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- Question 12 & 13, Exercise 3.3
- n 12===== Prove that the angle in a semicircle is right angle. ====Solution==== We are considering a tria... side a semicircle as shown. We have to show $\overrightarrow{B A} \cdot \overrightarrow{A C}=0$. We see in figure that: $|\vec{a}|=\vec{b}|=| \vec{c} \mid=$ ra... From $\triangle A B O$, we have \begin{align}\overrightarrow{O B}+\overrightarrow{A B}&=\overrightarrow{O
- Question 5 Exercise 3.4
- {align}\text{Area of triangle}&=\dfrac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}| \\ \text { Since } \overrightarrow{P Q}&=(3,2)-(-2 ,-3) \\ \Rightarrow \overrightarrow{P Q}&=(5,5) \\ \overrightarrow{P R}&=(-1,-8)-(-
- Question 9 & 10, Exercise 3.2
- eshawar, Pakistan. =====Question 9===== If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, $$\overrightarrow{b}=4\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{c}=\hat{i}-2\hat{j}+\hat{k}$, find a vector of ... f $6$ unit which is parallel to the vector $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}.$
- Question 11, Exercise 3.3
- \hat{k}$ and $2 \hat{i}+\hat{j}-4 \hat{k}$ form a right angle triangle. ====Solution==== Let $\vec{a}=3 \... gin{align}|\vec{a}|&=\sqrt{(3)^2+(-2)^2+(1)^2}\\ \Rightarrow \quad |\vec{a}|&=\sqrt{14},\\ |\vec{b}|&=\sq... c}$ represent the sides of triangle and they form right angle triangle. Also if we see \begin{align}\vec{... }+\hat{k}) \cdot(2 \hat{i}+\hat{j}-4 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{c}&=6 - 2-4=0 .\end{align
- Question 9 Exercise 3.4
- point of both diagonals. Thus\\ \begin{align}\overrightarrow{A E}&=\overrightarrow{E C}\\ &=\dfrac{1}{2} \vec{a}\\ &=2 \hat{i}+\dfrac{1}{2} \hat{j}-\hat{k} \\ \overrightarrow{E D}&=\overrightarrow{B E}\\ &=\dfrac{1}{2} \vec{b}\\ &=-\hat{i}+\dfrac{3}{2} \hat{j}+2 \hat{k}\end
- Question 2 and 3 Exercise 3.3
- t{j}-5 \hat{k})+(2 \hat{i}+\hat{j}-7 \hat{k}) \\ \Rightarrow &=4 \hat{i}+3 \hat{j}-12 \hat{k}\\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{(4)^2+(3)^2+(-12)^2} \\ \Rightarrow &=\sqrt{16+9+144} \\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{169}=13\end{align} Now let say $\hat{c}$ b
- Question 2 Exercise 3.4
- \hat{k} \\ -1 & 2 & -3 \\ 2 & -4 & 6 \end{array}\right| \\ & =(12-12) \hat{i}-(-6+6) \hat{j}+(4-4) \hat{k} \\ \Rightarrow \vec{a} \times \vec{b}&=0 . \\ & \Rightarrow \vec{a} \| \vec{b} .\end{align} Second Way \begin{ali... \hat{k}) \cdot(2 \hat{i}-4 \hat{j}+6 \hat{k}) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-1(2)+2(-4)-3(6
- Question 5 & 6, Exercise 3.2
- estion 5===== Find the length of the vector $\overrightarrow{AB}$ from the point $\vec{A}(-3,5)$ to $\vec... so find the unit vector in the direction of $\overrightarrow{AB}$. ====Solution==== The position vector of $\vec{A}$ and $\vec{B}$ are $$\overrightarrow{OA}=-3\hat{i}+5\hat{j},$$ $$\overrightarrow{OB}=7\hat{i}+9\hat{j}.$$ Thus we have \begin{align}\ov
- Question 6 Exercise 3.5
- and $D(3,5.0)$ then Position vector of $A, \overrightarrow{O A}=4 \hat{i}-2 \hat{j}+\hat{k}$ Position vector of $B, \overrightarrow{O B}=5 \hat{i}+\hat{j}+6 \hat{k}$ Position vector of $C, \overrightarrow{O C}=2 \hat{i}+2 \hat{i}-5 \hat{k}$ Position vector of $D, \overrightarrow{O D}=3 \hat{i}+5 \hat{j}$. Then \begin{alig
- Question 12, 13 & 14, Exercise 3.2
- alue of $\alpha$ =====Question 13===== If $\overrightarrow{u}=2\hat{i}+3\hat{j}+4\hat{k}$,$\overrightarrow{v}=-\hat{i}+3\hat{j}-\hat{k}$ and $\overrightarrow{w}=\hat{i}+6\hat{j}-z\hat{k}$ represents the sid... }$ and $\hat{k}.$ we have,\\ $$3=-z$$ $$-z=3$$ $$\Rightarrow \,\,\,z=-3$$ =====Question 14===== The posi
- Question 8 Exercise 3.5
- ac{1}{6}[\vec{u} \cdot \vec{v} \times \vec{w}]\\ \Rightarrow V&=\dfrac{1}{6}\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 0 & 7 & 8\end{array}\right|\\ V&=\dfrac{1}{6} \cdot 1(40-42)-4(16-21) \\ \Rightarrow V&=\dfrac{1}{6}(-2+20)=3 \text { units. }\end{... ices. ====Solution==== Position vector of $A,\overrightarrow{O A}=2 \hat{i}+3 \hat{j}+\hat{k}$ Position
- Question 2, Exercise 3.2
- ==Solution==== Let $$\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 8 & 9 Review Exercise 3
- $, $B(-1,3,2)$ and $C(1,0,4)$. Let $\vec{a}=\overrightarrow{A B}=(-1,3,2)-(0,0,2)$ $\Rightarrow \vec{a}=(-1,3,0)$ $\vec{b}=\overrightarrow{B C}=(1,0,4)-(-1,3,2)$ $\Rightarrow \vec{b}=(2,-3,2)$. We know that area of triangle is half of th