====== 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.

=====Question 5(iii)=====
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 $(\vec{a}. \vec{b})^2,\quad|a|^2,\quad|b|^2$
====Solution====
\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}

=====Question 5(iv)=====
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 $|\vec{a} \times\vec{b}|^2=(\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b})-(\vec{a} \cdot \vec{b})^2$.
====Solution====
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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