hat{k})\times(-6\hat{i}+9\hat{j}+3\hat{k})$

$\begin{gathered}\vec{A}\times\vec{B}={\left|\begin{array}{ccc}{\hat{i}}&{\hat{j}}&{\hat{k}}\\2&{-3}&{-1}\\{-6}&9&3\end{array}\right|}\end{gathered}$

$\vec{A}\times\vec{B}=(-9+9)\hat{i}-(6-6)\hat{j}+(18-18)\hat{k}$

$\vec{A}\times\vec{B}=0\hat{i}-0\hat{j}+0\hat{k}$

$\vec{A}\times\vec{B}=\vec{0}$

Thus vector A and vector B are parallel.