Vector Algebra Question 13
Question 13 - 2024 (31 Jan Shift 1)
Let $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}, \quad \vec{b}=4 \hat{i}+\hat{j}+7 \hat{k} \quad$ and $\overrightarrow{c}=\hat{i}-3 \hat{j}+4 \hat{k}$ be three vectors. If a vectors $\overrightarrow{p}$ satisfies $\vec{p} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{p} \cdot \vec{a}=0$, then $\vec{p} \cdot(\hat{i}-\hat{j}-\hat{k})$ is equal to
(1) 24
(2) 36
(3) 28
(4) 32
Show Answer
Answer (4)
Solution
$\overrightarrow{p} \times \overrightarrow{b}-\overrightarrow{c} \times \overrightarrow{b}=\overrightarrow{0}$
$(\overrightarrow{p}-\overrightarrow{c}) \times \overrightarrow{b}=\overrightarrow{0}$
$\overrightarrow{p}-\overrightarrow{c}=\lambda \overrightarrow{b} \Rightarrow \overrightarrow{p}=\overrightarrow{c}+\lambda \overrightarrow{b}$
Now, $\overrightarrow{p} \cdot \overrightarrow{a}=0$ (given)
So, $\overrightarrow{c} \cdot \overrightarrow{a}+\lambda \overrightarrow{a} \cdot \overrightarrow{b}=0$
$(3-3-8)+\lambda(12+1-14)=0$
$\lambda=-8$
$\overrightarrow{p}=\overrightarrow{c}-8 \overrightarrow{b}$
$\vec{p}=-31 \hat{i}-11 \hat{j}-52 \hat{k}$
So, $\vec{p} \cdot(\hat{i}-\hat{j}-\hat{k})$
$=-31+11+52$
$=32$