Question

Find the area of triangle having vertices at p(1,3,2), q(2,-1,1), r(-1,2,3) is

Answer 1

Given: the vertices $P(1,3,2)$, $Q(2,-1,1)$ and $R(-1,2,3)$

To find: area of $\Delta PQR$

Solution:

• We find the area of the triangle using vector method.

• Position vector of $P(\vec{p})=(1,3,2)$

• Position vector of $Q(\vec{q})=(2,-1,1)$

• Position vector of $R(\vec{r})=(-1,2,3)$

• The side $\vec{PQ}=(1,-4,-1)$

• The side $\vec{PR}=(-2,-1,1)$

• Now, $\vec{PQ}\times \vec{PR}$

• $=\left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\ 1&-4&-1\\-2&-1&1\end{array}\right|$

• $=-5\hat{i}+\hat{j}-9\hat{k}$ [expanding along the first row]

• Now, $\frac{1}{2}|\vec{PQ}\times\vec{PR}|$

• $=\frac{1}{2}\sqrt{(-5)^{2}+1^{2}+(-9)^{2}}$

• $=\frac{1}{2}\sqrt{107}$

Answer: area of $\Delta PQR$ is $\frac{1}{2}\sqrt{107}$ square units.