Question 9 Exercise 3.4

Solutions of Question 9 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.

Find the area of parallelogram whose diagonals are $\vec{a}=4 \hat{i}+\hat{j}-2 \hat{k}\quad$ and $\quad\vec{b}=-2 \hat{i}+3 \hat{j}+4 \hat{k}$.

We are give the diagonal as shown in figure, instead of two adjacent sides $\vec{c}$ and $\vec{d}$ which are used to find the area of parallelogram.
$E$ is the intersection of two diagonals, so $E$ is the midpoint of both diagonals. Thus
$$\begin{align}\overrightarrow{A E}&=\overrightarrow{E C}\\ &=\dfrac{1}{2} \vec{a}\\ &=2 \hat{i}+\dfrac{1}{2} \hat{j}-\hat{k} \\ \overrightarrow{E D}&=\overrightarrow{B E}\\ &=\dfrac{1}{2} \vec{b}\\ &=-\hat{i}+\dfrac{3}{2} \hat{j}+2 \hat{k}\end{align}$$ From $\triangle A E B$, we have
$$\begin{align}\vec{c}&=\overrightarrow{A E}+\overrightarrow{E B} \\ \Rightarrow \vec{c}&=2 \hat{i}+\dfrac{1}{2} \hat{j}-\hat{k}-(-\hat{i}+\dfrac{3}{2} \hat{j}+2 \hat{k}) \because \overrightarrow{B E}=-\overrightarrow{E B} \\ \Rightarrow \vec{c}&=3 \hat{i}-\hat{j}-3 \hat{k} \ldots \ldots \ldots \ldots(1)\end{align}$$ From $\triangle A E D$. we have
$$\begin{align}\vec{d}&=\overrightarrow{A E}+\overrightarrow{E D}\\ \Rightarrow \bar{d}&=2 \hat{i}+\dfrac{1}{2} \hat{j}-\hat{k}+(-\hat{i}+\dfrac{3}{2} \hat{j}+2 \hat{k}) \\ \Rightarrow \vec{d}&=\hat{i}+2 \hat{j}+ \hat{k} \text {. }....(2) \end{align}$$ By using (1) and (2), we have,
$$\begin{align}\vec{c} \times \vec{d}&=\left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & -3 \\ 1 & 2 & 1 \end{array}\right| \\ & =(-1+6) \hat{i}-(3+3) \hat{j}+(6+1) \hat{k} \\ \Rightarrow \vec{c} \times \vec{d}&=5 \hat{i}-6 \hat{j}+7 \hat{k}\end{align}$$ Now
$$\text{area of parallelogram} =|\vec{c} \times \vec{d}|=\sqrt{5^2+(-12)^2+9^2}=\sqrt{110} \text { units. }$$

Find the area of parallelogram whose diagonals are $\vec{a}=3 \hat{i}+2 \hat{j} \quad 2 \hat{k}\quad$ and $\quad\vec{h}=\hat{i}-3 \hat{j}+4 \hat{k}$

We are given the diagonals as shown in figure instead of Iwo :djacent sides $\vec{c}$ and $\vec{d}$ which are used to find the area of parallelogram.
$E$ is the intersection of two diagonals, so $E$ is the midpoint of both diagonals.
Therefore,
$$\begin{align}\overrightarrow{A E}&=\overrightarrow{E C}\\ &=\dfrac{1}{2} \vec{a}\\ &=\dfrac{3}{2} \hat{i}+\hat{j}-\hat{k}\\ \overrightarrow{E D}&=\overrightarrow{B E}\\ &=\dfrac{1}{2} \vec{b}\\ &=\dfrac{1}{2} \hat{i}-\dfrac{3}{2} \hat{j}+2 \hat{k}\end{align}$$ Which from $\triangle A E B$, we have
$$\begin{align}\vec{c} & =\overrightarrow{A E}+\vec{E} \vec{B} \\ & =\dfrac{3}{2} \hat{i}+\hat{j}-\hat{k}-( \dfrac{1}{2} \hat{i}-\dfrac{3}{2} j+2 \hat{k}),\quad \because \overrightarrow{B E}=-\vec{E} \vec{B} \\ \quad \vec{c}&=\hat{i}+\dfrac{5}{2} \hat{j} - 3 \hat{k}....(1)\end{align}$$ and From .AED. we have
$$\begin{align} \vec{d}&=A \vec{E} \cdot F \overrightarrow{ } \\ & =\dfrac{3}{2} \hat{i}+\hat{j}-k+\left(\dfrac{1}{2} \hat{i}-\dfrac{3}{2} \hat{j}+2 \hat{k}\right) \\ \Rightarrow \vec{d}&=2 \hat{i}-\dfrac{1}{2} \hat{j}+\hat{k}....(2)\end{align}$$ By using (1) and (2), we have,
$$\begin{align}\vec{c} \times \vec{d}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & \dfrac{5}{2} & -3 \\ 2 & -\dfrac{1}{2} & 1 \end{array}\right|\\ \Rightarrow \vec{c} \times \vec{d}&=\left(\dfrac{5}{2}-\dfrac{3}{2}\right) \hat{i}-(1+6) \hat{j}+\left(-\dfrac{1}{2}-\dfrac{10}{2}\right) \hat{k} \\ \Rightarrow \vec{c} \times \vec{d}&=\hat{i}-7 \hat{j}-\dfrac{1}{2} \hat{k}\end{align}$$ $$\begin{align}\text{Area of parallelogram}& =|\bar{c} \times \vec{d}| \\ & =\sqrt{(1)^2+(-7)^2+\left(\dfrac{-11}{2}\right)^2} \\ & =\sqrt{1+41+\dfrac{121}{4}} \\ & =\sqrt{\dfrac{4+196+121}{4}} \\ & =\dfrac{\sqrt{321}}{2} \end{align}$$ Thus area of parallelogram is $\dfrac{\sqrt{321}}{2}$ unit square.

< Question 7 & 8