Question

Let $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3\hat{i}-5\hat{j}+4\hat{k}$ be two vectors, such that $\vec{a}\times \vec{b}=-\hat{i}+9\hat{i}+12k$. Then the projection of $\vec{b}-2\vec{a}$ on $\vec{b}+\vec{a}$ is equal to

• A. 2
• B. $\frac{39}{5}$
• C. 9
• D. $\frac{46}{5}$

Answer 1

Answer: (A), (B), (C), (D)

Let $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k},\vec{b}=3\hat{i}-5\hat{j}+4\hat{k}$
$\vec{a}\times \vec{b}=-\hat{i}+9\hat{j}+12\hat{k}$
$\Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \alpha & 1& \beta \\ 3& -5& 4\end{array}\right|$
$\Rightarrow (4+5\beta )\hat{i}+(3\beta -4\alpha )\hat{j}+(-5\alpha -3)\hat{k}$
$=-\hat{i}+9\hat{j}+12\hat{k}$
$\therefore 4+5\beta =-1,3\beta -4\alpha =9,-5\alpha -3=12$
$\beta =-1,\alpha =-3$
$\therefore \vec{a}=-3\hat{i}+\hat{j}-\hat{k},\vec{b}=3\hat{i}-5\hat{j}+4\hat{k}$
$\therefore \vec{a}+\vec{b}=-4\hat{j}+3\hat{k}$
$|\vec{a}{|}^2=11,|\vec{b}{|}^2=50$
$\vec{a}\cdot \vec{b}=-9+(-5)-4=-18$
$\therefore$ Projectile of $(\vec{b}-2\vec{a})$ on $\vec{a}+\vec{b}$ is
$\frac{(\vec{b}-2\vec{a})\cdot (\vec{a}+\vec{b})}{|\vec{a}+\vec{b}|}$
$=\frac{|\vec{b}{|}^2-2|\vec{a}{|}^2-(\vec{a}\cdot \vec{b})}{|\vec{a}+\vec{b}|}=\frac{50-22-(-18)}{5}=\frac{46}{5}$
Ans. $\left(\frac{46}{5}\right)$