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- Question 10 Review Exercise 3
- e get \begin{align} \vec{b}&=\vec{a}+\vec{c} \\ \Rightarrow \vec{a}&=\vec{b}-\vec{c} \\ \Rightarrow \vec{a} \cdot \vec{a}&=(\vec{b}-\vec{c}) \cdot(\vec{b}-\vec{c}) \\ \Rightarrow|\vec{a}|^2&=\vec{b} \cdot \vec{b}-\vec{b} \cdot \... }-\vec{c} \cdot \vec{b}+\vec{c} \cdot \vec{c} \\ \Rightarrow|\vec{a}|^2&=|\vec{b}|^2+|\vec{c}|^2-2 \vec{b} \cd
- Question 8 & 9 Review Exercise 3
- $\vec{a}=\overrightarrow{A B}=(-1,3,2)-(0,0,2)$ $\Rightarrow \vec{a}=(-1,3,0)$ $\vec{b}=\overrightarrow{B C}=(1,0,4)-(-1,3,2)$ $\Rightarrow \vec{b}=(2,-3,2)$. We know that area of triangle... \vec{b} &=(\hat{i}+2 \hat{j} \cdot 3 \hat{k} \\ \Rightarrow | \vec{a} \times \vec{b} |&=\sqrt{(6)^2+(2)^2+(-3)^2} \\ \Rightarrow | \vec{a} \times \vec{b}|&=\sqrt{49}= 7 .\end{ali
- Question 6 & 7 Review Exercise 3
- align}\vec{a} \cdot \vec{b} \times \vec{c}&=0 \\ \Rightarrow\left|\begin{array}{ccc} 1 & 3 & 1 \\ 2 & -1 & -1 \\ 0 & \lambda & 3 \end{array}\right|&=0 \\ \Rightarrow \quad 1(-3+\lambda)-3(6+0)+1(2 \lambda-0)&=0\\ \Rightarrow \quad-3+\lambda-18+2 \lambda&=0 \\ \Rightarrow \quad 3 \lambda - 21&=0 \\ \Rightarrow \quad \lambda&=\dfrac{2
- Question 2 & 3 Review Exercise 3
- \hat{i}-\lambda \hat{j}+\mu \hat{k})&=\vec{O} \\ \Rightarrow\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{... & -\lambda & \mu \end{array}\right|&=\vec{O} \\ \Rightarrow(3 \mu+9 \lambda)\hat{i}-(\mu-27) \hat{j}+(-\lambda-9) \hat{k}&=\overrightarrow{0} \\ \Rightarrow \mu-27=0 \text { and }-\lambda-9&=0 \\ \Rightarrow \mu=27 \text { and } \lambda&=-9 .\end{align} =====Questi
- Question 4 & 5 Review Exercise 3
- & =(0-0) \hat{i}-(0-z) \hat{j}+(0-y) \hat{k} \\ \Rightarrow \vec{r} \times \hat{i}&=z \hat{j}-y \hat{k} \ldot... \hat{i}-(0-0) \hat{j}+(x-0) \hat{k}\end{align} $$\Rightarrow \vec{r} \times \hat{j}=-z \hat{i}+x \hat{k}$$ Tak... hat{j}-y \hat{k}) \cdot(-z \hat{i}+x \hat{k}) \\ \Rightarrow(\vec{r} \times \hat{i}) \cdot(\vec{r} \times \hat{j})&=0+0-x y \\ \Rightarrow(\vec{r} \times \hat{i}) \cdot(\vec{r} \times \vec
- Question 7 Exercise 3.5
- }&=0\\ \vec{u} \cdot \vec{v} \times \vec{w}&=0\\ \Rightarrow\left|\begin{array}{ccc}1 & 2 & 3 \\ 2 & -3 & 4 \\... ray}\right|&=0\\ 1(-3 c-4)-2(2 c-12)+3(2+9)&=0\\ \Rightarrow-3 c-4-4 c+24+33&=0\\ \Rightarrow \quad-7 c+53&=0\\ \Rightarrow c&=\dfrac{53}{7}.\end{align} which is required value of $c$ for which the given
- Question 8 Exercise 3.5
- ac{1}{6}[\vec{u} \cdot \vec{v} \times \vec{w}]\\ \Rightarrow V&=\dfrac{1}{6}\left|\begin{array}{lll}1 & 2 & 3 ... ht|\\ V&=\dfrac{1}{6} \cdot 1(40-42)-4(16-21) \\ \Rightarrow V&=\dfrac{1}{6}(-2+20)=3 \text { units. }\end{ali... t{i}-2 \hat{j})-(2 \hat{i}-3 \hat{j}+\hat{k}) \\ \Rightarrow \vec{a}&=-3 \hat{i}-5 \hat{j}-\hat{k} \\ \vec{b}&... at{j}-2 \hat{k}-(2 \hat{i}+3 \hat{j}+\hat{k}) \\ \Rightarrow \vec{c}&=-2 \hat{i}-2 \hat{j}-3 \hat{k}\end{align
- Question 6 Exercise 3.5
- A B}=\overrightarrow{O B}-\overrightarrow{O A}\\ \Rightarrow \vec{a}&=(5 \hat{i}+\hat{j}+6 \hat{k})-(4 \hat{i}-2 \hat{j}+\hat{k}) \\ \Rightarrow \vec{a}&=\hat{i}+3 \hat{j}+5 \hat{k} \ldots \ldot... C}=\overrightarrow{O C}-\overrightarrow{O A} \\ \Rightarrow \vec{b}&=2 \hat{i}+2 \hat{j} \quad 5 \hat{k}-(4 \hat{i}-2 \hat{j}-\hat{k}) \\ \Rightarrow \vec{b}&=-2 \hat{i}+4 \hat{j}-6 \hat{k} ....(2)\\
- Question 5(iii) & 5(iv) Exercise 3.5
- in{align}|\vec{a}|^2&=(a_1)^2+(a_2)^2+(a_3)^2 \\ \Rightarrow | \vec{a}|^2&=a_1^2+a_2^2+a_3^2\quad \text { and ... \cdot \vec{b})^2&=(a_1 b_1+a_2 b_2+a_3 b_3)^2 \\ \Rightarrow(\vec{a} \cdot \vec{b})^2&=a_1^2 b_1^2+a_2^2 b_2^2
- Question 5(i) & 5(ii) Exercise 3.5
- =0\quad \because \text{two rows are identical}\\ \Rightarrow \quad \vec{a} \cdot \vec{a} \times \vec{b}&=0\end... ad\because \quad\text{ two rows are identical}\\ \Rightarrow \vec{a} \cdot(\vec{a} \times \vec{b})&=0.\end{ali... b_3-a_3 b_1) \hat{j}+(a_1 b_2-a_2 b_1)\hat{k} \\ \Rightarrow|\vec{a} \times \vec{b}|&=\sqrt{ (a_2 b_3-a_3 b_2)
- Question 3 & 4 Exercise 3.5
- } \cdot \vec{b} \times \vec{c}&=3(8+1)+2(-1-0)\\ \Rightarrow \vec{a} \cdot \vec{b} \times \vec{c}&=25 \ldots \... \cdot \vec{c} \times \vec{a}&=1(-2-0)+3(8+1) \\ \Rightarrow \vec{b} \cdot \vec{c} \times \vec{a}&=25 \ldots \... ec{c}\cdot\vec{a}\times\vec{b}&=1(3-2)+4(6-0) \\ \Rightarrow \vec{c} \cdot \vec{a} \times \vec{b}&=1+24=25 \ld... 4 \hat{j}+\hat{k}) \cdot(3 \hat{i}+2 \hat{k}) \\ \Rightarrow \vec{c} \times \vec{b} \cdot \vec{a}&=-9 \cdot 3+
- Question 9 Exercise 3.4
- c}&=\overrightarrow{A E}+\overrightarrow{E B} \\ \Rightarrow \vec{c}&=2 \hat{i}+\dfrac{1}{2} \hat{j}-\hat{k}-(... se \overrightarrow{B E}=-\overrightarrow{E B} \\ \Rightarrow \vec{c}&=3 \hat{i}-\hat{j}-3 \hat{k} \ldots \ldot... {d}&=\overrightarrow{A E}+\overrightarrow{E D}\\ \Rightarrow \bar{d}&=2 \hat{i}+\dfrac{1}{2} \hat{j}-\hat{k}+(-\hat{i}+\dfrac{3}{2} \hat{j}+2 \hat{k}) \\ \Rightarrow \vec{d}&=\hat{i}+2 \hat{j}+ \hat{k} \text {. }...
- Question 1 & 2 Exercise 3.5
- \ -1 & 2 & 1 \\ 3 & 1 & 2 \end{array} \right| \\ \Rightarrow V& =2(4-1)-1(-2-3)+3(-1-6) \\ \Rightarrow V&=6+5-21=-10 \text {. }\text{unit cub}\end{align} =====Question ... \ 2 & -3 & 1 \\ 1 & -3 & -4 \end{array}\right|\\ \Rightarrow V&=3(12+3)-1(-8-1)-(-6,3) \\ V&=45+9+3=57 \text {
- Question 7 & 8 Exercise 3.4
- imes(\vec{A}+\vec{B}+\vec{C})=0$$\\ \begin{align}\Rightarrow \vec{A} \times \vec{A}+\vec{A} \times \vec{B}+\vec{A} \times \vec{C}&=\vec{O}...(1) \\ \Rightarrow \vec{A} \times \vec{B}+\vec{A} \times \vec{C} &= \vec{O} \quad \because \vec{A} \| \vec{A} \\ \Rightarrow \vec{A} \times \vec{B}&=-\vec{A} \times \vec{C} \\ \Rightarrow \vec{A} \times \vec{B}&=\vec{C} \times \vec{A}...
- Question 6 Exercise 3.4
- r}&=\overrightarrow{O P}\\ &=(1,-2,2)-(0,0,0) \\ \Rightarrow \vec{r}&=(1,-2,2).\\ \text { Hence } \vec{M}-\vec... 1 & -2 & 2 \\ 3 & -2 & 5 \end{array} \right| \\ \Rightarrow \vec{M}&=(-10+4) \hat{i}-(5-6) \hat{j}+(-2+6) \hat{k} \\ \Rightarrow \vec{M}&=-6 \hat{i}+\hat{j}+4 \hat{k}.\end{align}... r}&=\overrightarrow{A P}\\ &=(1,-2,2)-(1,2 ,1)\\ \Rightarrow \vec{r}&=(0 ,-4,1) \\ \vec{M}&=\vec{r} \times \ve