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  • Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C(4, 5, – 1).

Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C(4, 5, – 1).

Answers (1)

We have,

The coordinates of points A, B and C are (1, 2, 3), (2, -1, 4) and (4, 5, -1) respectively.

We need to find the area of this triangle ABC.

We have the formula given as,

\\ \begin{aligned} &\text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|_{\ldots \text { (i) }}\\ &\text { Let us find out } \overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}} \text { first. } \end{aligned}

\\ \begin{aligned} &\text { We can say, }\\ &\text { Position vector of } \mathrm{A}=\hat{i}+2 \hat{\jmath}+3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OA}}=\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}\\ &\text { Position vector of } \mathrm{B}=2 \hat{\mathrm{l}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \end{aligned}

\\ \begin{aligned} &\text { Position vector of } \mathrm{c}=4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}}\\ &\overrightarrow{\mathrm{AB}}\\ &\overrightarrow{\mathrm{AB}}=\text { position vector of B-Position vector of } \mathrm{A} \end{aligned}

\\ \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=(2 \hat{\imath}-\hat{\jmath}+4 \hat{\mathrm{k}})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{AB}}=2 \hat{\imath}-\hat{\imath}-\hat{\jmath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}-3 \hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=\hat{i}-3 \hat{\jmath}+\hat{\mathrm{k}} \\ \text { For } \overrightarrow{\mathrm{AC}}

\\ \begin{aligned} &\overrightarrow{\mathrm{AC}}=\text { position vector of } \mathrm{C} \text { -Position vector of } \mathrm{A}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=(4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}})\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=4 \hat{\imath}-\hat{\imath}+5 \hat{\jmath}-2 \hat{\jmath}-\hat{\mathrm{k}}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=3 \hat{\imath}+3 \hat{\jmath}-4 \hat{\mathrm{k}} \end{aligned}

\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\hat{\imath}(12-3)-\hat{\jmath}(-4-3)+\hat{\mathrm{k}}(3+9)\\ &\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=9 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\\ &\text { And, }\\ &|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=|9 \hat{\imath}+7 \hat{\jmath}+12 \hat{\mathrm{k}}|\\ &\Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{9^{2}+7^{2}+12^{2}} \end{aligned}

\\ \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{81+49+144} \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{274}

$ From equation (i), we get $\\ Area of \triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$ \\$\Rightarrow$ Area of $\Delta \mathrm{ABC}=\frac{1}{2} \times \sqrt{274}$

Thus, area of yriangle ABC is \frac{\sqrt{274}}{2}{} $ sq units .

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