Question

The value of b such that the 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 one, is

• A. -2
• B. -1
• C. 1

Answer 1

The correct option is C 1

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
$\hat{n}=\frac{(2+b)\hat{i}+6\hat{j}-2\ddot{k}}{\sqrt{(2+b{)}^2}+6^2+(-2{)}^2}=\frac{(2+b)\hat{i}+6\hat{j}-2\ddot{k}}{\sqrt{b^2+4b+44}}$
Now,$(\hat{i}+\hat{j}+\hat{k})$.$\hat{n}$=1
$\Rightarrow 2+b+6-2=\sqrt{b^2+4b+44}\Rightarrow b=1$