Question

The value of b such that the scalar product of the vector $i+j+k$ with the unit vector parallel to the sum of the vector $2i+4j-5k$, and $bi+2j+3k$ is one is

• A. -2
• B. -1
• C. 0
• D. 1

Answer 1

Answer: (D)

1

Let $r=(2i+4j-5k)+(bi+2j+3k)=(2+b)i+6j-2k$
Thus $\hat{r}=\frac{r}{|r|}=\frac{(2+b)i+6j-2k}{\sqrt{(2+b{)}^2+36+4}}$
Also given $\hat{r}\cdot (i+j+k)=1$
$\Rightarrow \frac{(2+b)+6-2}{\sqrt{(2+b{)}^2+36+4}}=1\Rightarrow b+6=\sqrt{b^2+4b+44}$
$\Rightarrow b^2+12b+36=b^2+4b+44\Rightarrow b=1$