# Question 1, Exercise 3.3

Solutions of Question 1 of Exercise 3.3 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.

If $\vec{a}=3 \hat{i}+4 \hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}-5 \hat{k}$ then find $\vec{a}\cdot \vec{b}$

\begin{align}\vec{a} \cdot \vec{b}&=(3 \hat{i}+4 \hat{j}-\hat{k}) \cdot(\hat{i}-\hat{j}+3 \hat{k})\\ \Rightarrow &=(3 \times 1)+(4 \times-1)+(-1 \times 3)\\ & =3-4-3=-4\end{align}.

If $\vec{a}=3 \hat{i}+4 \hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}-5 \hat{k}$ then find $\vec{a} \cdot \vec{c}$.

\begin{align}\vec{a} \cdot \vec{c}&=(3 \hat{i}+4 \hat{j}-\hat{k}) \cdot(2 \hat{i}+\hat{j}-5 \hat{k})\\ \Rightarrow \vec{a} \cdot \vec{c}&=(3 \times 2)+(4 \times 1)+(-1 \times-5)\\ & =6+4+5=15\end{align}.

If $\vec{a}=3 \hat{i}+4 \hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}-5 \hat{k}$ then find $\vec{a} \cdot(\vec{b}+\vec{c})$

\begin{align}\vec{b}+\vec{c}&=(\hat{i}-\hat{j}+3 \hat{k})+ (2 i+\hat{j}-5 \hat{k}) \\ \Rightarrow \vec{b}+\vec{c}&=3 \hat{i}-2 \hat{k}\end{align} Taking dot product with $\vec{a}$ \begin{align}\vec{a} \cdot(\vec{b}+\vec{c})&=(3 \hat{i}+4 \hat{j}-\hat{k}) \cdot(3 \hat{i}-2 \hat{k}) \\ \Rightarrow \vec{a} \cdot(\vec{b}+\vec{c})&=3.3+4.0+-1 .-2 \\ \Rightarrow \vec{a} \cdot(\vec{b}+\vec{c})&=11\end{align}.

If $\vec{a}=3 \hat{i}+4 \hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}-5 \hat{k}$ then find $(\vec{a}-\vec{b})\cdot\vec{c}$

\begin{align}\vec{a}-\vec{b}&=3 \hat{i}+4 \hat{j}-\hat{k}-(\hat{i}-\hat{j}+3 \hat{k})\\ \Rightarrow \vec{a}-\vec{b}&=2 \hat{i}+5 \hat{j}-4 \hat{k}\end{align} taking dot product with $\vec{c}$ \begin{align}(\vec{a}-\vec{b}) \cdot \vec{c}& =(2 \hat{i}+5 \hat{j}-4 \hat{k}) \cdot(2 \hat{i}+\hat{j}-5 \hat{k}) \\ \Rightarrow &=2 \cdot 2+5 \cdot 1+-4 \cdot-5 \\ \Rightarrow(\vec{a}-\vec{b}) \cdot \vec{c}&=-11\end{align}.