Question

Using vectors, find the area of the triangle with vertices: A(1, 1, 2), B(2, 3, 5), and C(1 , 5, 5).

Answer 1

Hint: The coordinate of the points A, B, and C are \[\left( 1,1,2 \right)\] , \[\left( 2,3,5 \right)\] , and \[\left( 1,5,5 \right)\] . Convert the points A, B, and C in vector form by adding \[\widehat{i}\] , \[\widehat{j}\] , and \[\widehat{k}\] in x, y, and z coordinates. Now, use the formula that if we have three vectors \[\overrightarrow{A}\] , \[\overrightarrow{B}\] , and \[\overrightarrow{C}\] . Then, the area of \[\Delta ABC\] is given by the half of the vector product of \[\left( \overrightarrow{A}-\overrightarrow{C} \right)\] and \[\left( \overrightarrow{B}-\overrightarrow{C} \right)\] i.e., The area of \[\Delta ABC\] = \[\dfrac{1}{2}\left[ ~\left( \overrightarrow{A}-\overrightarrow{C} \right)\times \left( \overrightarrow{B}-\overrightarrow{C} \right) \right]\] and calculate the area of the triangle. At last calculate its magnitude.

Complete step by step answer:
According to the question, we are given the coordinates of three points of the triangle and we are asked to find the area of the triangle using vector form.
The coordinate of point A = \[\left( 1,1,2 \right)\] ………………………………….(1)
The coordinate of point B = \[\left( 2,3,5 \right)\] ……………………………………..(2)
The coordinate of point C = \[\left( 1,5,5 \right)\] ………………………………………..(3)
Now, converting the above points into position vectors by adding \[\widehat{i}\] , \[\widehat{j}\] , and \[\widehat{k}\] in x, y, and z coordinates.
The position vector of A = \[\left( 1\widehat{i}+1\widehat{j}+2\widehat{k} \right)\] …………………………………..(4)
The position vector of B = \[\left( 2\widehat{i}+3\widehat{j}+5\widehat{k} \right)\] …………………………………..(5)
The position vector of C = \[\left( 1\widehat{i}+5+5\widehat{k} \right)\] …………………………………..(6)

We know the formula that if we have three vectors \[\overrightarrow{A}\] , \[\overrightarrow{B}\] , and \[\overrightarrow{C}\] . Then, the area of \[\Delta ABC\] is given by the half of the vector product of \[\left( \overrightarrow{A}-\overrightarrow{C} \right)\] and \[\left( \overrightarrow{B}-\overrightarrow{C} \right)\] i.e.,
The area of \[\Delta ABC\] = \[\dfrac{1}{2}\left[ ~\left( \overrightarrow{A}-\overrightarrow{C} \right)\times \left( \overrightarrow{B}-\overrightarrow{C} \right) \right]\] ………………………………..(7)
From equation (4), and equation (6), we get
\[\left( \overrightarrow{A}-\overrightarrow{C} \right)=\left( 1\widehat{i}+1\widehat{j}+2\widehat{k} \right)-\left( 1\widehat{i}+5\widehat{j}+5\widehat{k} \right)=\left( 1-1 \right)i+\left( 1-5 \right)+\left( 2-5 \right)k=-4\widehat{j}-3\widehat{k}\] ………………………………………….(8)
Similarly, from equation (2), and equation (3), we get
\[\left( \overrightarrow{B}-\overrightarrow{C} \right)=\left( 2\widehat{i}+3\widehat{j}+5\widehat{k} \right)-\left( 1\widehat{i}+5\widehat{j}+5\widehat{k} \right)=\left( 2-1 \right)i+\left( 3-5 \right)j+\left( 5-5 \right)k=1i-2j\] ………………………………………….(9)
Now, from equation (7), equation (8), and equation (9), we get
The area of \[\Delta ABC\] = \[\dfrac{1}{2}\times \left[ \left( -4\widehat{j}-3\widehat{k} \right)\times \left( 1\widehat{i}-2\widehat{j} \right) \right]\] ……………………………………….(10)
We know the property that \[\widehat{i}\times \widehat{i}=0\] , \[\widehat{j}\times \widehat{j}=0\] , \[\widehat{k}\times \widehat{k}=0\] , \[\widehat{i}\times \widehat{j}=\widehat{k}\] , \[\widehat{i}\times \widehat{k}=-\widehat{j}\] , \[\widehat{j}\times \widehat{i}=-\widehat{k}\] ,
\[\widehat{j}\times \widehat{k}=\widehat{i}\] , \[\widehat{k}\times \widehat{i}=\widehat{j}\] , and \[\widehat{k}\times \widehat{j}=\widehat{-i}\] ……………………………………..(11)
Now, using equation (11) and on simplifying equation (10), we get
The area of \[\Delta ABC\] = \[\dfrac{1}{2}\times \left[ -6\widehat{i}-3\widehat{j}+4\widehat{k} \right]\] = \[-3\widehat{i}-\dfrac{3}{2}\widehat{j}+2\widehat{k}\] ……………………………………….(12)
We know the formula for the magnitude of a vector \[xi+yj+zk\] , Magnitude = \[\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\] …………………………………………………(13)
Now, from equation (12) and equation (13), we get
The area of \[\Delta ABC\] = \[\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -\dfrac{3}{2} \right)}^{2}}+{{\left( 2 \right)}^{2}}}=\sqrt{9+\dfrac{9}{4}+4}=\sqrt{\dfrac{36+9+16}{4}}=\dfrac{\sqrt{61}}{2}\] sq units.
Therefore, the area of the triangle is \[\dfrac{\sqrt{61}}{2}\] sq units.

Note:
We can also solve this question by calculating position vector form of the sides AB and AC. Now, use the formula, that the area of the triangle whose position vector of two sides are \[{{x}_{1}}\widehat{i}+{{y}_{1}}\widehat{j}+{{z}_{1}}\widehat{k}\] and \[{{x}_{2}}\widehat{i}+{{y}_{2}}\widehat{j}+{{z}_{2}}\widehat{k}\] is given by \[\dfrac{1}{2}\times \left| \begin{align}
  & \begin{matrix}
   \widehat{i} & \,\,\,\widehat{j} & \widehat{\,k} \\
\end{matrix} \\
 & \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & {{z}_{1}} \\
\end{matrix} \\
 & \begin{matrix}
   {{x}_{2}} & {{y}_{2}} & {{z}_{2}} \\
\end{matrix} \\
\end{align} \right|\] .