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Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

Question:

Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

Solution:

The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5), and C (1, 5, 5).

The adjacent sides $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{BC}}$ of $\triangle \mathrm{ABC}$ are given as:

$\overrightarrow{\mathrm{AB}}=(2-1) \hat{i}+(3-1) \hat{j}+(5-2) \hat{k}=\hat{i}+2 \hat{j}+3 \hat{k}$

$\overrightarrow{\mathrm{BC}}=(1-2) \hat{i}+(5-3) \hat{j}+(5-5) \hat{k}=-\hat{i}+2 \hat{j}$

Area of $\triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{BC}}|$

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

$\therefore|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{BC}}|=\sqrt{(-6)^{2}+(-3)^{2}+4^{2}}=\sqrt{36+9+16}=\sqrt{61}$

Hence, the area of $\triangle \mathrm{ABC}$ is $\frac{\sqrt{61}}{2}$ square units.

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