Let $\mathbf{m}$ be a vector of magnitude $\sqrt{3}$ and perpendicular to the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\hat{\mathbf{j}}-\hat{\mathbf{k}}$. Let $\mathbf{n}$ be another vector of magnitude $2 \sqrt{6}$ and perpendicular to the vectors \(2…

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Let $\mathbf{m}$ be a vector of magnitude $\sqrt{3}$ and perpendicular to the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\hat{\mathbf{j}}-\hat{\mathbf{k}}$. Let $\mathbf{n}$ be another vector of magnitude $2 \sqrt{6}$ and perpendicular to the vectors $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}$ and $\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$. The area (in sq. units) of the triangle formed with $\mathbf{m}$ and $\mathbf{n}$ as sides is

Aicharya · Accepted Answer

Given, $\mathbf{m}=\sqrt{3} \times \text { unit vectors of }[(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}-\hat{\mathbf{k}})]$ $\begin{aligned} \because(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}-\hat{\mathbf{k}}) & =\left[\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ 1 & 1 & 0 \ 0 & 1 & -1 \end{array}\right] \ & =\hat{\mathbf{i}}(-1-0)-\hat{\mathbf{j}}(-1-0)+\hat{\mathbf{k}}(\mathbf{l}) \ & =-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \ \therefore \quad \mathbf{m} & =\sqrt{3} \times \frac{-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{1+1+1}} \ & =\sqrt{3} \times \frac{-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned}$ and $\mathbf{n}=2 \sqrt{6} \times$ unit vectors of $[(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}+2 \hat{\mathbf{k}})]$ $\begin{aligned} & \therefore(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \ &=\left[\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ 2 & -1 & 0 \ 0 & 1 & 2 \end{array}\right] \ &=\hat{\mathbf{i}}(-2-0)-\hat{\mathbf{j}}(4-0)+\hat{\mathbf{k}}(2+0)=-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \ & \mathbf{n}=2 \sqrt{6} \times \frac{-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{\sqrt{4+16+4}}=2 \sqrt{6} \times \frac{-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{2 \sqrt{6}} \end{aligned}$ $=-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ Now, required area $\begin{aligned} & =\frac{1}{2}(\mathbf{m} \times \mathbf{n})=\frac{1}{2}|(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \times(-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})| \ & =\frac{1}{2}\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ -1 & 1 & 1 \ -2 & -4 & 2 \end{array}\right| \ & =\frac{1}{2} \times \mid \hat{\mathbf{i}}(2+4)-\hat{\mathbf{j}}(-2+2)+\hat{\mathbf{k}}(4+2|| \ & =\frac{1}{2} \times|6 \hat{\mathbf{i}}+6 \hat{\mathbf{k}}|=\frac{1}{2} \sqrt{6^2+6^2}=\frac{1}{2} \times 6 \times \sqrt{2}=3 \sqrt{2} \end{aligned}$

Let \(\mathbf{m}\) be a vector of magnitude \(\sqrt{3}\) and perpendicular to the vectors…

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