Question 5(iii) & 5(iv) Exercise 3.5

Solutions of Question 5(iii) & 5(iv) of Exercise 3.5 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.

Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\quad$ and $\quad\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\quad$. Find $(\vec{a}. \vec{b})^2,\quad|a|^2,\quad|b|^2$

\begin{align}\vec{a} \cdot \vec{b}&=(a_1 \hat{i}+a_2 \hat{j} + a_3 \hat{k}) \cdot(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}) \\ \vec{a} \cdot \vec{b}&=a_1 b_1+ a_2 b_2+a_3 b_3 \end{align} Taking square of the both sides \begin{align}(\vec{a} \cdot \vec{b})^2&=(a_1 b_1 + a_2 b_2+a_3 b_3)^2 . \\ (\vec{a} \cdot \vec{b})^2&=a_1^2 b_1^2 + a_2^2 b_2^2+a_3^2 b_3^2+2(a_1a_2b_1b_2+a_1a_3b_1b_3+a_2a_3b_2b_3) . \\ |\vec{a}|&=\sqrt{(a_1)^2-(a_2)^2+(a_3)^2}\end{align} Taking square of the both sides

\begin{align}|\vec{a}|^2&=(a_1)^2+(a_2)^2+(a_3)^2 \\ \Rightarrow | \vec{a}|^2&=a_1^2+a_2^2+a_3^2\quad \text { and } \\ |\vec{b} |&=\sqrt{(b_1)^2+(b_2)^2-(b_3)^2} .\end{align} Taking square of the both sides

\begin{align}|\vec{b}|^2&=(b_1)^2+(b_2)^2+(b_3)^2 \\ |\vec{b}|^2&=b_1^2+b_2^2+b_3^2 .\end{align}

Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\quad$ and $\quad\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\quad$. Show that $|\vec{a} \times\vec{b}|^2=(\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b})-(\vec{a} \cdot \vec{b})^2$.

We have already calculated

L.H.S in (ii) that
\begin{align}|\vec{a} \times \vec{b}|^2&=(a_2 b_3-a_3 b_2)^2+(a_1 b_3 -a_3 b_1)^2+(a_1 b_2-a_2 b_1)^2 \\ & =a_1^2 b_2^2+a_2^2 b_1^2-2 a_1 a_2 b_1 b_2+a_2^2 b_3^2+a_3^2 b_2^2-2 a_2 a_3 b_2 b_3+a_1^2 b_3^2+a_3^2 b_1^2 -2 a_1 a_3 b_1 b_3 \end{align}
Adding and subtracting $a_1^2 b_1^2, a_2^2 b_2^2$ and $ a_3^2 b_3^2$ in the above we get
\begin{align}| \vec{a} \times \vec{b}|^2&=a_1^2 b_1^2+a_1^2 b_2^2+a_1^2 b_3^2+a_2^2 b_1^2+a_2^2 b_2^2+\\ &a_2^2 b_3^2+ a_3^2 b_1^2+a_3^2 b_2^2+a_3^2 b_3^2- a_1^2 b_1^2-a_2^2 b_2^2-a_3^2 b_2^2-\\ & 2 a_1 a_2 b_1 b_2 -2 a_2^2 a_3^3 b_2 b_3-2 a_1 a_3 b_1 b_3^3 \\ |\vec{a} \times \vec{b}|^2&=(a_1^2 b_1^2+a_1^2 b_2^2+a_1^2 b_3^2+a_2^2 b_1^2+a_2^2 b_2^2+\\ & a_2^2 b_3^2+a_3^2 b_1^2+a_3^2 b_2^2+ a_3^2 b_3^2)-(a_1^2 b_1^2+a_2^2 b_2^2+a_3^2 b_3^2+ \\ &2 a_1 a_2 b_1 b_2+2 a_2 a_3 b_2 b_3+2 a_1 a_3 b_1 b_3)....(1)\end{align}
Now we know that
\begin{align}\vec{a} . \vec{a}&=|\vec{a}|^2\quad \text{and}\\ \vec{b} \cdot \vec{b}&=|\vec{b}|^2 \\ (\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b}) &=(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2) \\ (\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b})& =a_1^2 b_1^2+a_1^2 b_2^2+a_1^2 b_3^2+a_2^2 b_1^2+\\ &a_2^2 b_2^2+ a_2^2 b_3^2+a_3^2 b_1^2+a_3^2 b_2^2+a_3^2 b_3^2 \cdots \cdots \cdots \cdot . .(2) \\ (\vec{a} \cdot \vec{b})^2&=(a_1 b_1+a_2 b_2+a_3 b_3)^2 \\ \Rightarrow(\vec{a} \cdot \vec{b})^2&=a_1^2 b_1^2+a_2^2 b_2^2+a_3^2 b_3^2+2 a_1 a_2 b_1 b_2+2 a_2 a_3 b_2 b_3+2 a_1 a_3 b_1 b_3\end{align}
Putting (3) and (2) in (1), we get the desired that is: $$| \vec{a} \times \vec{b}|^2=(\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b})-(\vec{a} \cdot \vec{b})^2 . $$

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