Question

Using vectors, find the area of the triangle with vertices A(2, 3, 5), B(3, 5, 8) and C(2, 7, 8).

Answer 1

Given three points A(2, 3, 5), B(3, 5, 8) and C(2, 7, 8) forming a triangle.

Let position vectors of the vertices A, B and C of ΔABC be  \(\vec a,\,\vec b\) and  \(\vec c\)respectively.

We know position vector of a point (x, y, z) is given by  \(\text x\hat i+y\hat j+z\hat k\), where \(\hat i\), \(\hat j\) and \(\hat k\) are unit vectors along X, Y and Z directions.

To find area of ΔABC, we need to find at least two sides of the triangle. So, we will find vectors \(\vec{AB}\) and \(\vec{AC}\).

Recall the vector \(\vec{AB}\) is given by

\(\vec{AB}\) = position vector of B  - position vector of A

Similarly, the vector \(\vec{AC}\)  is given by

\(\vec{AC}\) = position vector of C  - position vector of A

Recall the area of the triangle whose adjacent sides are given by the two vectors

Here, we have (a₁, a₂, a₃) = (1, 2, 3) and (b₁, b₂, b₃) = (0, 4, 3)

Recall the magnitude of the vector  \(\text x\hat i+y\hat j+z\hat k\)  is

Thus, area of the triangle is  \(\cfrac{\sqrt{61}}2\) square units.