Question 3 Exercise 3.4
Solutions of Question 3 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.
Question 3(i)
Find a unit vector that is orthogonal to the given vector $\vec{a}=\hat{i}- 2 \hat{j}+3 \hat{k}, \quad \vec{b}=2 \hat{i}+\hat{j}-\hat{k}$.
Solution
Let $\hat{n}$ be unit vector orthogonal to both $\vec{a}$ and $\vec{b}$. then by cross product
\begin{align}\hat{n}&=\dfrac{\vec{a} \times \vec{b}}{\mid \vec{a} \times \vec{b}} \\
\text { Now } \vec{a} \times \vec{b}&=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 3 \\
2 & 1 & -1
\end{array}\right| \\
\Rightarrow \vec{a} \times \vec{b}&=(2+3) \hat{i}-(-1-6) \hat{j}+(1-4) \hat{h} \\
\rightarrow \vec{a} \times \vec{b}&=-\hat{i}+7 \hat{j}+5 \hat{k} \\
\Rightarrow | \vec{a} \times \vec{b}|&=\sqrt{(-1)^2+(7)^2+(5)^2} \\
\Rightarrow \quad|\vec{a} \times \vec{b}|&=\sqrt{75}=5\sqrt{3}\end{align}
Putting (2) and (3) in (1), we get
$$\hat{n}=\dfrac{\vec{a} \times \vec{b}}{\vec{a} \times \vec{b}}=\dfrac{1}{5\sqrt{3}}(\hat{i}-7 \hat{j}+ 5 \hat{k}) .$$
Question 3(ii)
Find a unit vector that is orthogonal to the given vector $\vec{a}=3 \hat{i}-\hat{j}+6 \hat{k}, \quad \vec{b}=\hat{i}+4 \hat{j}+\hat{k}$
Solution
Let $\hat{n}$ be unit vector orthogonal to buth $\vec{a}$ and $\vec{b}$. then by cross product we have
\begin{align}\hat{n}&=\dfrac{\vec{a} \times \bar{b}}{\bar{a} \times \bar{b}} \ldots \ldots \ldots \ldots(1) \\
\text { Now } \vec{a} \times \vec{b}&=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -1 & 6 \\
1 & 4 & 1
\end{array}\right| \\
\Rightarrow \vec{a} \times \vec{b}&=(-1-24) \hat{i}-(3-6) \hat{j}+ (12+1) \hat{k} \\
\Rightarrow \vec{a} \times \vec{b}&=-25 \hat{i}+3 \hat{j}+13 \hat{k} \ldots \ldots \ldots \\
\Rightarrow|\vec{a} \times \vec{b}|&=\sqrt{\left(-\left.25\right|^2+(3)^2+(13)^2\right.} \\
\Rightarrow|\vec{a} \times \vec{b}|&=\sqrt{803} \ldots \ldots \ldots(3)\end{align}
Pulting (2) and (3) in (1), we get
\begin{align}\hat{n}&=\dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \\
\Rightarrow \hat{n}&=\dfrac{1}{\sqrt{803}}(-25 \hat{i}+3 \hat{j}+13 \hat{k}) .\end{align}
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