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- Question 2 and 3 Exercise 3.3
- }-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} \\ \Rightarr... b}}{|\vec{a}+\vec{b}|}=\dfrac{4 \hat{i}+3 \hat{j}-12 k}{13} \\ \Rightarrow &=\dfrac{1}{13}(4 \hat{i}+3 \hat{j}-12 \hat{k})\\ \Rightarrow &=\dfrac{4}{13}\hat{i}+\df
- Question 7 Exercise 3.5
- 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\\ \Ri... 1 \\ c & 2 & 6 \end{array}\right|&=0\\ 1(18-2)-1(12-c)+2(4-3 c)&=0 \\ 16-12+c+8-6 c&=0 \\ \Rightarrow -5 c+12&=0 \\ \Rightarrow c&=\dfrac{-12}{-5}=\dfrac{12}{5}\end{align} which
- Question 2 Exercise 3.4
- & 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 ... (3)^2+(6)^2+(-9)^2}\\ \Rightarrow|\vec{a}|&=\sqrt{12 \overline{6}}\text{ and}\\ |\vec{b}|&=\sqrt{(1)^2... t \vec{b}}{i|\vec{b}|}=\dfrac{42}{\sqrt{14} \sqrt{126}} \\ \Rightarrow \theta&=\cos^{-1}\left(\sqrt{\d
- Question 7 & 8 Exercise 3.4
- -4 \end{array}\right| \\ \vec{a} \times \vec{b}&=(12+8) \hat{i}-(-8-16) \hat{j}+ (-4+12) \hat{k} \\ \vec{a} \times \vec{b}&=2 \hat{i}-24 \hat{j}+8 \hat{k}... {65}} . \\ \Rightarrow \hat{n}&=\dfrac{10 \hat{i}+12 \hat{j}+4 \hat{k}}{\sqrt{65}} .\end{align} Now le... = \vec{c} \cdot \hat{n}=10\left(\dfrac{10 \hat{i}+12 \hat{j}+4 \hat{k}}{\sqrt{65}}\right)$, is the req
- Question 12, 13 & 14, Exercise 3.2
- ====== Question 12, 13 & 14, Exercise 3.2 ====== Solutions of Question 12, 13 & 14 of Exercise 3.2 of Unit 03: Vectors. Thi... KPTB or KPTBB) Peshawar, Pakistan. =====Question 12===== Find $\alpha ,$ so that $|\alpha \hat{i}+(\a
- Question 7 & 8 Exercise 3.3
- )+2(5) \\ \Rightarrow \vec{a} \cdot \vec{b}&=2+10=12 \ldots \ldots \ldots \ldots(1)\\ \mid \bar{a} i&=... }|}$\\ Projection of $\vec{a}$ on $\vec{b}=\dfrac{12}{\sqrt{27}}=\dfrac{4}{\sqrt{3}}$.\\ Projection of... }|}$\\ Projection of $\vec{b}$ on $\vec{a}=\dfrac{12}{\sqrt{14}}$\\ =====Question 8===== What is the
- Question 12 & 13, Exercise 3.3
- ====== Question 12 & 13, Exercise 3.3 ====== Solutions of Question 12 & 13 of Exercise 3.3 of Unit 03: Vectors. This is ... KPTB or KPTBB) Peshawar, Pakistan. =====Question 12===== Prove that the angle in a semicircle is righ
- Question 9 Exercise 3.4
- allelogram} =|\vec{c} \times \vec{d}|=\sqrt{5^2+(-12)^2+9^2}=\sqrt{110} \text { units. }$$ =====Quest... (\dfrac{-11}{2}\right)^2} \\ & =\sqrt{1+41+\dfrac{121}{4}} \\ & =\sqrt{\dfrac{4+196+121}{4}} \\ & =\dfrac{\sqrt{321}}{2} \end{align} Thus area of parallel
- Question 11, Exercise 3.2
- (2\hat{i}-3\hat{j})-4(3\hat{i}+2\hat{j})]\\ &=-(6-12)\hat{i}-(-9-8)\hat{j}\\ \implies \overrightarrow{... n="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3-2-p9|Question 12, 13 & 14 >]]</btn></text>
- Question 4 Exercise 3.4
- 1 & 4\\ \end{array}\right| \\ & =(-24+5) \hat{i}-(12-10) \hat{j}+(-3+12) \hat{k}\\ \Rightarrow \vec{a} \times \vec{b}&=-19 \hat{i}-2 \hat{j}+9 \hat{k} .\e
- Question 6 Exercise 3.4
- htarrow \vec{M}&=(-20+2) \hat{i}-(0-3) \hat{j}+(0+12) \hat{k} \\ \Rightarrow \vec{M}&=-18 \hat{i}+3 \hat{j}+12 \hat{k} .\end{align} ====Go To==== <text ali
- Question 8 Exercise 3.5
- & -3 \end{array} \right|\\ V& =\dfrac{1}{6}[ -3(3-12)+5(6-12)-1(4-2)]\\ & =\dfrac{1}{6}[ 27-30-2]\\ & \Rightarrow V=-\frac{5}{6} \text { units. } \end{ali
- Question 9 & 10, Exercise 3.3
- ghtarrow W&=2.7+3 .-1+1.1 \\ \therefore W&=14-3+1=12 \text { units. }\end{align} ====Go To==== <t
- Question 11, Exercise 3.3
- align="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3-3-p8|Question 12 & 13 >]]</btn></text>
- Question 3 Exercise 3.4
- } \times \vec{b}&=(-1-24) \hat{i}-(3-6) \hat{j}+ (12+1) \hat{k} \\ \Rightarrow \vec{a} \times \vec{b}&