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- Question 9 Exercise 3.4
- so $E$ is the midpoint of both diagonals. Thus\\ \begin{align}\overrightarrow{A E}&=\overrightarrow{E C}\... {k}\end{align} From $\triangle A E B$, we have\\ \begin{align}\vec{c}&=\overrightarrow{A E}+\overrightarr... (1)\end{align} From $\triangle A E D$. we have\\ \begin{align}\vec{d}&=\overrightarrow{A E}+\overrightarr... (2) \end{align} By using (1) and (2), we have,\\ \begin{align}\vec{c} \times \vec{d}&=\left| \begin{array
- 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 \\ 2 &... rrow \vec{a} \| \vec{b} .\end{align} Second Way \begin{align}\vec{a} \cdot \vec{b}&=(-\hat{i}+2 \hat{j}-... uad \vec{a} \cdot \vec{b}&=-28 .\end{align} Also \begin{align}|\vec{a}|&=\sqrt{(-1)^2+(2)^2+(-3)^2}\\ \Ri
- 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{A} ... ng cross product of $\vec{B}$ with (1), we get\\ \begin{align}\vec{B} \times(\vec{A}+\vec{B}+\vec{C})&=0 ... endicular to both $\vec{a}$ and $\vec{b}$ then\\ \begin{align}\hat{n}&=\dfrac{\vec{a} \times \vec{b}}{\mi... ldots \ldots(1) \\ \vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1
- 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 & 4\... ) \\ \vec{b} \cdot \vec{c} \times \vec{a}&=\left|\begin{array}{ccc} 1 & 2 & 1 \\ 0 & -1 & 4 \\ 3 & 0 & 2 ... ) \\ \vec{c} \cdot \vec{a} \times \vec{b}&=\left|\begin{array}{ccc} 0 & -1 & 4 \\ 3 & 0 & 2 \\ 1 & 2 & 1
- Question 2 and 3 Exercise 3.3
- hat{k}$$. ====Solution==== We first find the sum \begin{align}\vec{a}+\vec{b}&=(2 \hat{i}+2 \hat{j}-5 \ha... ctor $x$ the sum of $\vec{a}$ and $\vec{b}$ then \begin{align}\hat{c}&=\dfrac{\vec{a}+\vec{b}}{|\vec{a}+\... ta$ be the angle hetween $\vec{a}$ and $\vec{b}$ \begin{align}\text { then } \cos \theta&=\dfrac{\vec{a} ... ta$ be the angle between $\vec{a}$ and $\vec{b}$ \begin{align}\text { then } \quad \cos \theta &=\dfrac{\
- Question 12, 13 & 14, Exercise 3.2
- {j}+2\hat{k}|=3$. ====Solution==== We are given \begin{align}|\alpha \hat{i}+(\alpha +1)\hat{j}+2\hat{k}|&=3.\end{align} This gives \begin{align}\sqrt{(\alpha )^2+(\alpha +1)^2+(2)^2}&=3.\... end{align} Taking square on both sides, we have, \begin{align}&{\alpha ^2+(\alpha +1)^2}+4=9\\ \implies &... ion in $\alpha $. $$a=1, \quad b=1,\quad c=-2$$ \begin{align}\alpha &=\dfrac{-1\pm \sqrt{(1)^2-4(1)(-2)}
- 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 \ha... b_3 \end{align} Taking square of the both sides \begin{align}(\vec{a} \cdot \vec{b})^2&=(a_1 b_1 + a_2 b... )^2}\end{align} Taking square of the both sides \begin{align}|\vec{a}|^2&=(a_1)^2+(a_2)^2+(a_3)^2 \\ \Ri... 2} .\end{align} Taking square of the both sides \begin{align}|\vec{b}|^2&=(b_1)^2+(b_2)^2+(b_3)^2 \\ |\v
- 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{j}... n find $\vec{a} \cdot \vec{c}$. ====Solution==== \begin{align}\vec{a} \cdot \vec{c}&=(3 \hat{i}+4 \hat{j}... \vec{a} \cdot(\vec{b}+\vec{c})$ ====Solution==== \begin{align}\vec{b}+\vec{c}&=(\hat{i}-\hat{j}+3 \hat{k}... {k}\end{align} Taking dot product with $\vec{a}$ \begin{align}\vec{a} \cdot(\vec{b}+\vec{c})&=(3 \hat{i}+
- 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 \overrightarrow{... $D E$ and $F$ are the midpoints of sides shown\\ \begin{align}\therefore \quad \overrightarrow{O D}&=\dfr
- Question 1 Exercise 3.4
- times(2 \hat{j}+3 \hat{k})$ ====Solution==== Let \begin{align}\vec{a}=\hat{j}&=0 \hat{i}+\hat{j}+0 \hat{k... }&=\hat{j} \times(2 \hat{j}+3 \hat{k})\\ &=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 &... -3 \hat{j}) \times \hat{k}$ ====Solution==== Let \begin{align}\vec{a}&=2 \hat{i}-3 \hat{j}\\ \vec{b}&=\ha... t{k} \\ \therefore \vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3
- Question 4 Exercise 3.4
- $ find $\vec{a} \times \vec{b}$ ====Solution==== \begin{align}\vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & 5\\ 2 & -... $ find $\vec{b} \times \vec{c}$ ====Solution==== \begin{align}\vec{b} \times \vec{c}&=\left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 4 \\ 1 &
- 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 \\... \vec{a} \times \vec{b}$ orthogonal to $\vec{b}$ \begin{align}\vec{b} \cdot \vec{a} \times \vec{b}&=\left|\begin{array}{lll} b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \\
- 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 & c\e... on==== The given vectors are coplanar, therefore \begin{align}\vec{u} \cdot \vec{v} \times \vec{w}&=0\\ \Rightarrow\left|\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -2 & 1 \\ c & 1 & -
- Question 8 Exercise 3.5
- etrahedron with the Vectors as coterminous edges \begin{align}\vec{a}&=\hat{i}+2 \hat{j}+3 \hat{k},\\ \v... n} ====Solution==== The volume of tetrahedron is \begin{align}V&=\dfrac{1}{6}[\vec{u} \cdot \vec{v} \times \vec{w}]\\ \Rightarrow V&=\dfrac{1}{6}\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 0 & 7 & 8\en... D}=\hat{j}-2 \hat{k}$ We find the edges vectors \begin{align}\vec{a}&=\overrightarrow{A B}=\overrightarr
- Question 1, Exercise 3.2
- then find $\vec{a}+2\vec{b}$. ====Solution==== \begin{align}\vec{a}+2\vec{b}&=3\hat{i}-5\hat{j}+2(-2\ha... hen find $3\vec{a}-2\vec{b}$. ====Solution==== \begin{align}3\vec{a}-2\vec{b}&=3(3\hat{i}-5\hat{j})-2(-... {a}-\vec{b})$. ====Solution==== First we have, \begin{align}\vec{a}-\vec{b}&=3\hat{i}-5\hat{j}-(-2\hat{... $|\vec{a}+\vec{b}|$. ====Solution==== We have, \begin{align}\vec{a}+\vec{b}&=3\hat{i}-5\hat{j}+(-2\hat{