Question

The area of triangle whose vertices are (1, 2, 3), (2, 5, −1) and (−1, 1, 2) is
150 sq. units
145 sq. units

$\dfrac{\sqrt{155}}{2}$ sq. units

$\dfrac{155}{2}$ sq. units

Answer 1

Hint: All three vertices of the triangle are given. We have to find the vector AB and AC which represents the side vectors. Using the concept of cross product, we have to find cross product of the vector AB and AC. By using the formula of area of triangle, we can simply find the area of triangle ABC. The formula of area of triangle is given below:
Area of triangle $=\dfrac{1}{2}\times \left| \overrightarrow{AB}\times \overrightarrow{AC} \right|$

Complete step-by-step solution -

Given: The vertices of the triangle are:
A (1, 2, 3), B (2, 5, -1) and C (-1, 1, 2)
Consider the origin to be at (0, 0). Now, $\overrightarrow{AB}$ is the difference of $\overrightarrow{OA}$ from $\overrightarrow{OB}$.

& \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} \\\ & \overrightarrow{AB}=(2-1)\widehat{i}+(5-2)\widehat{j}+(-1-3)\widehat{k} \\\ & \overrightarrow{AB}=\widehat{i}+3\widehat{j}-4\widehat{k} \\\ \end{aligned}$$ Similarly, the $\overrightarrow{AC}$ is the difference of $\overrightarrow{OA}$ from $\overrightarrow{OC}$ $$\begin{aligned} & \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} \\\ & \overrightarrow{AC}=\left( -1-1 \right)\widehat{i}+\left( 1-2 \right)\widehat{j}+\left( 2-3 \right)\widehat{k} \\\ & \overrightarrow{AC}=-2\widehat{i}-\widehat{j}-\widehat{k} \\\ \end{aligned}$$ Then, the cross product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is: $$\begin{aligned} & \overrightarrow{AB}\times \overrightarrow{AC}=\left| \begin{matrix} i & j & k \\\ 1 & 3 & -4 \\\ -2 & -1 & -1 \\\ \end{matrix} \right| \\\ & \overrightarrow{AB}\times \overrightarrow{AC}=i\left( 3\times \left( -1 \right)-\left( -4 \right)\times \left( -1 \right) \right)-j\left( 1\times \left( -1 \right)-\left( -4 \right)\times \left( -2 \right) \right)+k\left( 1\times \left( -1 \right)-\left( 3 \right)\times \left( -2 \right) \right) \\\ & \overrightarrow{AB}\times \overrightarrow{AC}=-7i+9j+5k \\\ \end{aligned}$$ The magnitude of a resultant vector $$\overrightarrow{A}=x\widehat{i}+y\widehat{j}+z\widehat{k}$$, can be expanded as: $\left| \overrightarrow{A} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$ Now, by using the above-mentioned expression, the magnitude of both the vectors, $\begin{aligned} & \left| \overrightarrow{AB}\times \overrightarrow{AC} \right|=\sqrt{{{\left( -7 \right)}^{2}}+{{\left( 9 \right)}^{2}}+{{\left( 5 \right)}^{2}}} \\\ & \overrightarrow{AB}\times \overrightarrow{AC}=\sqrt{49+81+25} \\\ & \overrightarrow{AB}\times \overrightarrow{AC}=\sqrt{155} \\\ \end{aligned}$ Now, area of the triangle ABC $=\dfrac{1}{2}\times \left| \overrightarrow{AB}\times \overrightarrow{AC} \right|=\dfrac{1}{2}\times \sqrt{155}$ Area of the triangle ABC $=\dfrac{\sqrt{155}}{2}$sq. units. Hence, the area of the triangle ABC is $\dfrac{\sqrt{155}}{2}$ sq. units. Therefore, option (c) is correct. Note: The key concept involved in solving this problem is the expression of sides in the form of vectors. Students must be careful while calculating the cross product and must be aware of the intricacies of signs involved in cross product. General mistake is done in the calculation of cross product.