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- Question 5 Exercise 3.4
- ea of the parallelogram, that is:\\ \begin{align}\text{Area of triangle}&=\dfrac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}| \\ \text { Since } \overrightarrow{P Q}&=(3,2)-(-2 ,-3) \\... } \times \overrightarrow{P R}|&=30 \\ \therefore \text { Area of triangle }& =\dfrac{1}{2}|\overrightarr... times \overrightarrow{P R}|\\ &=\dfrac{30}{2}=15 \text { units square. }\end{align} =====Queswtion 5(ii
- Question 2 and 3 Exercise 3.3
- gle hetween $\vec{a}$ and $\vec{b}$ \begin{align}\text { then } \cos \theta&=\dfrac{\vec{a} \cdot \vec{b... gle between $\vec{a}$ and $\vec{b}$ \begin{align}\text { then } \quad \cos \theta &=\dfrac{\vec{a} \cdot... \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=8 \\ \text { Also }|\vec{a}|&=\sqrt{(3)^2+(4)^2}=\sqrt{9+19} \\ \Rightarrow \quad|\vec{a}|&=\sqrt{25}=5\\ \text { and }|\vec{b}|&=\sqrt{(2)^2+(-5)^2}\\ \Rightarr
- Question 7 & 8 Exercise 3.3
- }{5}=\dfrac{7}{5} \ldots \ldots \ldots . .(1) \\ \text { Also }|\vec{a}|&=\sqrt{\left(-\dfrac{3}{2}\righ... {\sqrt{287}}{10}=\dfrac{17}{10} \ldots . .(2) \\ \text { and }|\vec{b}|&=\sqrt{(1)^2+(-2)^2+(-2)^2} \\ \... begin{align}\Rightarrow \vec{a} \cdot \vec{b}&=1 \text { and } \\ \vec{a} \mid &=\sqrt{(\sqrt{2})^2+(1)^2+(1)^2} \\ \Rightarrow|\vec{a}|&=\sqrt{4}=2 . \text { and } \\ \vec{b} \mid&=\sqrt{(1)^2}=1.\end{alig
- Question 4 & 5 Review Exercise 3
- cdot(\vec{r} \times \hat{j})+x y $$ \begin{align}\text { Now } \vec{r} \times \hat{i}&=\left|\begin{arra... -y \hat{k} \ldots \ldots \ldots \ldots . .(1) \\ \text { and } \vec{r} \times \hat{j}&=\left|\begin{arra... \hat{i}) \cdot(\vec{r} \times \vec{j})&=-x y \\ \text { Now }(\vec{r} \times \hat{i}) \cdot(\vec{r} \ti... \\ \Rightarrow \vec{a} \cdot \vec{b}&=14+6-12=8 \text { and } \\ |\vec{b} |&=\sqrt{(2)^2+(6)^2+(3)^2} \
- Question 2 & 3 Review Exercise 3
- -\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0} \text {. }$$\\ ====Solution==== We are given\\ \begin{a... t{k}&=\overrightarrow{0} \\ \Rightarrow \mu-27=0 \text { and }-\lambda-9&=0 \\ \Rightarrow \mu=27 \text { and } \lambda&=-9 .\end{align} =====Question 3=====... }\\ &=\dfrac{11 \hat{i}-3 \hat{j}}{\sqrt{202}}\\ \text { Now } \hat{n}&=\dfrac{\vec{a}+\vec{b}}{|\vec{a}
- Question 8 & 9 Review Exercise 3
- angle is half of the area of parallelogram.\\ $$ \text{Area of triangle} =\dfrac{1}{2}|\vec{a} \times \v... $|\vec{a} \times \vec{b}|=7$ in (1), we get\\ $$\text{Area of triangle} =\dfrac{7}{2}\text{ units square}$$. =====Question 9===== Find the area of parallelo... uad D(0,-8,-2)$.\\ ====Solution==== \begin{align}\text { Let } \vec{a} &=\overrightarrow{A B}=(5,8 ,1)-(
- Question 9 & 10, Exercise 3.3
- \div 3(1)+1(-1) \\ \Rightarrow W &=4+3 \quad 1=6 \text { units.}\end{align} =====Question 10===== Find... verrightarrow{O B}=5 \hat{i}+0 \hat{j}+3 \hat{k} \text {. }$$ Displacement vector from $A$ to $B$ \begin... rrow W&=2.7+3 .-1+1.1 \\ \therefore W&=14-3+1=12 \text { units. }\end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit0
- Question 11, Exercise 3.3
- {i}+\hat{j})-(\hat{i}-\hat{k})=\hat{j}-\hat{k}\\ \text { Now }|\overrightarrow{P Q}|&=\sqrt{(1)^2}=1 \\ |\overrightarrow{Q R}|&=\sqrt{(-1)^2}=1, \quad \text { and } \\ |\overrightarrow{P R}|&=\sqrt{(1)^2+(-... n}\\ $\Rightarrow \quad P(1.0 .1) \quad Q(1,1,1) \text { and } R(1.1,0)$ forms a right angle triangle.\\... triangle is also isosceles. ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:so
- Question 7 & 8 Exercise 3.4
- We are given\\ $$\vec{A}+\vec{B}+\vec{C}=\vec{O} \text {. }$$\\ Taking cross product of $\vec{A}$, of bo... B}=\vec{B} \times \vec{C}=\vec{C} \times \vec{A} \text {. }$$\\ =====Question 8 (i)===== Find a unit ve... k} . \quad \vec{b}=4 \hat{i}-2 \hat{j}-4 \hat{k} \text {. }$$ =====Solution===== Let $\hat{n}$ be unil v... ght)$, is the required vector.\\ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:so
- Question 1 & 2 Exercise 3.5
- t{k}$, $\vec{b}=-\hat{i}+2 \hat{j}+\hat{k} \quad \text { and }\quad \vec{c}=3 \hat{i}+\hat{j}+2 \hat{k} \text {. }$ ====Solution==== We know that \begin{align}... -1)-1(-2-3)+3(-1-6) \\ \Rightarrow V&=6+5-21=-10 \text {. }\text{unit cub}\end{align} =====Question 2===== Find the volume of the parallelopiped whose edges
- Question 5 & 6, Exercise 3.2
- \hat{j}\end{align} By comparison, we have $$3=-x \text{ and } 7=5-y$$ This gives $$x=-3 \text{ and } y=-2.$$ Hence coordinates of $D$ are $(-3,-2)$. ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit03:ex3-2-p3 |< Question 3 & 4 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 12 & 13, Exercise 3.3
- tarrow{O C}-\overrightarrow{O A}=\vec{c}-\vec{a} \text {...(2) }\end{align} Now \begin{align}\overright... \overrightarrow{O E}=\dfrac{\vec{a}+\vec{c}}{2} \text { and } \\ \overrightarrow{O F}&=\dfrac{\vec{a}+\vec{b}}{2} . \\ &\text { Now } \overrightarrow{O D} \perp \overrightarro... }{2}&=0 \ldots \ldots \ldots \ldots \ldots(1) \\ \text { Also } \overrightarrow{O E} \perp \overrightarr
- Question 3 Exercise 3.4
- \times \vec{b}}{\mid \vec{a} \times \vec{b}} \\ \text { Now } \vec{a} \times \vec{b}&=\left|\begin{arra... times \bar{b}} \ldots \ldots \ldots \ldots(1) \\ \text { Now } \vec{a} \times \vec{b}&=\left|\begin{arra... \hat{j}+13 \hat{k}) .\end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit03:ex3-4-p2 |< Question 2]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 9 Exercise 3.4
- \Rightarrow \vec{d}&=\hat{i}+2 \hat{j}+ \hat{k} \text {. }....(2) \end{align} By using (1) and (2), we ... \hat{i}-6 \hat{j}+7 \hat{k}\end{align} Now \\ $$\text{area of parallelogram} =|\vec{c} \times \vec{d}|=\sqrt{5^2+(-12)^2+9^2}=\sqrt{110} \text { units. }$$ =====Question 9(ii)===== Find the a... {j}-\dfrac{1}{2} \hat{k}\end{align} \begin{align}\text{Area of parallelogram}& =|\bar{c} \times \vec{d}|
- Question 5(i) & 5(ii) Exercise 3.5
- _2 & b_3 \end{array}\right|\\ &=0\quad \because \text{two rows are identical}\\ \Rightarrow \quad \vec{... & b_3 \end{array}\right|\\ &=0\quad\because \quad\text{ two rows are identical}\\ \Rightarrow \vec{a} \c... b_1)^2 +(a_1 b_2-a_2 b_1)^2.$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit03:ex3-5-p2 |< Question 3 & 4]]</btn></text> <text align="right"><btn type="success">[[math-1