Question

The value of $b$ such that scalar product of the vector $(\hat{i}+\hat{j}+\hat{k})$ with the unit vector parallel to the sum of the vectors $(2\hat{i}+4\hat{j}-5\hat{k})$ and $(b\hat{i}+2\hat{j}+3\hat{k})$ is $1$, is

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is D $1$

Parallel vector $=(2+b)i+6j-2k$
Unit vector $=\frac{(2+b)i+6j-2k}{\sqrt{b^2+4b+44}}$
According to the question,
$1=\frac{(2+b)+6-2}{\sqrt{b^2+4b+44}}$
$\Rightarrow b^2+4b+44=b^2+12b+36$
$\Rightarrow 8b=8\Rightarrow b=1$