If the vectors $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}},…

If the vectors $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\mathrm{p} \overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}$ are coplanar, then $|\overline{\mathrm{a}} \times \overline{\mathrm{c}}|=$
  1. $\sqrt{14}$
  2. $\frac{3 \sqrt{10}}{2}$
  3. $\sqrt{26}$
  4. $\frac{\sqrt{90}}{4}$

Solution

$\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{c}}=\mathrm{p} \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ $\because \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} \& \overrightarrow{\mathrm{c}}$ are coplanar $ \begin{aligned} & \therefore a(\vec{b} \times \vec{c})=[\vec{a}, \vec{b}, \vec{c}]=0 \\ & \Rightarrow\left|\begin{array}{ccc} 2 & 3 & -1 \\ 4 & -1 & 3 \\ p & 1 & -1 \end{array}\right| \\ & \Rightarrow 2(1-3)-3(-4-3 p)-1(4+p)=0 \\ & \Rightarrow-4+12+9 p-4-p=0 \\ & \Rightarrow 8 p=-6 \Rightarrow p=\frac{-3}{4} \Rightarrow 8 p=-4 \Rightarrow p=-\frac{1}{2} \end{aligned} $ $\begin{aligned} & \text { Now, } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & -1 \\ -\frac{1}{2} & 1 & -1\end{array}\right| \\ & \Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=-2 \hat{\mathrm{i}}+\frac{5}{2} \hat{\mathrm{j}}+\frac{7}{2} \hat{\mathrm{k}} \\ & \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|=\sqrt{4+\frac{25}{4}+\frac{49}{4}}=\sqrt{\frac{90}{4}}=\frac{3 \sqrt{10}}{2}\end{aligned}$

Asked in: AP EAMCET 2023 (18 May Shift 2)

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