Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=2…

Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{c}}=\mathrm{pi}+\mathrm{q} \hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{d}}=\mathrm{p} \hat{\mathrm{j}}-\mathrm{q} \hat{\mathrm{k}}$ be four vectors. If $(\vec{a} \times \vec{b}) \cdot \vec{c}=3=(\vec{a} \times \vec{b}) \cdot \vec{d}$, then $3 \mathrm{p}+\mathrm{q}=$
  1. 0
  2. 3
  3. $-2$
  4. 6

Solution

$\vec{a}=\hat{i}-2 \hat{j}, \vec{b}=2 \hat{j}+3 \hat{k}, \vec{c}=p \hat{i}+q \hat{j}$ and $\vec{d}=p \hat{j}-q \hat{k}$ Also, we have $(\vec{a} \times \vec{b}) \cdot \vec{c}=3=(\vec{a} \times \vec{b}) \cdot \vec{d}$ $(\vec{a} \times \vec{b})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 0 \\ 0 & 2 & 3\end{array}\right|$ $\begin{aligned} & =\hat{i}(-6-0)-\hat{j}(3-0)+\hat{k}(2-0) \\ & =-6 \hat{i}-3 \hat{j}+2 \hat{k} \\ & \Rightarrow \quad \vec{a} \times \vec{b}=-6 \hat{i}-3 \hat{j}+2 \hat{k}\end{aligned}$ Then, $(\vec{a} \times \vec{b}) \cdot \vec{c}=(-6 \hat{i}-3 \hat{j}+2 \hat{k}) \cdot(p \hat{i}+q \hat{j})$ $\Rightarrow 3=0-6 p+2 q \Rightarrow-6 p-3 q=3$ ...(i) and $(\vec{a} \times \vec{b}) \cdot \vec{d}=(-6 \hat{i}-3 \hat{j}+2 \hat{k}) \cdot(p \hat{j}-q \hat{k})=-3 p-2 q$ $\Rightarrow 3=-3 p-2 q$ ...(ii) Solving eqns. (i) and (ii), we get : $p=1$ and $q=-3$ Now, $3 p+q=3 \times 1+(-3)=3-3=0$ $\Rightarrow 3 p+q=0$

Asked in: AP EAMCET 2023 (16 May Shift 1)

Practice more Vectors questions on Aicharya