If the vertices of a $\triangle A B C$ are $A=(2,3,5)$, $B=(-1,3,2), C=(3,5,-2)$, then the area of the…
If the vertices of a $\triangle A B C$ are $A=(2,3,5)$, $B=(-1,3,2), C=(3,5,-2)$, then the area of the $\triangle A B C$ (in sq. units) is
- $6 \sqrt{2}$
- $8 \sqrt{3}$
- $9 \sqrt{2}$
- $8 \sqrt{2}$
Solution
The vertices of $\triangle A B C$ are
$
\begin{aligned}
& A=(2,3,5), B=(-1,3,2), C=(3,5,-2) \\
& \text { So, } \mathbf{A B}=(-1-2) \hat{\mathbf{i}}+(3-3) \hat{\mathbf{j}}+(2-5) \hat{\mathbf{k}}=-3 \hat{\mathbf{i}}-3 \hat{\mathbf{k}} \\
& \mathbf{A C}=(3-2) \hat{\mathbf{i}}+(5-3) \hat{\mathbf{j}}+(-2-5) \hat{\mathbf{k}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-7 \hat{\mathbf{k}} \\
& \text { Now, }|\mathbf{A B} \times \mathbf{A C}|=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
-3 & 0 & -3 \\
1 & 2 & -7
\end{array}\right| \\
& =\hat{\mathbf{i}}(0+6)-\hat{\mathbf{j}}(21+3)+\hat{\mathbf{k}}(-6+0) \\
& =6 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \\
& \text { Now, }|\mathbf{A B} \times \mathbf{A C}|=\sqrt{(6)^2+(-24)^2+(-6)^2} \\
& =\sqrt{36+576+36} \\
& =\sqrt{648}=18 \sqrt{2} \\
& \therefore \text { area of } \triangle A B C=\frac{1}{2}|\mathbf{A B} \times \mathbf{A C}| \\
& =\frac{1}{2} \times 18 \sqrt{2} \\
& =9 \sqrt{2} \text { sq. units. } \\
&
\end{aligned}
$
Asked in: AP EAMCET 2018 (23 Apr Shift 1)
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