Question:
The area of the triangle whose vertices are $A = (1, -1, 2), B = (2, 1, -1)$ and $C = (3, -1, 2)$ is
The area of the triangle whose vertices are $A = (1, -1, 2), B = (2, 1, -1)$ and $C = (3, -1, 2)$ is
Updated On: May 12, 2024
- $4 \sqrt{5}$ s units
- $2 \sqrt{3}$ s units
- $ \sqrt{13}$ s units
- $ \sqrt{15}$ s units
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The Correct Option is C
Solution and Explanation
Area of $\Delta\: ABC = \frac{1}{2} | \overrightarrow{BA} \times \overrightarrow{BC} |$
$\overrightarrow{BA} = - \hat{i} - 2 \hat{j} + 3 \hat{k} $
$\overrightarrow{BC} = \hat{i} - 2 \hat{j} + 3 \hat{k} $
$\therefore$ Area = $ \frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1& - 2&3\\ 1& -2 & 3 \end{vmatrix}$
$ = \frac{1}{6} | 6 \, \hat{j} + 4 \hat{k} | $
$ = |3 \hat{j} + 2 \hat{k} | = \sqrt{9 + 4} = \sqrt{13}$ s units
$\overrightarrow{BA} = - \hat{i} - 2 \hat{j} + 3 \hat{k} $
$\overrightarrow{BC} = \hat{i} - 2 \hat{j} + 3 \hat{k} $
$\therefore$ Area = $ \frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1& - 2&3\\ 1& -2 & 3 \end{vmatrix}$
$ = \frac{1}{6} | 6 \, \hat{j} + 4 \hat{k} | $
$ = |3 \hat{j} + 2 \hat{k} | = \sqrt{9 + 4} = \sqrt{13}$ s units
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Concepts Used:
Angle between Two Lines
The two straight lines, whenever intersects, form two sets of angles. The angles so formed after the intersection are;
- A pair of acute angle
- Another pair of an obtuse angle
The absolute values of angles created depend on the slopes of the intersecting lines.
It is also worth taking note, that the angle so formed by the intersection of two lines cannot be calculated if any of the lines is parallel to the y-axis as the slope of a line parallel to the y-axis is an indeterminate.
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