Question 6 Exercise 3.3

Solutions of Question 6 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.

Let $\vec{a}=\hat{i}+3 \hat{j}-4 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$. Find the value of $m$ so that $\vec{a}+m \vec{b}$ is orthogonal to $\vec{a}$

We know that \begin{align} \vec{a}+m \vec{b}& =\hat{i}+3 \hat{j}-4 \hat{k}+m(2 \hat{i}-3 \hat{j}+5 \hat{k}) \\ & =(1+2 m) \hat{i}+(3-3 m) \hat{j}+(5 m-4) \hat{k}\end{align}. If $\vec{a}+m \vec{b}$ is orthogonal to $\vec{a}$, then \begin{align}(\vec{a}+m \vec{b}) \cdot \vec{a}&=0 \\ \Rightarrow \quad[(1+2 m) \hat{j}+(3-3 m) \hat{j}+(5 m- 4) \hat{k}] \cdot [\hat{i}+3 \hat{j}-4 \hat{k}]&=0 \\ \Rightarrow \quad 1(1+2 m)+3(3-3 m)-4(5 m-4)&=0 \\ \Rightarrow \quad 2 m-9 m-20 m+1+9+16&=0 \\ \Rightarrow \quad-27 m&=-26 \\ \Rightarrow \quad m&=\dfrac{26}{27}\end{align}

Let $\vec{a}=\hat{i}+3 \hat{j}-4 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$. Find the value of $m$ so that $\vec{a}+m \vec{b}$ is orthogonal to $\vec{b}$.

If $\vec{a}+n \vec{b}$ is orthogonal to $\vec{b}$, then \begin{align}(\vec{a}+m \vec{b}) \cdot \vec{b}&=0 \\ \Rightarrow[(i 1+2 m) \hat{i}+(3-3 m) \hat{j}+(5 m-4) \hat{k}] \cdot [2 \hat{i}-3 \hat{j}+5 \hat{k}]&=0 \\ \Rightarrow \quad 2(1-2 m)-3(3-3 m) +5(5 m-4)&=0 \\ \Rightarrow \quad 4 m+9 m+25 m+2-9-20&=0 \\ \Rightarrow \quad 38 m-27&=0 \\ \Rightarrow \quad 38 m &=27 \\ \Rightarrow m&=\dfrac{27}{38} \end{align}

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