Question

Using vectors, find the area of the triangle with vertices:
(i) A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)
(ii) A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1) [CBSE 2011, NCERT EXEMPLAR]

Answer 1

(i) The vertices of the triangle are A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Position vector of A = $\hat{i}+\hat{j}+2\hat{k}$

Position vector of B = $2\hat{i}+3\hat{j}+5\hat{k}$

Position vector of C = $\hat{i}+5\hat{j}+5\hat{k}$

$\overrightarrow{\mathrm{AB}}=\left(2\hat{i}+3\hat{j}+5\hat{k}\right)-\left(\hat{i}+\hat{j}+2\hat{k}\right)=\hat{i}+2\hat{j}+3\hat{k}$

$\overrightarrow{\mathrm{AC}}=\left(\hat{i}+5\hat{j}+5\hat{k}\right)-\left(\hat{i}+\hat{j}+2\hat{k}\right)=4\hat{j}+3\hat{k}$

Now,

$\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 0& 4& 3\end{array}\right|=-6\hat{i}-3\hat{j}+4\hat{k}$

∴ Area of ∆ABC = $\frac{1}{2}\left|\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}\right|$

$$\begin{gathered} =\frac{1}{2}\left|-6\hat{i}-3\hat{j}+4\hat{k}\right| \\ =\frac{1}{2}\sqrt{{\left(-6\right)}^2+{\left(-3\right)}^2+4^2} \\ =\frac{1}{2}\sqrt{36+9+16} \\ =\frac{\sqrt{61}}{2}\mathrm{square}\mathrm{units} \end{gathered}$$

(ii) The vertices of the triangle are A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1).

Position vector of A = $\hat{i}+2\hat{j}+3\hat{k}$

Position vector of B = $2\hat{i}-\hat{j}+4\hat{k}$

Position vector of C = $4\hat{i}+5\hat{j}-\hat{k}$

$\overrightarrow{\mathrm{AB}}=\left(2\hat{i}-\hat{j}+4\hat{k}\right)-\left(\hat{i}+2\hat{j}+3\hat{k}\right)=\hat{i}-3\hat{j}+\hat{k}$

$\overrightarrow{\mathrm{AC}}=\left(4\hat{i}+5\hat{j}-\hat{k}\right)-\left(\hat{i}+2\hat{j}+3\hat{k}\right)=3\hat{i}+3\hat{j}-4\hat{k}$

Now,

$\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 1\\ 3& 3& -4\end{array}\right|=9\hat{i}+7\hat{j}+12\hat{k}$

∴ Area of ∆ABC = $\frac{1}{2}\left|\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}\right|$

$$\begin{gathered} =\frac{1}{2}\left|9\hat{i}+7\hat{j}+12\hat{k}\right| \\ =\frac{1}{2}\sqrt{9^2+7^2+{12}^2} \\ =\frac{1}{2}\sqrt{81+49+144} \\ =\frac{\sqrt{274}}{2}\mathrm{square}\mathrm{units} \end{gathered}$$