# Question 8 & 9 Review Exercise 3

Solutions of Question 8 & 9 of Review Exercise 3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find the area of triangle whose vertices are $(0,0,2),(-1,3,2),(1,0,4)$.

Lel $A(0,0,2)$, $B(-1,3,2)$ and $C(1,0,4)$.

Let $\vec{a}=\overrightarrow{A B}=(-1,3,2)-(0,0,2)$ $\Rightarrow \vec{a}=(-1,3,0)$

$\vec{b}=\overrightarrow{B C}=(1,0,4)-(-1,3,2)$ $\Rightarrow \vec{b}=(2,-3,2)$.

We know that area of triangle is half of the area of parallelogram.
$$\text{Area of triangle} =\dfrac{1}{2}|\vec{a} \times \vec{b}|....(1)$$. \begin{align}\vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 0 \\ 2 & -3 & 2 \end{array}\right| \\ \therefore \vec{a} \times \vec{b} &=(\hat{i}+2 \hat{j} \cdot 3 \hat{k} \\ \Rightarrow | \vec{a} \times \vec{b} |&=\sqrt{(6)^2+(2)^2+(-3)^2} \\ \Rightarrow | \vec{a} \times \vec{b}|&=\sqrt{49}= 7 .\end{align} Putting $|\vec{a} \times \vec{b}|=7$ in (1), we get
$$\text{Area of triangle} =\dfrac{7}{2}\text{ units square}$$.

Find the area of parallelogram with vertices $A(1,2, 3),\quad B(5,8,1),\quad$ $C(4,-2,2)\quad$ and $\quad D(0,-8,-2)$.

\begin{align}\text { Let } \vec{a} &=\overrightarrow{A B}=(5,8 ,1)-(1,2 ,-3) \\ \Rightarrow \vec{a} &=(4,6 ,4) \\ \vec{b}&=\overrightarrow{A C}=(4,2 ,2)- (1,2 ,3)\\ \Rightarrow \vec{b}&=(3,-4,5)\end{align} Now \begin{align}\vec{a} \times \vec{b}&=\left|\begin{array}{lll} i & j & k \\ 4 & 6 & 4 \\ 3 & 4 & 5\end{array}\right|\\ \Rightarrow \vec{a} \times \vec{b}& =(30-16) \hat{i}-(20-12) \hat{j}+(-16- 18) \hat{k} \\ \Rightarrow \vec{a} \times \vec{b}&=46 \hat{i}-8 \hat{j} - 34 \hat{k} . \\ \Rightarrow|\vec{a} \times \vec{b}| & =\sqrt{(46)^2+(-8)^2+(-34)^2} \\ & =\sqrt{3336} .\end{align} Thus the area of parallelogram is: $|\vec{a} \times \vec{b}|=\sqrt{3336}$ units square.