How to Find the Area of a Triangle using Vectors

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.

We know that in a parallelogram when the two adjacent sides are given by AB\vec {AB} and AC\vec {AC} and the angle between the two sides are given by θ then the area of the parallelogram will be given by AB×AC|\vec {AB} \times \vec {AC}| and the value will be given by AB×AC|\vec {AB} | \times |\vec {AC}|× sin⁡ θ.

A triangle divides a parallelogram into two equal parts, so the area of the triangle will be given by 1/2 x AB×AC|\vec {AB} | \times |\vec {AC}|× sin⁡θ

Formula

When two vectors are given:

Below are the expressions used to find the area of a triangle when two vectors are known.

1/2 x AB×AC|\vec {AB} \times \vec {AC}|

or 1/2 x BC×CA|\vec {BC} \times \vec {CA}|

or 1/2 x CB×CA|\vec {CB} \times \vec {CA}|

When three vectors are given:

Expression to find the area of a triangle when three vectors will be given.

Let the sides of ∆ABC be represented by a,b and c\vec a, \vec b\ and\ \vec c. 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

1/2 × a×b+b×c+c×a|\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a |.

Some important properties:

(i) If the three vertices of a ∆ABC be presented by a,b and c\vec a, \vec b\ and\ \vec c then the centroid of triangle is a+b+c3\frac{\vec a + \vec b + \vec c}{3}

(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: (x1+x2+x3)3,(y1+y2+y3)3,(z1+z2+z3)3\frac{(x_{1}+ x_{2}+ x_{3})}{3},\frac{(y_{1}+ y_{2}+ y_{3})}{3},\frac{(z_{1}+ z_{2}+ z_{3})}{3} and

the area will be given by

Area_(yz) = 1/2y1z11y2z21y3z31\begin{vmatrix} y_1& z_1& 1\\ y_2&z_2 &1 \\y_3 &z_3 &1 \end{vmatrix}

Or Area_(zx) = 1/2z1x11z2x21z3x31\begin{vmatrix} z_1& x_1& 1\\ z_2&x_2 &1 \\z_3 &x_3 &1 \end{vmatrix}

Or Area_(xy) = 1/2x1y11x2y21x3y31\begin{vmatrix} x_1& y_1& 1\\ x_2&y_2 &1 \\x_3 &y_3 &1 \end{vmatrix}

(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. The values of OA,OB and OC\vec {OA}, \vec {OB}\ and\ \vec {OC} are given by

OA=i^+3j^+5k^\vec {OA}= \hat i + 3 \hat j+ 5 \hat k OB=2i^+5j^+7k^\vec {OB}= 2 \hat i + 5 \hat j+ 7 \hat k and

OC=3i^+10j^k^\vec {OC}= -3 \hat i + 10 \hat j – \hat k

Find the area of the triangle.

Solution:

We know, area of ∆ABC = 1/2 x AB×AC|\vec {AB} \times \vec {AC}|

So, AB\vec {AB} = OBOA\vec {OB} – \vec {OA} that is given by the difference of the position vectors of A and B.

Now, OBOA=(2i^+5j^+7k^)(i^+3j^+5k^)=i^+2j^+2k^\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 AC\vec {AC} = OCOA\vec {OC} – \vec {OA} that is given by the difference of the position vectors of A and C.

OCOA=(3i^+10j^k^)(i^+3j^+5k^)=4i^+7j^6k^\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

Now, determine AB×AC\vec {AB} \times \vec {AC} which will be given by:

i^j^k^122476\begin{vmatrix} \hat i & \hat j& \hat k \\ 1&2 &2 \\-4 &7 &-6 \end{vmatrix}

= i^(1214)j^(6(8)+k^(7(8)=2i^14j^+15k^\hat i(12-14) – \hat j (6-(-8) + \hat k (7-(-8) = -2\hat i – 14 \hat j + 15 \hat k

Now, AB×AC=4+169+225=517|\vec {AB} \times \vec {AC}| = \sqrt{4 + 169 + 225}=5 \sqrt{17}

Therefore, area of triangle = 5/2 (√17) 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 to be 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 DE\vec {DE} and DF\vec {DF} then both vectors lie on the plane of the ∆DEF. So the vector DE×DF|\vec {DE} \times \vec {DF}| 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: If x,y and z\vec x, \vec y\ and\ \vec zare the position vectors of the vertices X,Y and Z of ∆XYZ, find the perpendicular distance of vertex X from the base YZ of ∆XYZ.

Solution:

We know, yz×xy=x×y+y×z+z×y|\vec {yz} \times \vec {xy}| = |\vec x \times \vec y + \vec y \times \vec z + \vec z \times \vec y|

=> yzxysin Y=x×y+y×z+z×y|\vec {yz}|| \vec {xy}| sin\ Y = |\vec x \times \vec y + \vec y \times \vec z + \vec z \times \vec y|

=> zy(XYsin Y)=x×y+y×z+z×y|\vec z – \vec y|(XY sin\ Y) = |\vec x \times \vec y + \vec y \times \vec z + \vec z \times \vec y|

So, the length of the perpendicular from X on YZ will be given by:

Area of Triangle Using Vectors Example

Example 5: Let us take a triangle ∆XYZ where the position vectors of X,Y and Z is given by

x=i^j^+3k^\vec x = \hat i – \hat j+ 3 \hat k y=2i^5j^+4k^\vec y = -2\hat i – 5\hat j+ 4 \hat k and

z=3i^+j^4k^\vec z = 3 \hat i + \hat j -4 \hat k

Find the centroid of the triangle. And find the length of the median from X upon YZ.

Solution: The position vectors of the three points X,Y and Z are given by (i^j^+3k^),(2i^5j^+4k^),(3i^+j^4k^)(\hat i – \hat j+ 3 \hat k), (-2\hat i – 5\hat j+ 4 \hat k), (3 \hat i + \hat j -4 \hat k) respectively.

We know, the formula for the centroid of a triangle is given by

(x1+x2+x3)3,(y1+y2+y3)3,(z1+z2+z3)3\frac{(x_{1}+ x_{2}+ x_{3})}{3},\frac{(y_{1}+ y_{2}+ y_{3})}{3},\frac{(z_{1}+ z_{2}+ z_{3})}{3}

by putting the values we get, the coordinates will be:

(12+3)3,(15+1)3,(3+44)/3)3=23,53,1\frac{(1-2+3)}{3},\frac{(-1-5+1)}{3},\frac{(3+4-4 )/3)}{3}=\frac{2}{3},\frac{-5}{3},1

Let us represent this point by G. We have to find the value of XT. We know that a centroid divides a median in the ratio 2 : 1. So, the ratio of XG to GT will be 2 : 1. Take the co-ordinate of T to be (a,b,c).

By section formula we can write,

mx2+nx1m+n coordinate of x,my2+ny1m+n coordinate of y,mz2+nz1m+n coordinate of z,\frac{mx_{2} + nx_{1}}{m+n} \rightarrow \text \ co-ordinate \ of \ x,\\ \frac{my_{2} + ny_{1}}{m+n} \rightarrow \text \ co-ordinate \ of \ y,\\ \frac{mz_{2} + nz_{1}}{m+n} \rightarrow \text \ co-ordinate \ of \ z,\\

Putting the respective values we get,

2[a+1]3=232a+1=2a=122[b1]3=132b1=1b=02[c+3]3=12c+3=3c=0\frac{2* [a + 1]}{3} = \frac{2}{3} \Rightarrow 2a+1 = 2 \Rightarrow a = \frac{1}{2}\\ \frac{2*[b-1]}{3} = \frac{-1}{3} \Rightarrow 2b – 1 = -1 \Rightarrow b=0\\ \frac{2*[c+3]}{3}= 1 \Rightarrow 2c + 3 = 3 \Rightarrow c = 0\\

So, the co-ordinates of T will be (1, 0, 0).

The median length will then be given as:

AT=((11/2)2+(10)2+(30)2)=1/4+1+9=41/4 units.AT = \sqrt{((1 – 1 / 2)^2+ (- 1 – 0)^2 + (3 – 0)^2 )} = \sqrt{1 / 4 + 1 + 9} = \sqrt{41 / 4} \text \ units.

Example 6: 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 vector by initial and terminal points.

AB=BxAx;ByAy;BzAz=21;5(1);43=3;4;1AC=CxAx;CyAy;CzAz=31;1(1);43=2;2;7A=12ABAC\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}|\\

Calculate the cross product of the vectors

c=ABACABAC=ijkABxAByABzACxACyACz=i((4)(7)12)j((3)(7)12)+k((3)2(4)2)=i(282)j(212)+k(6+8)=26;19;2\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}\\

Calculate the magnitude of the vector

c=cx2+cy2+cz2=262+[192]+22=676+361+4=1041\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}

Calculate triangle area:

A=12104116.132265804901678A=\frac{1}{2}\sqrt{1041}\approx 16.132265804901678

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