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 $\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ be three vectors. If a vectors $\overrightarrow{\mathrm{p}}$ 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
$\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{0}$
$(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=\overrightarrow{0}$
$\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}+\lambda \overrightarrow{\mathrm{b}}$
Now, $\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{a}}=0$ (given)
So, $\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0$
$(3-3-8)+\lambda(12+1-14)=0$
$\lambda=-8$
$\overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}-8 \overrightarrow{\mathrm{b}}$
$\vec{p}=-31 \hat{i}-11 \hat{j}-52 \hat{k}$
So, $\vec{p} \cdot(\hat{i}-\hat{j}-\hat{k})$
$=-31+11+52$
$=32$
