Question

Using vectors, find the area of the triangle with vertices A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1)

Answer 1

Given three points A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1) 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 \(\text x_1\hat i+y_1\hat j+z_1\hat k \), 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

\(\vec{AC}\) = 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, –3, 1) and (b₁, b₂, b₃) = (3, 3, –4)

Thus, area of the triangle is  \(\cfrac{\sqrt{274}}2\) square units.