Question

Calculate the area of a triangle whose vertices are given by (3, 1, 2), (1, 1, 2) and (4, 2, 1).

Answer 1

The area of a triangle whose vertices are given by (3, 1, 2), (1, 1, 2) and

(4, 2, 1) is $\sqrt{2}$  square units

• Given vertices are

          A(3, 1, 2) B(1, 1, 2) C(4, 2, 1)

• Area of triangle ABC is given by

                 $\frac{1}{2}|ABXAC|$

• AB = (1, 1, 2) - (3, 1, 2) = (-2, 0, 0)
• AC = (4, 2, 1) - (3, 1, 2) = (1, 1, -1)

           $ABXAC=\left|\begin{array}{ccc}i&j&k\\-2&0&0\\1&1&-1\end{array}\right|$

                          = i(0-0)-j(2-0)+k(-2-0)

                          = -2j-2k

• Ar(ΔABC) = $\frac{1}{2}|ABXAC|$

                         = $\frac{1}{2}\sqrt{(-2)^{2}+(-2)^{2}}$

                          = $\frac{1}{2}\sqrt{4+4}$

                          = $\frac{1}{2}2\sqrt{2}$

                          = $\sqrt{2}$ square units