# Question 7 & 8 Exercise 3.4

Solutions of Question 7 & 8 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.

If $\vec{A}+\vec{B}+\vec{C}=\vec{O}$, show that $$\vec{A} \times \vec{B}=\vec{B} \times \vec{C}=\vec{C} \times \vec{A}.$$

We are given
$$\vec{A}+\vec{B}+\vec{C}=\vec{O} \text {. }$$
Taking cross product of $\vec{A}$, of both sides with above. we get
$$\vec{A} \times(\vec{A}+\vec{B}+\vec{C})=0$$
\begin{align}\Rightarrow \vec{A} \times \vec{A}+\vec{A} \times \vec{B}+\vec{A} \times \vec{C}&=\vec{O}...(1) \\ \Rightarrow \vec{A} \times \vec{B}+\vec{A} \times \vec{C} &= \vec{O} \quad \because \vec{A} \| \vec{A} \\ \Rightarrow \vec{A} \times \vec{B}&=-\vec{A} \times \vec{C} \\ \Rightarrow \vec{A} \times \vec{B}&=\vec{C} \times \vec{A}...(2)\end{align} $\because \quad$ cross product is anti-commutative
$$\Rightarrow \vec{A} \times \vec{B}=\vec{C} \times \vec{A}$$
Taking cross product of $\vec{B}$ with (1), we get
\begin{align}\vec{B} \times(\vec{A}+\vec{B}+\vec{C})&=0 \\ \Rightarrow \vec{B} \times \vec{A}+\vec{B} \times \vec{B}+\vec{B} \times \vec{C}&=\vec{O} \\ \Rightarrow \vec{B} \times \vec{A}+\vec{B} \times \vec{C}&=\vec{O} \quad \because \vec{B} \| \vec{B} \\ \Rightarrow \vec{B} \times \vec{C}&=-\vec{B} \times \vec{A} \\ \Rightarrow \vec{B} \times \vec{C}&=\vec{A} \times \vec{B}....(3)\end{align} $\because$ crass product is anti-commuative
$$\vec{B} \times \vec{C}=\vec{A} \times \vec{B}$$
From (2) and (3), we get
$$\vec{A} \times \vec{B}=\vec{B} \times \vec{C}=\vec{C} \times \vec{A} \text {. }$$

Find a unit vector perpendicular to both $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=-2 \hat{i}+\hat{j}-3 \hat{k}$

Let $\hat{n}$ be unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ then
\begin{align}\hat{n}&=\dfrac{\vec{a} \times \vec{b}}{\mid \vec{a} \times \vec{b}} \ldots \ldots \ldots(1) \\ \vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & 1 & -3 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{b}&=(-3-2) \hat{i}-(-3-4) \hat{j}+(1-2) \hat{k} \\ \Rightarrow \vec{a} \times \vec{b}&=-5 \hat{i}+7 \hat{j}-\hat{k}\end{align} \begin{align}\vec{a} \times \vec{b} \mid&=\sqrt{(-5)^2+(7)^2+(-1)^2} \\ |\vec{a} \times \vec{b}|&=\sqrt{75} .\end{align} Putting in (1), we have
\begin{align}\hat{n}&=\dfrac{\vec{a} \times \vec{b}}{\vec{a} \times \vec{b}}\\ &=\dfrac{-5 \hat{i}+7 \hat{j}-\hat{k}}{\sqrt{75}}.\end{align} Which is the required unit vector perpendicular to both $\vec{a}$ and $\vec{b}$.

Find a unit vector perpendicular to both Find a vector of magnitude 10 and perpendicular to both $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k} . \quad \vec{b}=4 \hat{i}-2 \hat{j}-4 \hat{k} \text {. }$$

Let $\hat{n}$ be unil vector perpendicular to both $\vec{a}$ and $\vec{b}$ then
\begin{align} \hat{n}=\dfrac{\vec{a} \times \vec{b}}{\vec{a} \times \vec{b}} \\ \vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 4 \\ 4 & -2 & -4 \end{array}\right| \\ \vec{a} \times \vec{b}&=(12+8) \hat{i}-(-8-16) \hat{j}+ (-4+12) \hat{k} \\ \vec{a} \times \vec{b}&=2 \hat{i}-24 \hat{j}+8 \hat{k} \\ |\vec{a} \times \vec{b} |&=\sqrt{(20)^2+(24)^2+(8)^2} \\ | \vec{a} \times \vec{b}|=\sqrt{1040} \\ \Rightarrow|\vec{a} \times \vec{b}|&=2 \sqrt{65} .\end{align} Putting in (1), we have
\begin{align}\hat{n}&=\dfrac{\vec{a} \times \vec{b}}{\vec{a} \times \vec{b}}\\ &=\dfrac{20 \hat{i}+24 \hat{j}+8 \hat{k}}{2 \sqrt{65}} . \\ \Rightarrow \hat{n}&=\dfrac{10 \hat{i}+12 \hat{j}+4 \hat{k}}{\sqrt{65}} .\end{align} Now let $\vec{c}$ be a vector perpendicular to both and having magnitude
$\vec{c} =10$ then
$\vec{c}= \vec{c} \cdot \hat{n}=10\left(\dfrac{10 \hat{i}+12 \hat{j}+4 \hat{k}}{\sqrt{65}}\right)$, is the required vector.