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question 29
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consider the points
a(1,0, - 1), b(0,2,1) and c(1,1,1).
the area of the triangle (widehat{abc}) is given by:
none of the given options
(\frac{7}{3}\text{ units}^2)
(\frac{3}{2}\text{ units}^2)
2 units²
5 units²

Explanation:

Step1: Find vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$

$\overrightarrow{AB}=(0 - 1,2 - 0,1-(-1))=(-1,2,2)$
$\overrightarrow{AC}=(1 - 1,1 - 0,1-(-1))=(0,1,2)$

Step2: Calculate the cross - product $\overrightarrow{AB}\times\overrightarrow{AC}$

$\overrightarrow{AB}\times\overrightarrow{AC}=

$$\begin{vmatrix} \vec{i}&\vec{j}&\vec{k}\\ -1&2&2\\ 0&1&2 \end{vmatrix}$$

=\vec{i}(4 - 2)-\vec{j}(- 2-0)+\vec{k}(-1 - 0)=2\vec{i}+2\vec{j}-\vec{k}=(2,2,-1)$

Step3: Find the magnitude of the cross - product

$|\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{2^{2}+2^{2}+(-1)^{2}}=\sqrt{4 + 4+1}=\sqrt{9}=3$

Step4: Calculate the area of the triangle

The area of triangle $ABC$ is $S=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|=\frac{3}{2}\text{ units}^2$

Answer:

$\frac{3}{2}\text{ units}^2$