Vectors 2 Question 19
19. The area of the triangle whose vertices are
$A(1,-1,2), B(2,1,-1) C(3,-1,2)$ is … .
(1983, 2M)
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Answer:
Correct Answer: 19. (a)
Solution:
- Area of $\triangle A B C=\frac{1}{2}|\overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{A C}}|$
$$ \begin{aligned} & \quad \overrightarrow{\mathbf{A B}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{A C}}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ & \therefore \quad \overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{A C}}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & -3 \\ 2 & 0 & 0 \end{array}\right|=2(-3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & \Rightarrow \text { Area of triangle }=\frac{1}{2}|\overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{A C}}| \\ &=\frac{1}{2} \cdot 2 \cdot \sqrt{9+4}=\sqrt{13} \text { sq units } \end{aligned} $$
