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 and AC and the angle between the two sides are given by θ then the area of the parallelogram will be given by ∣AB×AC∣ and the value will be given by ∣AB∣×∣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∣× sinθ
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∣
or 1/2 x ∣BC×CA∣
or 1/2 x ∣CB×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,bandc. 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∣.
Some important properties:
(i) If the three vertices of a ∆ABC be presented by a,bandc then the centroid of triangle is 3a+b+c
(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: 3(x1+x2+x3),3(y1+y2+y3),3(z1+z2+z3) and
Or Area_(zx) = 1/2∣∣∣∣∣∣∣z1z2z3x1x2x3111∣∣∣∣∣∣∣
Or Area_(xy) = 1/2∣∣∣∣∣∣∣x1x2x3y1y2y3111∣∣∣∣∣∣∣
(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.
Example 1: Consider a ∆ABC. O is any point inside it. The values of OA,OBandOC are given by
Find the area of the triangle.
We know, area of ∆ABC = 1/2 x ∣AB×AC∣
So, AB = OB–OA that is given by the difference of the position vectors of A and B.
Now, OB–OA=(2i^+5j^+7k^)–(i^+3j^+5k^)=i^+2j^+2k^AC = OC–OA that is given by the difference of the position vectors of A and C.
Now, determine AB×AC which will be given by:
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?
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|
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.
If the vectors of two sides of the triangle ∆DEF be given by DE and DF then both vectors lie on the plane of the ∆DEF. So the vector ∣DE×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,yandzare 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.
We know, ∣yz×xy∣=∣x×y+y×z+z×y∣
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).