Question

Find the value of a for which the vectors 3i + 3j+9k and i+aj+3k are parallel (using cross product and plzz detailed answer)

Answer 1

Answer:

When the given vectors are parallel to each other, the value of a is 1.

Explanation:

Given two vectors, $\vec A=3\hat i + 3\hat j+9\hat k$

                       and $\vec B=\hat i + a\hat j+3\hat k$

The cross product or vector product of two vectors is given by

$\vec A\times \vec B=ABsin\theta$      

where θ is the angle between the vectors A and B .

Since A and B are parallel , the angle between the vectors is 0.

Therefore $sin \theta = sin 0 = 0$,

and thus the cross product will be $\vec A\times \vec B=ABsin\theta=0$

The cross product of

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

          $=(3\times 3-9\times a)\hat i-(3\times 3-9\times 1)\hat j+(3\times a-3\times 1)\hat k$

          $=(9-9a)\hat i-(9-9)\hat j+(3a-3)\hat k$

          $=(9-9a)\hat i+(3a-3)\hat k$

Since the vector product is 0 , its magnitude should be equal to zero.

i.e., $|\vec A\times \vec B|=0$

Therefore, $\sqrt{(9-9a)^2+(3a-3)^2} =0$

$\implies (9-9a)^2+(3a-3)^2} =0$

$\implies 81+81a^2-162a+9a^2+9-18a=0$

$\implies 90+90a^2-180a=0$

$\implies a^2-2a+1=0$

$\implies a^2-a-a+1=0$

$\implies a(a-1)-1(a-1)=0$

$\implies a-1=0$

$\implies a=1$

Therefore, when the given vectors are parallel to each other, the value of a is 1.