Question

Let $\vec{a}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}$ be a vector such that $\vec{a}\times \vec{b}=2\hat{i}-\hat{k}$ and $\vec{a}\cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is :-

• A. $\frac{2}{\sqrt{21}}$
• B. $2\sqrt{\frac{3}{7}}$
• C. $\frac{2}{3}\sqrt{\frac{7}{3}}$
• D. $\frac{2}{3}$

Answer 1

Answer: (A)

Projection of $\vec{b}$ on vector $\vec{a}-\vec{b}$ is
$=\frac{\vec{b}\cdot (\vec{a}-\vec{b})}{|\vec{a}-\vec{b}|}$
$=\frac{\vec{a}\cdot \vec{b}-|b{|}^2}{\sqrt{a^2+b^2-2a\cdot b}}=\frac{3-b^2}{\sqrt{6+b^2-6}}=\frac{3-b^2}{b}$
$|\vec{a}\times \vec{b}{|}^2=5$
$a^2{b}^2-(a\cdot b{)}^2=5$
$6b^2=14\Rightarrow b^2=\frac{7}{3}$
$\therefore \frac{3-b^2}{b}=\frac{3-\frac{7}{3}}{\sqrt{\frac{7}{3}}}=2\times \sqrt{21}$