In vector theory, vectors are visualized as directed line segments whose lengths are their magnitudes. We will use this concept well in this concept explanation, the area of a triangle formed by vectors. Normally when we try to find out the area of a triangle, we usually find out the value by the formula of Heron’s Formula. We can express the area of a triangle by vectors also.
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We know that in a parallelogram when the two adjacent sides are given by
\(\begin{array}{l}\vec {AB}\ \text{and}\ \vec{AC}\end{array} \)
and the angle between the two sides are given by θ then the area of the parallelogram will be given by \(\begin{array}{l}|\vec {AB} \times \vec {AC}|\end{array} \)
and the value will be given by; \(\begin{array}{l}|\vec {AB} | \times |\vec {AC}|\times \sin\ \theta\end{array} \)
A triangle divides a parallelogram into two equal parts, so the area of the triangle will be given by;
\(\begin{array}{l}\frac{1}{2}\times |\vec {AB} | \times |\vec {AC}|\times \sin\ \theta\end{array} \)
Formula
When two vectors are given:
Below are the expressions used to find the area of a triangle when two vectors are known.
\(\begin{array}{l}\frac{1}{2}\times |\vec {AB} \times \vec {AC}|\end{array} \)
or
\(\begin{array}{l}\frac{1}{2}\times |\vec {BC} \times \vec {CA}|\end{array} \)
or
\(\begin{array}{l}\frac{1}{2}\times |\vec {CB} \times \vec {CA}|\end{array} \)
When three vectors are given:
Expression to find the area of a triangle when three vectors will be given.
\(\begin{array}{l}\text{Let the sides of}\ \triangle ABC\ \text{be represented by}\ \vec a, \vec b\ \text{and}\ \vec c.\end{array} \)
Basically, they will give us the position vectors of the corresponding sides. If they are the position vectors of the ∆ABC, then the area of the triangle will be written as:
\(\begin{array}{l}\frac{1}{2}\times |\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a |\end{array} \)
Some important properties:
(i) If the three vertices of a ∆ABC be presented by
\(\begin{array}{l}\vec a, \vec b\ and\ \vec c\end{array} \)
then the centroid of triangle is \(\begin{array}{l}\frac{\vec a + \vec b + \vec c}{3}\end{array} \)
(ii) If the three vertices of a ∆ABC be presented by (x1, y1, z1 ),(x2, y2, z2) and (x3, y3, z3) then in that case centroid co-ordinate will be given by:
\(\begin{array}{l}\frac{(x_{1}+ x_{2}+ x_{3})}{3},\frac{(y_{1}+ y_{2}+ y_{3})}{3},\frac{(z_{1}+ z_{2}+ z_{3})}{3}\end{array} \)
and the area will be given by
\(\begin{array}{l}\text{Area}_{(yz)} = \frac{1}{2}\begin{vmatrix} y_1& z_1& 1\\ y_2&z_2 &1 \\y_3 &z_3 &1 \end{vmatrix}\end{array} \)
Or
\(\begin{array}{l}\text{Area}_{(zx)} = \frac{1}{2}\begin{vmatrix} z_1& x_1& 1\\ z_2&x_2 &1 \\z_3 &x_3 &1 \end{vmatrix}\end{array} \)
Or
\(\begin{array}{l}\text{Area}_{(xy)} = \frac{1}{2}\begin{vmatrix} x_1& y_1& 1\\ x_2&y_2 &1 \\x_3 &y_3 &1 \end{vmatrix}\end{array} \)
(iii) If given vectors are collinear, the angle between them will be 0° and the value of sin θ = sin(0°) = 0. So, the value of the area will be zero.
Solved Examples
Example 1: Consider a ∆ABC. O is any point inside it.
\(\begin{array}{l}\text{The values of}\ \vec {OA}, \vec {OB}\ and\ \vec {OC}\ \text{are given by}\end{array} \)
\(\begin{array}{l}\vec {OA}= \hat i + 3 \hat j+ 5 \hat k\end{array} \)
\(\begin{array}{l}\vec {OB}= 2 \hat i + 5 \hat j+ 7 \hat k\end{array} \)
and
\(\begin{array}{l}\vec {OC}= -3 \hat i + 10 \hat j – \hat k\end{array} \)
Find the area of the triangle.
Solution:
We know,
\(\begin{array}{l}\text{area of}\ \triangle ABC = \frac{1}{2}\times |\vec {AB} \times \vec {AC}|\end{array} \)
So,
\(\begin{array}{l}\vec {AB}=\vec {OB} – \vec {OA}\end{array} \)
that is given by the difference of the position vectors of A and B.
Now,
\(\begin{array}{l}\vec {OB} – \vec {OA} = (2 \hat i + 5 \hat j+ 7 \hat k) – (\hat i + 3 \hat j+ 5 \hat k) = \hat i + 2 \hat j+ 2 \hat k\end{array} \)
\(\begin{array}{l}\vec {AC}=\vec {OC} – \vec {OA}\end{array} \)
that is given by the difference of the position vectors of A and C.
\(\begin{array}{l}\vec {OC} – \vec {OA} = (-3 \hat i + 10 \hat j – \hat k) – (\hat i + 3 \hat j+ 5 \hat k) = -4 \hat i + 7 \hat j- 6 \hat k\end{array} \)
\(\begin{array}{l}\text{Now, determine}\ \vec {AB} \times \vec {AC}\ \text{which will be given by}:\end{array} \)
\(\begin{array}{l}\begin{vmatrix} \hat i & \hat j& \hat k \\ 1&2 &2 \\-4 &7 &-6 \end{vmatrix}\end{array} \)
\(\begin{array}{l}=\hat i(12-14) – \hat j (6-(-8) + \hat k (7-(-8) = -2\hat i – 14 \hat j + 15 \hat k\end{array} \)
Now,
\(\begin{array}{l}|\vec {AB} \times \vec {AC}| = \sqrt{4 + 169 + 225}=5 \sqrt{17}\end{array} \)
Therefore, area of triangle = 5(√17)/2 sq. units
Example 2: If x, y and z are the position vectors for three vertices of the ∆DEF. How to represent the area of the triangle in vector form?
Solution:
The cross products of the position vectors are given by |xy + yz + zx|, and the area will be given by:
(1/2) |xy + yz + zx|
So, the answer will be: 1/2 |xy + yz + zx|
Example 3:
If x, y and z are the position vectors for three vertices of the ∆DEF, then show the vector form of the unit vector perpendicular to the plane of the triangle.
Solution:
If the vectors of two sides of the triangle ∆DEF be given by
\(\begin{array}{l}\vec {DE}\ \text{and}\ \vec {DF},\end{array} \)
then both vectors lie on the plane of the ∆DEF. \(\begin{array}{l}\text{So the vector}\ |\vec {DE} \times \vec {DF}|\end{array} \)
will be the unit vector which will lie perpendicular to the plane of the ∆DEF.
So, the unit vector can be represented as:
Example 4:
\(\begin{array}{l}\text{If}\ \vec x, \vec y\ and\ \vec z\ \text{are the position vectors of the vertices X, Y and Z}\end{array} \)
of ∆XYZ, find the perpendicular distance of vertex X from the base YZ of ∆XYZ.
Solution:
We know,
\(\begin{array}{l}|\vec {yz} \times \vec {xy}| = |\vec x \times \vec y + \vec y \times \vec z + \vec z \times \vec y|\end{array} \)
\(\begin{array}{l}\Rightarrow |\vec {yz}|| \vec {xy}| sin\ Y = |\vec x \times \vec y + \vec y \times \vec z + \vec z \times \vec y|\end{array} \)
\(\begin{array}{l}\Rightarrow |\vec z – \vec y|(XY sin\ Y) = |\vec x \times \vec y + \vec y \times \vec z + \vec z \times \vec y|\end{array} \)
So, the length of the perpendicular from X on YZ will be given by:
Example 5: Find the area of the triangle with coordinates of points A (1, -1, 3), B (-2, -5, 4) and C (3, 1, -4).
Solution:
Calculate the vector by initial and terminal points.
\(\begin{array}{l}\overline {AB} = {B_{x} – A_{x}; B_{y} – A_{y}; B_{z} – A_{z}} = {-2 – 1; -5 – (-1); 4 – 3} = {-3; -4; 1}\\ \overline {AC} = {C_{x} – A_{x}; C_{y} – A_{y}; C_{z} – A_{z}} = {3 – 1; 1 – (-1); -4 – 3} = {2; 2; -7}\\ A=\frac{1}{2}|\overline{AB}*\overline{AC}|\\\end{array} \)
Calculate the cross-product of the vectors
\(\begin{array}{l}\overline {c}=\overline{AB}*\overline{AC}\\ \overline{AB}*\overline{AC}=\begin{vmatrix} i &j &k \\ AB_{x} &AB_{y} &AB_{z} \\ AC_{x}&AC_{y} &AC_{z} \end{vmatrix}\\ = i ((-4)\cdot (-7) – 1\cdot 2) – j ((-3)\cdot (-7) – 1\cdot 2) + k ((-3)\cdot 2 – (-4)\cdot 2)\\ = i (28 – 2) – j (21 – 2) + k (-6 + 8) = {26; -19; 2}\\\end{array} \)
Calculate the magnitude of the vector
\(\begin{array}{l}\overline {c}=\sqrt{{c_{x}^{2}}+{c_{y}^{2}}+{c_{z}^{2}}}\\=\sqrt{{26^{2}+[-19^{2}]+2^{2}}}\\=\sqrt{{676 + 361 + 4}}\\=\sqrt{1041}\end{array} \)
Calculate triangle area:
\(\begin{array}{l}A=\frac{1}{2}\sqrt{1041}\approx 16.132265804901678\end{array} \)
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Frequently Asked Questions
Q1
How do you find the area of a triangle when two vectors are known?
If vector AB and vector AC are given, then the area of triangle ABC = ½ |AB × AC|. We find the cross product of the vectors and divide by 2.
Q2
How do you find the area of a triangle when three vectors are given?
If vector a, vector b and vector c are the position vectors of the ∆ABC, then the area of the triangle = ½ |a×b + b×c + c×a|.
Q3
Give the formula for the Centroid of a triangle.
If the three vertices of the triangle are (x1, y1), (x2, y2), (x3, y3), the centroid of a triangle is given by ((x1+x2+x3)/3, (y1+y2+y3)/3).
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