Question 1, Exercise 3.2

Solutions of Question 1 of Exercise 3.2 of Unit 03: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

If $\vec{a}=3\hat{i}-5\hat{j}$ and $\vec{b}=-2\hat{i}+3\hat{j}$, then find $\vec{a}+2\vec{b}$.

\begin{align}\vec{a}+2\vec{b}&=3\hat{i}-5\hat{j}+2(-2\hat{i}+3\hat{j})\\ &=3\hat{i}-5\hat{j}-4\hat{i}+6\hat{j}\\ &=-\hat{i}+\hat{j}\end{align}

If $\vec{a}=3\hat{i}-5\hat{j}$ and $\vec{b}=-2\hat{i}+3\hat{j}$, then find $3\vec{a}-2\vec{b}$.

\begin{align}3\vec{a}-2\vec{b}&=3(3\hat{i}-5\hat{j})-2(-2\hat{i}+3\hat{j})\\ &=9\hat{i}-15\hat{j}+4\hat{i}-6\hat{j}\\ &=13\hat{i}-21\hat{j}\end{align}

If $\vec{a}=3\hat{i}-5\hat{j}$ and $\vec{b}=-2\hat{i}+3\hat{j}$, then find $2(\vec{a}-\vec{b})$.

First we have, \begin{align}\vec{a}-\vec{b}&=3\hat{i}-5\hat{j}-(-2\hat{i}+3\hat{j})\\ &=3\hat{i}-5\hat{j}+2\hat{i}-3\hat{j}\\ &=5\hat{i}-8\hat{j}\end{align} Multiply both sides by $2$. We have, $$2(\vec{a}-\vec{b})=10\hat{i}-16\hat{j}$$

If $\vec{a}=3\hat{i}-5\hat{j}$ and $\vec{b}=-2\hat{i}+3\hat{j}$, then find $|\vec{a}+\vec{b}|$.

We have, \begin{align}\vec{a}+\vec{b}&=3\hat{i}-5\hat{j}+(-2\hat{i}+3\hat{j})\\ &=3\hat{i}-5\hat{j}-2\hat{i}+3\hat{j}\\ &=\hat{i}-2\hat{j}\end{align} Taking modulus of both sides. We have, $$|\vec{a}+\vec{b}|=\sqrt{(1)^2+(-2)^2}=\sqrt{5}$$

If $\vec{a}=3\hat{i}-5\hat{j}$ and $\vec{b}=-2\hat{i}+3\hat{j},$then find $|\vec{a}|-|\vec{b}|.$

First, we find \begin{align}|\vec{a}|&=\sqrt(3)^2+(-5)^2=\sqrt{34} …(i)\\ |\vec{b}|&=\sqrt(-2)^2+(3)^2=\sqrt{13} …(ii)\end{align} Subtracting (i) from (ii). We get $$|\hat{a}|-|\hat{b}|=\sqrt{34}-\sqrt{13}$$

If $\vec{a}=3\hat{i}-5\hat{j}$ and $\vec{b}=-2\hat{i}+3\hat{j}$, then find $\dfrac{|\vec{a}|}{|\vec{b}|}$. $

Solve yourself.

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