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.

Use the vector product to compute the area of the triangle with the given vertices $P(-2,-3), \quad Q(3,2)\quad$ and $\quad R(-1,-8)$

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}

Use the vector product to compute the area of the triangle with the given vertices $P(-2,-1,3), Q(1,2,-1)$ and $R(4.3,-3)$

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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