Vector Algebra Question 1
Question 1 - 24 January - Shift 1
Let $m \vec{u}=\hat{i}-\hat{j}-2 \hat{k}, \vec{v}=2 \hat{i}+\hat{j}-\hat{k}, \vec{v} \cdot \vec{w}=2$
$\vec{v} \times \vec{w}=\vec{u}+\lambda \vec{v}$. Then $\vec{u} \cdot \vec{w}$ is equal to
(1) 1
(2) $\frac{3}{2}$
(3) 2
(4) $-\frac{2}{3}$
Show Answer
Answer: (1)
Solution:
Formula: Scalar product of vectors, Vactor product of vectors
$\overrightarrow{{}u}=(1,-1,-2), \overrightarrow{{}v}=(2,1,-1), \overrightarrow{{}v} \cdot \overrightarrow{{}w}=2$
$\overrightarrow{{}v} \times \overrightarrow{{}w}=\overrightarrow{{}u}+\lambda \overrightarrow{{}v}$
Taking dot with $\overrightarrow{{}w}$ 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 $\overrightarrow{{}v}$ in (1)
$\vec{v} \cdot(\vec{v} \times \vec{w})=\vec{u} \cdot \vec{v}+\lambda \vec{v} \cdot \vec{v}$
$\Rightarrow 0=(2-1+2)+\lambda(6)$
$\lambda=-\frac{1}{2}$
$\Rightarrow \overrightarrow{{}u} \cdot \overrightarrow{{}w}=-2 \lambda=1$
