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

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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, $\vec{p} \cdot(\hat{i}-\hat{j}-\hat{k})$

$=-31+11+52$

$=32$