Question

Let $\vec{u}=\hat{i}-\hat{j}-2\hat{k},\vec{v}=2\hat{i}+\hat{j}-\hat{k},\vec{v}\cdot \vec{w}=2$ and $\vec{v}\times \vec{w}=\vec{u}+\lambda \vec{v}$. Then $\vec{u}\cdot \vec{w}$ is equal to

• A. 1
• B. 2
• C. $\frac{3}{2}$
• D. $-\frac{2}{3}$

Answer 1

Answer: (A)

$\overrightarrow{u}=(1,-1,-2),\overrightarrow{v}=(2,1,-1),\overrightarrow{v}\cdot \overrightarrow{w}=2$
$\overrightarrow{v}\times \overrightarrow{w}=\overrightarrow{u}+\lambda \overrightarrow{v}\dots \dots \dots \dots \dots \dots \dots (1)$
Taking dot with $\overrightarrow{w}\text{ in }(1)$
$\overrightarrow{w}\cdot (\overrightarrow{v}\times \overrightarrow{w})=\overrightarrow{u}\cdot \overrightarrow{w}+\lambda \overrightarrow{v}\cdot \overrightarrow{w}$
$\Rightarrow 0=\overrightarrow{u}\cdot \overrightarrow{w}+2\lambda$
Taking dot with $\vec{v}$ in (1)
$\overrightarrow{v}\cdot (\overrightarrow{v}\times \overrightarrow{w})=\overrightarrow{u}\cdot \overrightarrow{v}+\lambda \overrightarrow{v}\cdot \overrightarrow{v}$
$\Rightarrow 0=(2-1+2)+\lambda \cdot (6)$
$\lambda =-\frac{1}{2}$
$\Rightarrow \overrightarrow{u}\cdot \overrightarrow{w}=-2\lambda =1$