====== Question 3 & 4 Exercise 3.5 ======
Solutions of Question 3 & 4 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 3=====
For the vectors $\vec{a}=3 \hat{i}+2 \hat{k}$, $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}\quad$ and $\quad\vec{c}=-\hat{j}+4 \hat{k}$.
Verify that:\\
$\vec{a} \cdot \vec{b} \times \vec{c}=\vec{b} \cdot \vec{c} \times \vec{a}=\vec{c} \cdot \vec{a} \times \vec{b}$\\ but\\ $\vec{a} \cdot \vec{b} \times \vec{c}=-\vec{c} \times \vec{b} \cdot \vec{a}$\\
====Solution====
\begin{align}\vec{a} \cdot \vec{b} \times \vec{c}&=\left|\begin{array}{ccc}
3 & 0 & 2 \\ 1 & 2 & 1 \\ 0 & -1 & 4\end{array}\right|\\
\vec{a} \cdot \vec{b} \times \vec{c}&=3(8+1)+2(-1-0)\\
\Rightarrow \vec{a} \cdot \vec{b} \times \vec{c}&=25 \ldots \ldots \ldots . .(1) \\
\vec{b} \cdot \vec{c} \times \vec{a}&=\left|\begin{array}{ccc}
1 & 2 & 1 \\
0 & -1 & 4 \\
3 & 0 & 2
\end{array}\right|\\
\vec{b} \cdot \vec{c} \times \vec{a}&=1(-2-0)+3(8+1) \\
\Rightarrow \vec{b} \cdot \vec{c} \times \vec{a}&=25 \ldots \ldots \ldots . . .(2) \\
\vec{c} \cdot \vec{a} \times \vec{b}&=\left|\begin{array}{ccc}
0 & -1 & 4 \\
3 & 0 & 2 \\
1 & 2 & 1
\end{array}\right|\\
\vec{c}\cdot\vec{a}\times\vec{b}&=1(3-2)+4(6-0) \\
\Rightarrow \vec{c} \cdot \vec{a} \times \vec{b}&=1+24=25 \ldots \ldots (3) \end{align}
From (1), (2) and (3), we get that

\begin{align}\vec{a}\cdot \vec{b} \times \vec{c} =\vec{b} \cdot \vec{c} \times \vec{a}=\vec{c} \cdot \vec{a} \cdot \vec{b}  \\
\text { Now } \vec{c} \times \vec{b}&=\left|\begin{array}{ccc}
\hat{i}& \hat{j} & \hat{k} \\
0& -1 & 4 \\
1 & 2 & 1 
\end{array}\right|\\
\vec{c} \times \vec{b}&=(-1-8) \hat{i}-(0-4) \hat{j}+(0+1) \hat{k} \\
\vec{c} \times \vec{b}&=-9 \hat{i}+4 \hat{j}+\hat{k}\end{align}
Taking dut product with $\vec{a}$. we have

\begin{align}\vec{c} \times \vec{b} \cdot \vec{a}&=(-9 \hat{i}+4 \hat{j}+\hat{k}) \cdot(3 \hat{i}+2 \hat{k}) \\
\Rightarrow \vec{c} \times \vec{b} \cdot \vec{a}&=-9 \cdot 3+4 \cdot 0+1.2=-25\end{align}
Multiplying both sides by -1\\
$$-\vec{c} \times \vec{b} \cdot \vec{a}=25....(4)$$
From (1) and (4), we get that\\
$$\vec{a} \cdot \vec{b} \times \vec{c}=\vec{c} \times \vec{b} \cdot \vec{a} .$$

=====Question 4=====
Verify that the triple product of $\hat{i}-\hat{j}, \hat{j}-\hat{k}\quad$ and $\quad\hat{k}-\hat{i}$ is zero.
====Solution====
Let 
$$\vec{a}=\hat{i}-\hat{j}, \vec{b}=\hat{j}-\hat{k}$$
and\\ $$\vec{c}=\hat{k}-\hat{i}$$
Then
\begin{align}\vec{a} \cdot \vec{b} \times \vec{c}&=\left|\begin{array}{ccc}1 & -1 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 1\end{array}\right|\\
\vec{a} \cdot \vec{b} \times \vec{c}&=1(1-0)+1(0-1) \\
\Rightarrow \vec{a} \cdot \vec{b} \times \vec{c}&=1-1=0 .\end{align}
Hence the scalar triple product of the given vectors is zero.




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