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- Question 2 and 3 Exercise 3.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... hat{k}$$. ====Solution==== We first find the sum \begin{align}\vec{a}+\vec{b}&=(2 \hat{i}+2 \hat{j}-5 ... |&=\sqrt{169}=13\end{align} Now let say $\hat{c}$ be the unit vector $x$ the sum of $\vec{a}$ and $\vec{b}$ then \begin{align}\hat{c}&=\dfrac{\vec{a}+\vec{b}}{|\vec{a
- Question 7 & 8 Exercise 3.4
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... $$\vec{A} \times(\vec{A}+\vec{B}+\vec{C})=0$$\\ \begin{align}\Rightarrow \vec{A} \times \vec{A}+\vec{... \vec{B}+\vec{A} \times \vec{C} &= \vec{O} \quad \because \vec{A} \| \vec{A} \\ \Rightarrow \vec{A} \t... vec{B}&=\vec{C} \times \vec{A}...(2)\end{align} $\because \quad$ cross product is anti-commutative\\ $
- Question 9 Exercise 3.4
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... so $E$ is the midpoint of both diagonals. Thus\\ \begin{align}\overrightarrow{A E}&=\overrightarrow{E ... {k}\end{align} From $\triangle A E B$, we have\\ \begin{align}\vec{c}&=\overrightarrow{A E}+\overright... at{k}-(-\hat{i}+\dfrac{3}{2} \hat{j}+2 \hat{k}) \because \overrightarrow{B E}=-\overrightarrow{E B} \
- Question 12 & 13, Exercise 3.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... le in direction. From $\triangle A B O$, we have \begin{align}\overrightarrow{O B}+\overrightarrow{A B... \end{align} Also from $\triangle A C O$, we have \begin{align}\overrightarrow{O A}+\overrightarrow{A C... =\vec{c}-\vec{a} \text {...(2) }\end{align} Now \begin{align}\overrightarrow{B A} \cdot \overrightarr
- Question 2 Exercise 3.4
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... \hat{j}+$ $6 \hat{k}$ ====Solution==== First Way \begin{align}\vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & -3 ... rrow \vec{a} \| \vec{b} .\end{align} Second Way \begin{align}\vec{a} \cdot \vec{b}&=(-\hat{i}+2 \hat{
- Question 3 & 4 Exercise 3.5
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... \times \vec{b} \cdot \vec{a}$\\ ====Solution==== \begin{align}\vec{a} \cdot \vec{b} \times \vec{c}&=\left|\begin{array}{ccc} 3 & 0 & 2 \\ 1 & 2 & 1 \\ 0 & -1 &... ) \\ \vec{b} \cdot \vec{c} \times \vec{a}&=\left|\begin{array}{ccc} 1 & 2 & 1 \\ 0 & -1 & 4 \\ 3 & 0 &
- Question 7 Exercise 3.5
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... on==== The given vectors are coplanar, therefore \begin{align}\vec{u} \cdot \vec{v} \times \vec{w}&=0\... dot \vec{v} \times \vec{w}&=0\\ \Rightarrow\left|\begin{array}{ccc}1 & 2 & 3 \\ 2 & -3 & 4 \\ 3 & 1 & ... required value of $c$ for which the given vectors become coplanar. =====Question 7(ii)===== For what
- Question 1 Review Exercise 3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... $</collapse> vi. Find nun-zero scalar $\alpha . \beta$ for which $\alpha(\vec{a}+2 \vec{b})-\beta \vec{a}+(4 \vec{b}-\vec{a})=0$ for all vectors $\vec{a}$ and $\vec{b}$ * %%(a)%% $\alpha=-2, \beta=-3$ * (b) $\alpha=2 \cdot \beta=-3$
- Question 5(i) & 5(ii) Exercise 3.5
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... \vec{a} \times \vec{b}$ orthogonal to $\vec{a}$ \begin{align}\vec{a} \cdot \vec{a} \times \vec{b}&=\left|\begin{array}{lll} a_1 & a_2 & a_3 \\ a_1 & a_2 & a_3... \ b_1 & b_2 & b_3 \end{array}\right|\\ &=0\quad \because \text{two rows are identical}\\ \Rightarrow
- Question 12, 13 & 14, Exercise 3.2
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... {j}+2\hat{k}|=3$. ====Solution==== We are given \begin{align}|\alpha \hat{i}+(\alpha +1)\hat{j}+2\hat{k}|&=3.\end{align} This gives \begin{align}\sqrt{(\alpha )^2+(\alpha +1)^2+(2)^2}&=... end{align} Taking square on both sides, we have, \begin{align}&{\alpha ^2+(\alpha +1)^2}+4=9\\ \implie
- Question 3 Exercise 3.4
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... +\hat{j}-\hat{k}$. ====Solution==== Let $\hat{n}$ be unit vector orthogonal to both $\vec{a}$ and $\vec{b}$. then by cross product\\ \begin{align}\hat{n}&=\dfrac{\vec{a} \times \vec{b}}{... } \\ \text { Now } \vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 &
- Question 5 Exercise 3.4
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... 1,-8)$ ====Solution==== Let $P Q$ and $\bar{P} R$ be the adjacent sides of parallelogram determined, s... f hall the area of the parallelogram, that is:\\ \begin{align}\text{Area of triangle}&=\dfrac{1}{2}|\o... rightarrow{PQ}\times \overrightarrow{P R}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 5 &
- Question 5(iii) & 5(iv) Exercise 3.5
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... ec{b})^2,\quad|a|^2,\quad|b|^2$ ====Solution==== \begin{align}\vec{a} \cdot \vec{b}&=(a_1 \hat{i}+a_2 ... b_3 \end{align} Taking square of the both sides \begin{align}(\vec{a} \cdot \vec{b})^2&=(a_1 b_1 + a_... )^2}\end{align} Taking square of the both sides \begin{align}|\vec{a}|^2&=(a_1)^2+(a_2)^2+(a_3)^2 \\
- Question 8 Exercise 3.5
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... etrahedron with the Vectors as coterminous edges \begin{align}\vec{a}&=\hat{i}+2 \hat{j}+3 \hat{k},\\ ... n} ====Solution==== The volume of tetrahedron is \begin{align}V&=\dfrac{1}{6}[\vec{u} \cdot \vec{v} \t... imes \vec{w}]\\ \Rightarrow V&=\dfrac{1}{6}\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 0 & 7 & 8
- Question 3 & 4, Exercise 3.2
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... $\vec{b}=5\hat{i}-\hat{j}$, determine the real number $p$ and $q$ such that $\vec{r}=p\vec{a}+q\vec{b... $2$ by (i) and subtract (ii) from (i). We have \[\begin{array}{ccc} 2p&+10q&=2 \\ \mathop+\limits_{-... c{p}+\vec{q}|=5.$ ====Solution==== We calculate \begin{align}\vec{p}+\vec{q}&=2\hat{i}-\hat{j}+x\hat{