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- Question 7 & 8 Exercise 3.4
- perpendicular to both Find a vector of magnitude 10 and perpendicular to both $$\vec{a}=2 \hat{i}-3 \... +(24)^2+(8)^2} \\ | \vec{a} \times \vec{b}|=\sqrt{1040} \\ \Rightarrow|\vec{a} \times \vec{b}|&=2 \sqr... k}}{2 \sqrt{65}} . \\ \Rightarrow \hat{n}&=\dfrac{10 \hat{i}+12 \hat{j}+4 \hat{k}}{\sqrt{65}} .\end{al... dicular to both and having magnitude \\ $\vec{c} =10$ then\\ $\vec{c}= \vec{c} \cdot \hat{n}=10\left(\
- Question 7 & 8 Exercise 3.3
- 5}} \\ \Rightarrow|\vec{a}|&=\sqrt{\dfrac{225+64}{100}}=\dfrac{\sqrt{287}}{10}=\dfrac{17}{10} \ldots . .(2) \\ \text { and }|\vec{b}|&=\sqrt{(1)^2+(-2)^2+(-2)^2} \\ \Rightarrow|\v... (-1)+2(5) \\ \Rightarrow \vec{a} \cdot \vec{b}&=2+10=12 \ldots \ldots \ldots \ldots(1)\\ \mid \bar{a}
- Question 10 Review Exercise 3
- ====== Question 10 Review Exercise 3 ====== Solutions of Question 10 of Review Exercise 3 of Unit 03: Vectors. This is u... KPTB or KPTBB) Peshawar, Pakistan. =====Question 10(i)===== Prove that in any triangle $A B C$ that $... c{b}||\vec{c}| \cos A .\end{align} =====Question 10(ii)===== Prove that in any triangle $A B C$ that
- Question 5 & 6, Exercise 3.2
- \\ &=7\hat{i}+9\hat{j}-(-3\hat{i}+5\hat{j}) \\ &=10\hat{i}+4\hat{j}.\end{align} This gives \begin{align}|\overrightarrow{AB}|&=\sqrt{(10)^2+(4)^2}\\ &=\sqrt{116} = 2\sqrt{29}.\end{align}... rightarrow{AB}}{|\overrightarrow{AB}|}\\ &=\dfrac{10\hat{i}+4\hat{j}}{2\sqrt{29}} \\ &=\dfrac{5}{\sqrt
- Question 9 & 10, Exercise 3.2
- ====== Question 9 & 10, Exercise 3.2 ====== Solutions of Question 9 & 10 of Exercise 3.2 of Unit 03: Vectors. This is unit o... ired vector with given conditions. =====Question 10===== Find the position vector of a point $R$ whic
- Question 9 & 10, Exercise 3.3
- ====== Question 9 & 10, Exercise 3.3 ====== Solutions of Question 9 & 10 of Exercise 3.3 of Unit 03: Vectors. This is unit o... ad 1=6 \text { units.}\end{align} =====Question 10===== Find the work done by the force $\vec{F}=2 \
- Question 1, Exercise 3.2
- both sides by $2$. We have, $$2(\vec{a}-\vec{b})=10\hat{i}-16\hat{j}$$ =====Question.1(iv)===== If $
- Question 3 & 4, Exercise 3.2
- t (ii) from (i). We have \[\begin{array}{ccc} 2p&+10q&=2 \\ \mathop+\limits_{-}2p&\mathop-\limits_{+
- Question 11, Exercise 3.2
- [[math-11-kpk:sol:unit03:ex3-2-p7 |< Question 9 & 10 ]]</btn></text> <text align="right"><btn type="su
- Question 11, Exercise 3.3
- [[math-11-kpk:sol:unit03:ex3-3-p6 |< Question 9 & 10]]</btn></text> <text align="right"><btn type="suc
- Question 1 Exercise 3.4
- \right| \\ \Rightarrow \vec{a} \times \vec{b}&=(9-10) \hat{i}-(-6-30) \hat{j} +(4+18) \hat{k} \\ \Righ
- Question 4 Exercise 3.4
- 4\\ \end{array}\right| \\ & =(-24+5) \hat{i}-(12-10) \hat{j}+(-3+12) \hat{k}\\ \Rightarrow \vec{a} \t
- Question 6 Exercise 3.4
- & 5 \end{array} \right| \\ \Rightarrow \vec{M}&=(-10+4) \hat{i}-(5-6) \hat{j}+(-2+6) \hat{k} \\ \Right
- Question 9 Exercise 3.4
- {c} \times \vec{d}|=\sqrt{5^2+(-12)^2+9^2}=\sqrt{110} \text { units. }$$ =====Question 9(ii)===== Fin... \hat{i}-(1+6) \hat{j}+\left(-\dfrac{1}{2}-\dfrac{10}{2}\right) \hat{k} \\ \Rightarrow \vec{c} \times
- Question 1 & 2 Exercise 3.5
- =2(4-1)-1(-2-3)+3(-1-6) \\ \Rightarrow V&=6+5-21=-10 \text {. }\text{unit cub}\end{align} =====Quest