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- Question 9 Exercise 3.4
- n}\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} From $\triangle A E B$, we have\\ \
- Question 7 & 8 Exercise 3.3
- the vectors $\vec{a}$ and $\vec{b}$ as $\vec{a}=-\dfrac{3}{2} \hat{j}+\dfrac{4}{5} \hat{k} \cdot \vec{b}=\hat{i}-2 \hat{j}-2 \hat{k}$. Find in each case the pro... vec{b}$ on $\vec{a}$. ====Solution==== $\vec{a}=-\dfrac{3}{2} \hat{j}+\dfrac{4}{5} \hat{k}\quad$ $\vec{b}=\hat{i}-2 \hat{j}-2 \hat{k}$ We compute the dot produc
- Question 2, Exercise 3.2
- sqrt{{{(3)}^{2}}}=3$$ Now we know that $$\hat{a}=\dfrac{{\overset{\scriptscriptstyle\rightharpoonup}{a}}}... \overset{\scriptscriptstyle\rightharpoonup}{a}|}=\dfrac{3\hat{i}}{3}=\hat{i}$$ This is the required unit... 2}}+{{(-4)}^{2}}}=5$$ Now we know that $$\hat{a}=\dfrac{{\overset{\scriptscriptstyle\rightharpoonup}{a}}}... \overset{\scriptscriptstyle\rightharpoonup}{a}|}=\dfrac{3\hat{i}-4\hat{j}}{5}=\dfrac{3}{5}\hat{i}-\dfrac{
- Question 2 and 3 Exercise 3.3
- vec{a}$ and $\vec{b}$ then \begin{align}\hat{c}&=\dfrac{\vec{a}+\vec{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}+\dfrac{3}{13}\hat{j}-\dfrac{12}{13}
- Question 12 & 13, Exercise 3.3
- gin{align}\therefore \quad \overrightarrow{O D}&=\dfrac{\vec{b}+\vec{c}}{2}, \quad \overrightarrow{O E}=\dfrac{\vec{a}+\vec{c}}{2} \text { and } \\ \overrightarrow{O F}&=\dfrac{\vec{a}+\vec{b}}{2} . \\ &\text { Now } \overrigh... D} \cdot \overrightarrow{B C}&=0 \\ \Rightarrow \dfrac{\vec{b}-\vec{c}}{2} \cdot(\vec{c}-\vec{b})&=0 \qu
- Question 9 & 10, Exercise 3.2
- w{b}+3\overrightarrow{c}.$ \begin{align}\hat{a}&=\dfrac{2\overrightarrow{a}-\overrightarrow{b}+3\overrigh... w{a}\overrightarrow{b}+3\overrightarrow{c}|}\\ &=\dfrac{\hat{i}-2\hat{j}+2\hat{k}}{3}\end{align} Now vec... eorem, we have \begin{align}\overrightarrow{OR}&=\dfrac{1\cdot \overrightarrow{OP}+2\cdot \overrightarrow{OQ}}{1+2}\\ &=\dfrac{1(\hat{i}+2\hat{j}-\hat{k})+2(-\hat{i}+\hat{j}+\h
- Question 7 & 8 Exercise 3.4
- c{a}$ and $\vec{b}$ then\\ \begin{align}\hat{n}&=\dfrac{\vec{a} \times \vec{b}}{\mid \vec{a} \times \vec{... Putting in (1), we have \\ \begin{align}\hat{n}&=\dfrac{\vec{a} \times \vec{b}}{\vec{a} \times \vec{b}}\\ &=\dfrac{-5 \hat{i}+7 \hat{j}-\hat{k}}{\sqrt{75}}.\end{ali... c{a}$ and $\vec{b}$ then\\ \begin{align} \hat{n}=\dfrac{\vec{a} \times \vec{b}}{\vec{a} \times \vec{b}} \
- Question 8 Exercise 3.5
- === The volume of tetrahedron is \begin{align}V&=\dfrac{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{align} =====
- Question 6 & 7 Review Exercise 3
- 3 \lambda - 21&=0 \\ \Rightarrow \quad \lambda&=\dfrac{21}{3}=7 .\end{align} =====Question 7===== Vect... e such that $|\vec{a}|=\sqrt{3}$, and $|\vec{b}|=\dfrac{2}{3}$ and $\vec{a} \times \vec{b}$ is a unit vec... {b}|=1,|\vec{a}|=\sqrt{3} \text { and }|\vec{c}|=\dfrac{2}{3} \text {. }$$ We know that $$|\vec{a} \times... in above, we get \begin{align}1&=\sqrt{3} \cdot \dfrac{2}{3} \sin \theta \\ \Rightarrow \sqrt{3} \cdot \
- Question 1 Review Exercise 3
- $\hat{i}+2 \hat{j}+3 \hat{k}$ is: * (a) $\dfrac{1}{\sqrt{14}}$ * %%(b)%% $\dfrac{2}{\sqrt{14}}$ * %%(c)%% $\dfrac{3}{14}$ * (d) None of these \\ <btn type="link" collapse="a5">... er</btn><collapse id="a5" collapsed="true">(b): $\dfrac{2}{\sqrt{14}}$</collapse> vi. Find nun-zero scal
- Question 5 & 6, Exercise 3.2
- \overrightarrow{AB}$. Then \begin{align}\hat{r}&=\dfrac{\overrightarrow{AB}}{|\overrightarrow{AB}|}\\ &=\dfrac{10\hat{i}+4\hat{j}}{2\sqrt{29}} \\ &=\dfrac{5}{\sqrt{29}}\hat{i}+\dfrac{2}{\sqrt{29}}\hat{j}. \end{align} Hence length of $\overrightarrow{AB}$ is $
- Question 11, Exercise 3.2
- vector $H$ is: \begin{align}\overrightarrow{OH}&=\dfrac{5\overrightarrow{OC}+2\overrightarrow{OD}}{5+2}\\ &=\dfrac{5(5\hat{j})+2(4\hat{i}+\hat{j})}{7}\\ &=\dfrac{1}{7}(8\hat{i}+27\hat{j})\\ \implies \overrightarrow{OH}&=\dfrac{8}{7}\hat{i}+\dfrac{27}{7}\hat{j}\end{align} ===
- Question 2 Exercise 3.4
- \ \Rightarrow|\vec{b}|&=\sqrt{56}\\ \cos \theta&=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}|}=\dfrac{-28}{\sqrt{14} \sqrt{56}}\\ \Rightarrow \quad \theta&=\cos ^{-1... gn} Now we know that\\ \begin{align}\cos \theta&=\dfrac{\vec{a} \cdot \vec{b}}{i|\vec{b}|}=\dfrac{42}{\sqrt{14} \sqrt{126}} \\ \Rightarrow \theta&=\cos^{-1}\lef
- Question 3 Exercise 3.4
- $. then by cross product\\ \begin{align}\hat{n}&=\dfrac{\vec{a} \times \vec{b}}{\mid \vec{a} \times \vec{... } Putting (2) and (3) in (1), we get\\ $$\hat{n}=\dfrac{\vec{a} \times \vec{b}}{\vec{a} \times \vec{b}}=\dfrac{1}{5\sqrt{3}}(\hat{i}-7 \hat{j}+ 5 \hat{k}) .$$ ... by cross product we have\\ \begin{align}\hat{n}&=\dfrac{\vec{a} \times \bar{b}}{\bar{a} \times \bar{b}} \
- Question 5 Exercise 3.4
- that is:\\ \begin{align}\text{Area of triangle}&=\dfrac{1}{2}|\overrightarrow{P Q} \times \overrightarrow... |&=30 \\ \therefore \text { Area of triangle }& =\dfrac{1}{2}|\overrightarrow{PQ} \times \overrightarrow{P R}|\\ &=\dfrac{30}{2}=15 \text { units square. }\end{align} ===... elogram, that is:\\ $$\text { Area of triangle }=\dfrac{1}{2}|\overrightarrow{P Q} \times \overrightarrow