====== Question 9 & 10, Exercise 3.3 ======
Solutions of Question 9 & 10 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.

=====Question 9=====
A force $\vec{k}-2 \hat{i}+3 \hat{j}+\hat{k}$ acts through the displacement $\vec{S}=2 \hat{i}+\hat{j}-\hat{k}$. Find work done.
====Solution====
We know that
\begin{align}W &=\vec{F} \cdot s \\
\Rightarrow W &=(2 \hat{i}+3 \hat{j}+\hat{k}) \cdot(2 \hat{i}+\hat{j}-\hat{k}) \\
\Rightarrow W &=2(2) \div 3(1)+1(-1) \\
\Rightarrow W &=4+3 \quad 1=6 \text { units.}\end{align}


=====Question 10=====
Find the work done by the force $\vec{F}=2 \hat{i}+3 \hat{j}+\hat{k}$ in displacement of an object from a point $A(-2,1,2)$ to the poim $B(5.0,3)$.
====Solution====
Position vector of $A$ is: $\overrightarrow{O A}=2 \hat{i}+\hat{j}+2 \hat{k}$.

The position vector of $B$ is:
$$\overrightarrow{O B}=5 \hat{i}+0 \hat{j}+3 \hat{k} \text {. }$$
Displacement vector from $A$ to $B$
\begin{align}A \vec{B}&=O \vec{B}-O \hat{A} \\
&=5 \hat{i}+0 \hat{j}+3 \hat{k}-(-2 \hat{i}+\hat{j}+2 \hat{k}) \\
\Rightarrow \overrightarrow{A B}&=7 \hat{i}-\hat{j}+\hat{k} .\end{align}
Now work done is $W=\vec{F} \cdot \overrightarrow{A B}$
$$\Rightarrow W=(2 \hat{i}+3 \hat{j}+\hat{k}) \cdot(7 \hat{i}-\hat{j}+\hat{k})$$
\begin{align}\rightarrow W&=2.7+3 .-1+1.1 \\
\therefore W&=14-3+1=12 \text { units. }\end{align}





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