Find the area of triangle whose two sides are represented by the vectors 3i + 4j and 5i + 7j + k is 

Concept:

If a triangle is formed by three vectors, then the sum of the vectors should be zero.

AB + BC + CA = 0 

Cross product of the vectors:

For two vectors \(\bar a = ai+bj+ck\) and \(\bar b = di+ej+fk\) the cross, the product is given by: \(\bar a\times \bar b = \begin{vmatrix} i & j & k\\ a & b & c\\ d & e & f \end{vmatrix}=pi+qj+rk\)

The magnitude of the cross product is:

\(A = |a\times b| = \sqrt{p^2+q^2+r^2}\)

Area of a triangle:

If the vectors \(\bar a\mbox{ and } \bar b\) form adjacent sides of the triangle then the area of the triangle is given by: \(A = \dfrac{1}{2}|\bar a\times \bar b|\)

Calculation:

Given:

Let, AB = 3i + 4j and CA = 5i +7j + k

If a triangle is formed by three vectors, then the sum of the vectors should be zero.

AB + BC + CA = 0 ⇒ 3i + 4j  + BC + 5i +7j + k = 0

BC = - 8i - 11j - k

Let the adjacent vectors be AB (a), AC (b)  \(\bar a = 3i+4j\) , and \(\bar b = 5i+7j+k\).

First, we will calculate the cross product as follow:

\(\begin{align*} \bar a\times \bar b &= \begin{vmatrix} i & j & k\\ 3 & 4 & 0\\ 5 & 7 & 1 \end{vmatrix}\\ &= i(4-0) - j(3-0) + k(21-20)\\ &= 4i-3j+1k \end{align*}\)

Therefore, the magnitude of the cross product is:

\(\begin{align*} |\bar a\times \bar b| &= \sqrt{16+9+1}\\ &= \sqrt {26} \end{align*}\)

Using the formula for the area of the triangle, the area is given by:

\(A = {1\over 2}|a\times b| = {1\over 2}\times \sqrt {26} = {\sqrt {26}\over 2}\)