The area of the triangle with vertices $(1,2,0)$, $(1,0,2)$ and $(0,3,1)$ is.
The area of the triangle with vertices $(1,2,0)$, $(1,0,2)$ and $(0,3,1)$ is.
- $\sqrt{3}$ sq. units
- $\sqrt{6}$ sq. units
- $\sqrt{5}$ sq. units
- $\sqrt{7}$ sq. units
Solution
$\begin{aligned} & \text {Let } \mathrm{A} \equiv(1,2,0), \mathrm{B} \equiv(1,0,2) \text { and } C \equiv(0,3,1) \\ \therefore \quad & \overline{\mathrm{AB}}=-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overline{\mathrm{AC}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \therefore \quad & \text { Area of } \triangle \mathrm{ABC}=\frac{1}{2}|\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}| \\ & |\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}|=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & -2 & 2 \\ -1 & 1 & 1\end{array}\right| \\ & =\hat{\mathrm{i}}(-2-2)-\hat{\mathrm{j}}(0+2)+\hat{\mathrm{k}}(0-2) \\ & =-4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\end{aligned}$
\(\begin{aligned} & \frac{\sqrt{4^2+2^2+2^2}}{2} \\ & =\sqrt{6} \end{aligned}\)
Asked in: MHT CET 2024 (11 May Shift 1)
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