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- Question 2 and 3 Exercise 3.3
- hat{k}$$. ====Solution==== We first find the sum \begin{align}\vec{a}+\vec{b}&=(2 \hat{i}+2 \hat{j}-5 ... |&=\sqrt{169}=13\end{align} Now let say $\hat{c}$ be the unit vector $x$ the sum of $\vec{a}$ and $\vec{b}$ then \begin{align}\hat{c}&=\dfrac{\vec{a}+\vec{b}}{|\vec{a... d{align} =====Question 3(i)===== Find the angles between the pairs of vectors: $\hat{i}-\hat{j}+\hat{
- Question 7 & 8 Exercise 3.4
- $$\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\\ $... ng cross product of $\vec{B}$ with (1), we get\\ \begin{align}\vec{B} \times(\vec{A}+\vec{B}+\vec{C})&
- Question 9 Exercise 3.4
- 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} \... (1)\end{align} From $\triangle A E D$. we have\\ \begin{align}\vec{d}&=\overrightarrow{A E}+\overright
- Question 12 & 13, Exercise 3.3
- 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... htarrow{B A} \cdot \overrightarrow{A C}&=0 \quad \because \quad|\vec{a}|=|\vec{c}|\end{align} Triangle
- Question 2 Exercise 3.4
- \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{... uad \vec{a} \cdot \vec{b}&=-28 .\end{align} Also \begin{align}|\vec{a}|&=\sqrt{(-1)^2+(2)^2+(-3)^2}\\
- Question 3 & 4 Exercise 3.5
- \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 &... ) \\ \vec{c} \cdot \vec{a} \times \vec{b}&=\left|\begin{array}{ccc} 0 & -1 & 4 \\ 3 & 0 & 2 \\ 1 & 2 &
- Question 7 Exercise 3.5
- 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 ... on==== The given vectors are coplanar, therefore \begin{align}\vec{u} \cdot \vec{v} \times \vec{w}&=0\
- Question 1 Review Exercise 3
- $</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$ * %%(c)%% $\alpha=1 . \beta=-3$ * (d) $\alpha=-
- Question 5(i) & 5(ii) Exercise 3.5
- \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 ... \vec{a} \times \vec{b}$ orthogonal to $\vec{b}$ \begin{align}\vec{b} \cdot \vec{a} \times \vec{b}&=\l
- Question 12, 13 & 14, Exercise 3.2
- {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... ion in $\alpha $. $$a=1, \quad b=1,\quad c=-2$$ \begin{align}\alpha &=\dfrac{-1\pm \sqrt{(1)^2-4(1)(-
- Question 3 Exercise 3.4
- +\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 &... 4 \hat{j}+\hat{k}$ ====Solution==== Let $\hat{n}$ be unit vector orthogonal to buth $\vec{a}$ and $\ve
- Question 5 Exercise 3.4
- 1,-8)$ ====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 &... $\overrightarrow{P Q}$ and $\overrightarrow{P R}$ be the adjacent sides of parallelogram determined, s
- Question 5(iii) & 5(iv) Exercise 3.5
- 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 \\ ... 2} .\end{align} Taking square of the both sides \begin{align}|\vec{b}|^2&=(b_1)^2+(b_2)^2+(b_3)^2 \\
- Question 9 & 10, Exercise 3.2
- arrow{c}.$ ====Solution==== We compute that\\ \begin{align}2\overrightarrow{a}-\overrightarrow{b}+3... 2)^2}\\ &=\sqrt{9}=3\end{align} Let sat $\hat{a}$ be unit vector in direction of $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}.$ \begin{align}\hat{a}&=\dfrac{2\overrightarrow{a}-\ove... {a}-\overrightarrow{b}+3\overrightarrow{c}$is \\ \begin{align}6\hat{a}&=6[\frac{1}{3}(\hat{i}-2\hat{j}
- Question 1, Exercise 3.3
- hen find $\vec{a}\cdot \vec{b}$ ====Solution==== \begin{align}\vec{a} \cdot \vec{b}&=(3 \hat{i}+4 \hat... n find $\vec{a} \cdot \vec{c}$. ====Solution==== \begin{align}\vec{a} \cdot \vec{c}&=(3 \hat{i}+4 \hat... \vec{a} \cdot(\vec{b}+\vec{c})$ ====Solution==== \begin{align}\vec{b}+\vec{c}&=(\hat{i}-\hat{j}+3 \hat... {k}\end{align} Taking dot product with $\vec{a}$ \begin{align}\vec{a} \cdot(\vec{b}+\vec{c})&=(3 \hat{