Question 4 & 5 Review Exercise 3
Solutions of Question 4 & 5 of Review Exercise 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 4
If $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$, then find $(\vec{r} \times \hat{i}) \cdot(\bar{r} \times \hat{j})+x y$
Solution
We have to find
$$(\vec{r} \times \hat{i}) \cdot(\vec{r} \times \hat{j})+x y $$
\begin{align}\text { Now } \vec{r} \times \hat{i}&=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
x & y & z \\
1 & 0 & 0
\end{array}\right| \\
& =(0-0) \hat{i}-(0-z) \hat{j}+(0-y) \hat{k} \\
\Rightarrow \vec{r} \times \hat{i}&=z \hat{j}-y \hat{k} \ldots \ldots \ldots \ldots . .(1) \\
\text { and } \vec{r} \times \hat{j}&=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
x & y & z \\
0 & 1 & 0
\end{array}\right| \\
& =(0-z) \hat{i}-(0-0) \hat{j}+(x-0) \hat{k}\end{align}
$$\Rightarrow \vec{r} \times \hat{j}=-z \hat{i}+x \hat{k}$$
Taking dot product of (1) and (2)
\begin{align}(\vec{r} \times \hat{i}) \cdot(\vec{r} \times \hat{j})&=(z \hat{j}-y \hat{k}) \cdot(-z \hat{i}+x \hat{k}) \\ \Rightarrow(\vec{r} \times \hat{i}) \cdot(\vec{r} \times \hat{j})&=0+0-x y \\ \Rightarrow(\vec{r} \times \hat{i}) \cdot(\vec{r} \times \vec{j})&=-x y \\ \text { Now }(\vec{r} \times \hat{i}) \cdot(\vec{r} \times \hat{j})+x y & =-x y+x y=0 .\end{align}
Question 5
If $\vec{a}=7 \hat{i}-\hat{j}-4 \hat{k}$ and $\vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}$, then find the projection of $\vec{a}$ on $\vec{b}$.
Solution
We have to compute
Projection of $\vec{a}$ on $\vec{b}=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
\begin{align}\vec{u} \cdot \vec{b}&=(7 \hat{i}+\hat{j}-4 \hat{k}) \cdot(2 \hat{i}+6 \hat{j}+3 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{b}&=14+6-12=8 \text { and } \\ |\vec{b} |&=\sqrt{(2)^2+(6)^2+(3)^2} \\ \Rightarrow|\vec{b}|&=\sqrt{49}=7 .\end{align} Hence projection of $\vec{a}$ on $\vec{b}=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
Projection of $\vec{a} \text{on} \vec{b}=\dfrac{8}{7}$
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