# 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.

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
$\vec{a} \cdot \vec{b} \times \vec{c}=-\vec{c} \times \vec{b} \cdot \vec{a}$

\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
$$\vec{a} \cdot \vec{b} \times \vec{c}=\vec{c} \times \vec{b} \cdot \vec{a} .$$

Verify that the triple product of $\hat{i}-\hat{j}, \hat{j}-\hat{k}\quad$ and $\quad\hat{k}-\hat{i}$ is zero.

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.