Question

If the scalar projection of vector $x^2\hat{i}-2x\hat{j}+\hat{k}$ on vector $\hat{i}+3\hat{j}+6\hat{k}$ is $\frac{1}{\sqrt{46}},$ then the value of $x$ can be

• A. $1$
• B. $2$
• C. $3$
• D. $5$

Answer 1

Answer: (A), (D)

$5$

The scalar projection of vector $x^2\hat{i}-2x\hat{j}+\hat{k}$ on vector $\hat{i}+3\hat{j}+6\hat{k}$ is
$\frac{(x^2\hat{i}-2x\hat{j}+\hat{k})\cdot (\hat{i}+3\hat{j}+6\hat{k})}{\sqrt{1+9+36}}=\frac{1}{\sqrt{46}}\Rightarrow x^2-6x+6=1\Rightarrow x^2-6x+5=0\Rightarrow (x-1)(x-5)=0\Rightarrow x=1,5$