The area of the triangle, whose vertices are $\mathrm{A} \equiv(1,-1,2), \mathrm{B} \equiv(2,1,-1)$ and…

The area of the triangle, whose vertices are $\mathrm{A} \equiv(1,-1,2), \mathrm{B} \equiv(2,1,-1)$ and $\mathrm{C} \equiv(3,-1,2)$, is
  1. $2 \sqrt{3}$ sq.units
  2. $4 \sqrt{13}$ sq.units
  3. $\sqrt{13}$ sq.units
  4. $4 \sqrt{3}$ sq.units

Solution

$\begin{aligned} & \overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \overline{\mathrm{c}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \overline{\mathrm{AB}}=\overline{\mathrm{b}}-\overline{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\ & \overline{\mathrm{AC}}=\overline{\mathrm{c}}-\overline{\mathrm{a}}=2 \hat{\mathrm{i}}\end{aligned}$ $\begin{array}{ll} & \text { Area of triangle }=\frac{1}{2}|\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}| \\ \therefore & \overline{\mathrm{AB}} \times \overline{\mathrm{AC}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & -3 \\ 2 & 0 & 0\end{array}\right|=-6 \hat{\mathrm{j}}-4 \hat{\mathrm{k}} \\ \therefore \quad & |\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}|=\sqrt{36+16}=\sqrt{52}=2 \sqrt{13} \\ \therefore \quad & \text { Area of triangle }=\frac{1}{2}(2 \sqrt{13})=\sqrt{13} \text { sq. units }\end{array}$

Asked in: MHT CET 2024 (02 May Shift 2)

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