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- Question 12 & 13, Exercise 3.3 @math-11-kpk:sol:unit03
- side a semicircle as shown. We have to show $\overrightarrow{B A} \cdot \overrightarrow{A C}=0$. We see in figure that: $|\vec{a}|=\vec{b}|=| \vec{c} \mid=$ radius of ci... From $\triangle A B O$, we have \begin{align}\overrightarrow{O B}+\overrightarrow{A B}&=\overrightarrow{O A}\\ \Rightarrow \overrightarrow{B A}&=\overrightarrow{O A}-\ove
- Question 5 Exercise 3.4 @math-11-kpk:sol:unit03
- {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)-(-2 ,-3) \\
- Question 9 & 10, Exercise 3.2 @math-11-kpk:sol:unit03
- eshawar, Pakistan. =====Question 9===== If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, $$\overrightarrow{b}=4\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{c}=\hat{i}-2\hat{j}+\hat{k}$, find a vector of magnitude ... f $6$ unit which is parallel to the vector $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}.$ ==
- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- lign}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alph... gn}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} ... -\sqrt{1+\frac{9}{16}}=-\sqrt{\frac{25}{16}\,}\\ \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now ... }{\sec \alpha }=\frac{1}{-\tfrac{5}{4}}$$ $$\Rightarrow \quad \cos\alpha=-\dfrac{4}{5}$$ Now \begin{align
- Question 5, Exercise 10.1 @math-11-kpk:sol:unit10
- lign}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alph... gn}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} ... -\sqrt{1+\frac{9}{16}}=-\sqrt{\frac{25}{16}\,}\\ \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now ... }{\sec \alpha }=\frac{1}{-\tfrac{5}{4}}$$ $$\Rightarrow \quad \cos\alpha=-\dfrac{4}{5}$$ Now \begin{align
- Question 1 Exercise 6.3 @math-11-kpk:sol:unit06
- on==== We are given: \begin{align}&^n C_2=36\\ & \Rightarrow \dfrac{n !}{(n-2) ! 2 !}=36 \\ & \Rightarrow \dfrac{n(n-1)(n-2) !}{(n-2) ! \cdot 2}=36 \\ & \Rightarrow n(n-1)=72 \\ & \Rightarrow n^2-n-72=0 \\ & \Rightarrow n^2-9 n+8 n-72=0\\ & \Rightarrow n(n-9)+8(n-9)=0 \\ &
- Question 11, Exercise 3.3 @math-11-kpk:sol:unit03
- gin{align}|\vec{a}|&=\sqrt{(3)^2+(-2)^2+(1)^2}\\ \Rightarrow \quad |\vec{a}|&=\sqrt{14},\\ |\vec{b}|&=\sqrt{(1... }+\hat{k}) \cdot(2 \hat{i}+\hat{j}-4 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{c}&=6 - 2-4=0 .\end{align} $\t... angle from the ver-tices given. \begin{align}\overrightarrow{P Q}&=\overrightarrow{O Q} - \overrightarrow{O P} \\ \Rightarrow \overrightarrow{P Q}&=(\hat{i}+\hat{j}+\hat{
- Question 9 Exercise 3.4 @math-11-kpk:sol:unit03
- point of both diagonals. Thus\\ \begin{align}\overrightarrow{A E}&=\overrightarrow{E C}\\ &=\dfrac{1}{2} \vec{a}\\ &=2 \hat{i}+\dfrac{1}{2} \hat{j}-\hat{k} \\ \overrightarrow{E D}&=\overrightarrow{B E}\\ &=\dfrac{1}{2} \vec{b}\\ &=-\hat{i}+\dfrac{3}{2} \hat{j}+2 \hat{k}\end{align} Fr
- Question 5 & 6, Exercise 3.2 @math-11-kpk:sol:unit03
- estion 5===== Find the length of the vector $\overrightarrow{AB}$ from the point $\vec{A}(-3,5)$ to $\vec{B}(7... so find the unit vector in the direction of $\overrightarrow{AB}$. ====Solution==== The position vector of $\vec{A}$ and $\vec{B}$ are $$\overrightarrow{OA}=-3\hat{i}+5\hat{j},$$ $$\overrightarrow{OB}=7\hat{i}+9\hat{j}.$$ Thus we have \begin{align}\overrightarr
- Question 2 and 3 Exercise 3.3 @math-11-kpk:sol:unit03
- t{j}-5 \hat{k})+(2 \hat{i}+\hat{j}-7 \hat{k}) \\ \Rightarrow &=4 \hat{i}+3 \hat{j}-12 \hat{k}\\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{(4)^2+(3)^2+(-12)^2} \\ \Rightarrow &=\sqrt{16+9+144} \\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{169}=13\end{align} Now let say $\hat{c}$ be the unit
- Question 12, 13 & 14, Exercise 3.2 @math-11-kpk:sol:unit03
- alue of $\alpha$ =====Question 13===== If $\overrightarrow{u}=2\hat{i}+3\hat{j}+4\hat{k}$,$\overrightarrow{v}=-\hat{i}+3\hat{j}-\hat{k}$ and $\overrightarrow{w}=\hat{i}+6\hat{j}-z\hat{k}$ represents the sides of a tr... }$ and $\hat{k}.$ we have,\\ $$3=-z$$ $$-z=3$$ $$\Rightarrow \,\,\,z=-3$$ =====Question 14===== The position
- Question 6 Exercise 3.5 @math-11-kpk:sol:unit03
- 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{align}\ve
- Question 2 Exercise 3.4 @math-11-kpk:sol:unit03
- =(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{align}\vec{a}... \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) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-28 .\end{align} Als
- Question 8 Exercise 3.5 @math-11-kpk:sol:unit03
- 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... ices. ====Solution==== Position vector of $A,\overrightarrow{O A}=2 \hat{i}+3 \hat{j}+\hat{k}$ Position vector of $B, \overrightarrow{O B}=-\hat{i}-2 \hat{j}$ Position vector of $C,
- Question 1 and 2 Exercise 6.2 @math-11-kpk:sol:unit06
- We are given: \begin{align}^n P_5&=56(^n P_3) \\ \Rightarrow \dfrac{n !}{(n-5) !}&=56 \dfrac{n !}{(n-3) !} \\ \Rightarrow \dfrac{1}{(n-5) !}&=\dfrac{56}{(n-3) !} \\ \Rightarrow \dfrac{1}{(n-5) !}&=\dfrac{56}{(n-3)(n-4)(n-5) !} \\ \Rightarrow(n-3)(n-4)&=56 \\ \Rightarrow n^2-7 n+12-56&=0 \\