====== Question 5 Exercise 3.4 ======
Solutions of Question 5 of Exercise 3.4 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.

=====Question 5(i)=====
Use the vector product to compute the area of the triangle with the given vertices $P(-2,-3), \quad Q(3,2)\quad$ and $\quad R(-1,-8)$
====Solution====
Let $P Q$ and $\bar{P} R$ be the adjacent sides of parallelogram determined, so the required area of the triangle if hall the area of the parallelogram, that is:\\
\begin{align}\text{Area of triangle}&=\dfrac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}| \\
\text { Since } \overrightarrow{P Q}&=(3,2)-(-2 ,-3) \\
\Rightarrow \overrightarrow{P Q}&=(5,5) \\
\overrightarrow{P R}&=(-1,-8)-(-2 ,-3) \\
\Rightarrow \overrightarrow{P R}&=(1,-5) \\
\overrightarrow{PQ}\times \overrightarrow{P R}&=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
5 & 5 & 0\\
1 & -5 & 0
\end{array}\right|\\
\Rightarrow \overrightarrow{P Q} \times \overrightarrow{P R}&=(-25-5) \hat{k}=-30 \hat{k} \\
\Rightarrow|\overrightarrow{P Q} \times \overrightarrow{P R}|&=30 \\
\therefore \text { Area of triangle }& =\dfrac{1}{2}|\overrightarrow{PQ} \times \overrightarrow{P R}|\\
&=\dfrac{30}{2}=15 \text { units square. }\end{align}

=====Queswtion 5(ii)=====
Use the vector product to compute the area of the triangle with the given vertices $P(-2,-1,3), Q(1,2,-1)$ and $R(4.3,-3)$
====Solution====
Let $\overrightarrow{P Q}$ and $\overrightarrow{P R}$ be the adjacent sides of parallelogram determined, so the required area of the triangle if half the area of the parallelogram, that is:\\
$$\text { Area of triangle }=\dfrac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|$$\\
Since \begin{align}\overrightarrow{P Q}&=(1,2 .-1)-(-2,-1,3)\\
\Rightarrow \overrightarrow{P Q}&=(3,3,-4)\\
\overrightarrow{P R}&=(4,3,-3)-(1.2,-1)\\
\Rightarrow \overrightarrow{P R}&=(3,1,-2).\end{align}
\begin{align}\text { so } \overrightarrow{P Q} \times \overrightarrow{P R}&=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 3 & -4 \\
3 & 1 & -2
\end{array}\right| \\
\Rightarrow \overrightarrow{P Q} \times \overrightarrow{P R}& =(-6+4) \hat{i}-(-6-12) \hat{j}+(3-9) \hat{k} \\
\Rightarrow \overrightarrow{P Q} \times \overrightarrow{P R}&=-2 \hat{i}-6 \hat{j}-6 \hat{k} \\
\Rightarrow|\overrightarrow{P Q} \times \overrightarrow{P R}|& =\sqrt{(-2)^2+(-6)^2+(-6)^2} \\
\Rightarrow|\overrightarrow{P Q} \times \overrightarrow{P R}|&=\sqrt{76} \\
\therefore \text { Area of triangle }&=\dfrac{1}{2} | \vec{P} \hat{Q} \times \overrightarrow{P R}| \\
\text { Area of triangle }&=\dfrac{\sqrt{76}}{2}=\sqrt{19} \text { units square. }\end{align}


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