Question

Find the value of 'a' for which the vectors 3i+3j+9k and i+aj-3k are parallel.

Answer 1

Answer:

$\frac{25}{3}$

Explanation:

Let,

$\vec{a} = 3i+3j+9k$

$\vec{b} = i+aj-3k$

The given two vectors are parallel. Thus, their dot product is equal to one. ...(Given)

$\vec{a}.\vec{b} = 1$

$\vec{a}.\vec{b} =(3)(1)+(3)(a)+(9)(-3)$

$\vec{a}.\vec{b} = 3+3a-27$

3a-24=1

3a=25

a= $\frac{25}{3}$

Thus, the value of a in the given question is $\frac{25}{3}$