Question

If the vector $\overrightarrow{b}=3\hat{j}+4\hat{k}$ is written as the sum of a vector $\overrightarrow{b_1}$, parallel to $\overrightarrow{a}=\hat{i}+\hat{j}$ and a vector $\overrightarrow{b_2}$, perpendicular to $\overrightarrow{a}$, then $\overrightarrow{b_1}\times \overrightarrow{b_2}$ is equal to.

• A. $3\hat{i}-3\hat{j}+9\hat{k}$
• B. $-6\hat{i}+6\hat{j}-\frac{9}{2}\hat{k}$
• C. $6\hat{i}-6\hat{j}+\frac{9}{2}\hat{k}$
• D. $-3\hat{i}+3\hat{j}-9\hat{k}$

Answer 1

Answer: (C)

$6\hat{i}-6\hat{j}+\frac{9}{2}\hat{k}$

Given, $\overrightarrow{a}=\hat{i}+\hat{j}$

Since, $\overrightarrow{b_1}$ is parallel to $\overrightarrow{a}$, we can say $\overrightarrow{b_1}=k(\hat{i}+\hat{j})$

Let $\overrightarrow{b_2}=p\hat{i}+q\hat{j}+r\hat{k}$

Since $\overrightarrow{a}\perp \overrightarrow{b_2},\overrightarrow{a}.\overrightarrow{b_2}=0$

$\Rightarrow (p\hat{i}+q\hat{j}+r\hat{k})\cdot (\hat{i}+\hat{j})=0$

$\Rightarrow p+q=0$

Since, $\overrightarrow{b}=\overrightarrow{b_1}+\overrightarrow{b_2}$

$\therefore 3\hat{j}+4\hat{k}=k(\hat{i}+\hat{j})+(p\hat{i}-p\hat{j}+r\hat{k})$

Comparing the components, we get

$0=k+p$ ....$\left(1\right)$

$3=k-p$ ....$\left(2\right)$

$4=r$ ....$\left(3\right)$

Adding $\left(1\right)$ and $\left(2\right)$,

$2k=3\Rightarrow k=\frac{3}{2}$

From $\left(1\right)$, $p=-k=-\frac{3}{2}$

$\Rightarrow k=\frac{3}{2}$ and $p=-\frac{3}{2}$

Hence, $\overrightarrow{b_1}=\frac{3}{2}\hat{i}+\frac{3}{2}\hat{j}$ and $\overrightarrow{b_2}=-\frac{3}{2}\hat{i}+\frac{3}{2}\hat{j}+4\hat{k}$

$\Rightarrow \overrightarrow{b_1}\times \overrightarrow{b_2}$

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{3}{2}& \frac{3}{2}& 0\\ -\frac{3}{2}& \frac{3}{2}& 4\end{array}\right|=\hat{i}\left(6\right)-\hat{j}\left(6\right)+\hat{k}(\frac{9}{4}+\frac{9}{4})=6\hat{i}-6\hat{j}+\frac{9}{2}\hat{k}$