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 $\vec{a} \times \vec{b}=-\hat{i}+9 \hat{i}+12 k$. 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 \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & \beta \\ 3 & -5 & 4\end{vmatrix} $
$ \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. $\left(\frac{46}{5}\right)$