# Question 2 Exercise 3.4

Solutions of Question 2 of Exercise 3.4 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.

Show in two different ways that the vectors $\vec{a}$ and $\vec{b}$ are parallel to $\vec{a}=-\hat{i}+2 \hat{j}-3 \hat{k}, \quad \vec{b}=2 \hat{i}-4 \hat{j}+$ $6 \hat{k}$

First Way \begin{align}\vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & -3 \\ 2 & -4 & 6 \end{array}\right| \\ & =(12-12) \hat{i}-(-6+6) \hat{j}+(4-4) \hat{k} \\ \Rightarrow \vec{a} \times \vec{b}&=0 . \\ & \Rightarrow \vec{a} \| \vec{b} .\end{align}

Second Way \begin{align}\vec{a} \cdot \vec{b}&=(-\hat{i}+2 \hat{j}-3 \hat{k}) \cdot(2 \hat{i}-4 \hat{j}+6 \hat{k}) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-1(2)+2(-4)-3(6) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-28 .\end{align} Also \begin{align}|\vec{a}|&=\sqrt{(-1)^2+(2)^2+(-3)^2}\\ \Rightarrow|\vec{a}|&=\sqrt{14}\\ |\vec{b}|&=\sqrt{(2)^2+(-4)^2+(6)^2} \\ \Rightarrow|\vec{b}|&=\sqrt{56}\\ \cos \theta&=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}|}=\dfrac{-28}{\sqrt{14} \sqrt{56}}\\ \Rightarrow \quad \theta&=\cos ^{-1}\left(\frac{-28}{2 \sqrt{14} \sqrt{14}}\right) \\ \Rightarrow \quad \theta&=\cos ^{-1}(-1)=180^{\circ} . \\ \Rightarrow \quad \vec{a} \| \vec{b} .\end{align}

Show in two different ways that the vectors $\vec{a}$ and $\vec{b}$ are parallel to $\vec{a}=3 \hat{i}+6 \hat{j}-9 \hat{k}$, $\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}$

First Way \begin{align}\vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 6 & -9 \\ 1 & 2 & -3 \end{array}\right| \\ & =(-18+18) \hat{i}+(-9+9) \hat{j}+(6-6) \hat{k} \\ \Rightarrow \vec{a} \times \vec{b}&=0 . \\ & \Rightarrow \vec{a} \| \vec{b} . \end{align}

Second Way \begin{align}\vec{a} \cdot \vec{b}&=(3 \hat{i}+6 \hat{j}-9 \hat{k}) \cdot(\hat{i}+2 \hat{j}-3 \hat{k}) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=3(1)+6(2)-9(-3) \\ \Rightarrow \vec{a} \cdot \vec{b}&=42\end{align} Also \begin{align}|\vec{a}|&=\sqrt{(3)^2+(6)^2+(-9)^2}\\ \Rightarrow|\vec{a}|&=\sqrt{12 \overline{6}}\text{ and}\\ |\vec{b}|&=\sqrt{(1)^2+(2)^2+(-3)^2} \\ \Rightarrow|\vec{b}|&=\sqrt{14} .\end{align} Now we know that
\begin{align}\cos \theta&=\dfrac{\vec{a} \cdot \vec{b}}{i|\vec{b}|}=\dfrac{42}{\sqrt{14} \sqrt{126}} \\ \Rightarrow \theta&=\cos^{-1}\left(\sqrt{\dfrac{42 \times 42}{14 \times 126}}\right) \\ \Rightarrow \theta&=\cos^{-1}\left(\sqrt{\dfrac{1764}{1764}}\right) \\ \Rightarrow \theta&=\cos^{-1}(1)=0^{\circ} \\ \Rightarrow \quad \vec{a} \| \vec{b}.\end{align}