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- Question 2 and 3 Exercise 3.3
- t{j}-5 \hat{k})+(2 \hat{i}+\hat{j}-7 \hat{k}) \\ \Rightarrow &=4 \hat{i}+3 \hat{j}-12 \hat{k}\\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{(4)^2+(3)^2+(-12)^2} \\ \Rightarrow &=\sqrt{16+9+144} \\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{169}=13\end{align} Now let say $\hat{c}$ b
- Question 2 Exercise 3.4
- \hat{k} \\ -1 & 2 & -3 \\ 2 & -4 & 6 \end{array}\right| \\ & =(12-12) \hat{i}-(-6+6) \hat{j}+(4-4) \hat{k} \\ \Rightarrow \vec{a} \times \vec{b}&=0 . \\ & \Rightarrow \vec{a} \| \vec{b} .\end{align} Second Way \begin{ali... \hat{k}) \cdot(2 \hat{i}-4 \hat{j}+6 \hat{k}) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-1(2)+2(-4)-3(6
- Question 2, Exercise 3.2
- ==Solution==== Let $$\overset{\scriptscriptstyle\rightharpoonup}{a}=3\hat{i}$$ Then $$|\overset{\scriptscriptstyle\rightharpoonup}{a}|=\sqrt{{{(3)}^{2}}}=3$$ Now we know ... hat $$\hat{a}=\dfrac{{\overset{\scriptscriptstyle\rightharpoonup}{a}}}{|\overset{\scriptscriptstyle\rightharpoonup}{a}|}=\dfrac{3\hat{i}}{3}=\hat{i}$$ This is
- 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{
- Question 7 & 8 Exercise 3.3
- =\left(-\dfrac{3}{2} \hat{j}+\dfrac{4}{5} \hat{k}\right) \cdot(\hat{i}-2 \hat{j}-2 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{b}&=0(1)+\left(-\dfrac{3}{2}\right)(-2)+\dfrac{4}{5}(-2) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=3-\dfrac{8}{5}=\dfrac{7}{5} \ldots \ld
- 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 \\ 3 & 1 & c\end{array}\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 i
- Question 12 & 13, Exercise 3.3
- n 12===== Prove that the angle in a semicircle is right angle. ====Solution==== We are considering a tria... B}+\overrightarrow{A B}&=\overrightarrow{O A}\\ \Rightarrow \overrightarrow{B A}&=\overrightarrow{O A}-\... A}+\overrightarrow{A C}&=\overrightarrow{O C}\\ \Rightarrow \overrightarrow{A C}&=\overrightarrow{O C}-\... \vec{a}=|\vec{a}|^2 and\quad \vec{b}=-\vec{c}\\ \Rightarrow \overrightarrow{B A} \cdot \overrightarrow{A
- Question 3 Exercise 3.4
- & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{b}&=(2+3) \hat{i}-(-1-6) \hat{j}+(1-4) \hat{h} \\ \rightarrow \vec{a} \times \vec{b}&=-\hat{i}+7 \hat{j}+5 \hat{k} \\ \Rightarrow | \vec{a} \times \vec{b}|&=\sqrt{(-1)^2+(7)^
- Question 9 Exercise 3.4
- c}&=\overrightarrow{A E}+\overrightarrow{E B} \\ \Rightarrow \vec{c}&=2 \hat{i}+\dfrac{1}{2} \hat{j}-\hat... se \overrightarrow{B E}=-\overrightarrow{E B} \\ \Rightarrow \vec{c}&=3 \hat{i}-\hat{j}-3 \hat{k} \ldots ... {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 11, Exercise 3.3
- \hat{k}$ and $2 \hat{i}+\hat{j}-4 \hat{k}$ form a right angle triangle. ====Solution==== Let $\vec{a}=3 \... gin{align}|\vec{a}|&=\sqrt{(3)^2+(-2)^2+(1)^2}\\ \Rightarrow \quad |\vec{a}|&=\sqrt{14},\\ |\vec{b}|&=\sq... c}$ represent the sides of triangle and they form right angle triangle. Also if we see \begin{align}\vec{... }+\hat{k}) \cdot(2 \hat{i}+\hat{j}-4 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{c}&=6 - 2-4=0 .\end{align
- Question 5 Exercise 3.4
- Since } \overrightarrow{P Q}&=(3,2)-(-2 ,-3) \\ \Rightarrow \overrightarrow{P Q}&=(5,5) \\ \overrightarrow{P R}&=(-1,-8)-(-2 ,-3) \\ \Rightarrow \overrightarrow{P R}&=(1,-5) \\ \overrightar... } & \hat{k} \\ 5 & 5 & 0\\ 1 & -5 & 0 \end{array}\right|\\ \Rightarrow \overrightarrow{P Q} \times \overrightarrow{P R}&=(-25-5) \hat{k}=-30 \hat{k} \\ \Rightar
- Question 6 Exercise 3.3
- in{align}(\vec{a}+m \vec{b}) \cdot \vec{a}&=0 \\ \Rightarrow \quad[(1+2 m) \hat{j}+(3-3 m) \hat{j}+(5 m- ... at{k}] \cdot [\hat{i}+3 \hat{j}-4 \hat{k}]&=0 \\ \Rightarrow \quad 1(1+2 m)+3(3-3 m)-4(5 m-4)&=0 \\ \Rightarrow \quad 2 m-9 m-20 m+1+9+16&=0 \\ \Rightarrow \quad-27 m&=-26 \\ \Rightarrow \quad m&=\dfrac{26}{27}\
- 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 tri... & \hat{k} \\ -1 & 3 & 0 \\ 2 & -3 & 2 \end{array}\right| \\ \therefore \vec{a} \times \vec{b} &=(\hat{i}+2 \hat{j} \cdot 3 \hat{k} \\ \Rightarrow | \vec{a} \times \vec{b} |&=\sqrt{(6)^2+(2)^
- Question 3 & 4 Exercise 3.5
- } 3 & 0 & 2 \\ 1 & 2 & 1 \\ 0 & -1 & 4\end{array}\right|\\ \vec{a} \cdot \vec{b} \times \vec{c}&=3(8+1)+2(-1-0)\\ \Rightarrow \vec{a} \cdot \vec{b} \times \vec{c}&=25 \ld... 1 & 2 & 1 \\ 0 & -1 & 4 \\ 3 & 0 & 2 \end{array}\right|\\ \vec{b} \cdot \vec{c} \times \vec{a}&=1(-2-0)+3(8+1) \\ \Rightarrow \vec{b} \cdot \vec{c} \times \vec{a}&=25 \ld
- 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 \\ \Rightarr