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- Question 5 Exercise 3.4
- ercise 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. ==... 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) \\
- Question 2 and 3 Exercise 3.3
- ercise 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. ==... 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
- Question 7 & 8 Exercise 3.3
- ercise 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. ==... }{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} \\ \
- Question 4 & 5 Review Exercise 3
- 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. ==... cdot(\vec{r} \times \hat{j})+x y $$ \begin{align}\text { Now } \vec{r} \times \hat{i}&=\left|\begin{arra... -y \hat{k} \ldots \ldots \ldots \ldots . .(1) \\ \text { and } \vec{r} \times \hat{j}&=\left|\begin{arra
- Question 2 & 3 Review Exercise 3
- Exercise 3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... -\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0} \text {. }$$\\ ====Solution==== We are given\\ \begin{a... t{k}&=\overrightarrow{0} \\ \Rightarrow \mu-27=0 \text { and }-\lambda-9&=0 \\ \Rightarrow \mu=27 \text
- Question 8 & 9 Review Exercise 3
- 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. ==... 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 squar
- Question 9 & 10, Exercise 3.3
- ercise 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. ==... \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 {. }$$ Displacement vector from $A$ to $B$ \begin
- Question 11, Exercise 3.3
- ercise 3.3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... {i}+\hat{j})-(\hat{i}-\hat{k})=\hat{j}-\hat{k}\\ \text { Now }|\overrightarrow{P Q}|&=\sqrt{(1)^2}=1 \\ |\overrightarrow{Q R}|&=\sqrt{(-1)^2}=1, \quad \text { and } \\ |\overrightarrow{P R}|&=\sqrt{(1)^2+(-
- Question 7 & 8 Exercise 3.4
- ercise 3.4 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. ==... We are given\\ $$\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
- Question 1 & 2 Exercise 3.5
- ercise 3.5 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. ==... 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}
- Question 5 & 6, Exercise 3.2
- ercise 3.2 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. ==... \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 $(-3,-2)$. ====G
- Question 12 & 13, Exercise 3.3
- ercise 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. ==... 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}+\
- Question 3 Exercise 3.4
- ercise 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. ==... \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
- Question 9 Exercise 3.4
- ercise 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. =... \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}|=
- Question 5(i) & 5(ii) Exercise 3.5
- ercise 3.5 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. ==... _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