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- Question 3 & 4, Exercise 3.2
- Solution==== Given $$\vec{r}=p\vec{a}+q\vec{b}.$$ We put the values of $\vec{r},\vec{a}$ and $\vec{b}$ in the given equation. We get $$\hat{i}-9\hat{j}=p(\hat{i}+2\hat{j})+q(5\ha... paring the coeffients of $\hat{i}$ and $\hat{j}$, we have, $$p+5q=1…(i)$$ $$2p-q=-9 …(ii)$$ Multiply $2$ by (i) and subtract (ii) from (i). We have \[\begin{array}{ccc} 2p&+10q&=2 \\ \mathop
- Question 9 Exercise 3.4
- -2 \hat{i}+3 \hat{j}+4 \hat{k}$. ====Solution==== We are give the diagonal as shown in figure, instead... t{j}+2 \hat{k}\end{align} From $\triangle A E B$, we have\\ \begin{align}\vec{c}&=\overrightarrow{A E}... dots \ldots(1)\end{align} From $\triangle A E D$. we have\\ \begin{align}\vec{d}&=\overrightarrow{A E}... ext {. }....(2) \end{align} By using (1) and (2), we have,\\ \begin{align}\vec{c} \times \vec{d}&=\lef
- Question 12 & 13, Exercise 3.3
- in a semicircle is right angle. ====Solution==== We are considering a triangle inside 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}... ut opposile in direction. From $\triangle A B O$, we have \begin{align}\overrightarrow{O B}+\overright
- Question 1, Exercise 3.2
- nd $2(\vec{a}-\vec{b})$. ====Solution==== First we have, \begin{align}\vec{a}-\vec{b}&=3\hat{i}-5\ha... }-8\hat{j}\end{align} Multiply both sides by $2$. We have, $$2(\vec{a}-\vec{b})=10\hat{i}-16\hat{j}$$... then find $|\vec{a}+\vec{b}|$. ====Solution==== We have, \begin{align}\vec{a}+\vec{b}&=3\hat{i}-5\ha... 2\hat{j}\end{align} Taking modulus of both sides. We have, $$|\vec{a}+\vec{b}|=\sqrt{(1)^2+(-2)^2}=\sq
- Question 7 & 8 Exercise 3.4
- vec{C}=\vec{C} \times \vec{A}.$$ ====Solution==== We are given\\ $$\vec{A}+\vec{B}+\vec{C}=\vec{O} \te... s product of $\vec{A}$, of both sides with above. we get\\ $$\vec{A} \times(\vec{A}+\vec{B}+\vec{C})=0... }$$\\ Taking cross product of $\vec{B}$ with (1), we get\\ \begin{align}\vec{B} \times(\vec{A}+\vec{B}... c{C}=\vec{A} \times \vec{B}$$\\ From (2) and (3), we get\\ $$\vec{A} \times \vec{B}=\vec{B} \times \ve
- Question 9 & 10, Exercise 3.2
- rrow{b}+3\overrightarrow{c}.$ ====Solution==== We compute that\\ \begin{align}2\overrightarrow{a}-\... 1$ internally and externally. ====Solution==== We find the position vector of a point $R$ which div... ng the points internally. By using ratio theorem, we have \begin{align}\overrightarrow{OR}&=\dfrac{1\c... c{4}{3}\hat{j}+\dfrac{1}{3}\hat{k}\end{align} Now we find the position vector of a point $\vec{R}$, wh
- Question 12, 13 & 14, Exercise 3.2
- \alpha +1)\hat{j}+2\hat{k}|=3$. ====Solution==== We are given \begin{align}|\alpha \hat{i}+(\alpha +1... 2)^2}&=3.\end{align} Taking square on both sides, we have, \begin{align}&{\alpha ^2+(\alpha +1)^2}+4=9... on 13}} By head to tail rule of vectors addition, we have\\ \begin{align}\vec{u}+\vec{v}&=\vec{w}\\ (... } By comparison $\hat{i},\hat{j}$ and $\hat{k}.$ we have,\\ $$3=-z$$ $$-z=3$$ $$\Rightarrow \,\,\,z=-
- Question 2, Exercise 3.2
- le\rightharpoonup}{a}|=\sqrt{{{(3)}^{2}}}=3$$ Now we know that $$\hat{a}=\dfrac{{\overset{\scriptscrip... onup}{a}|=\sqrt{{{(3)}^{2}}+{{(-4)}^{2}}}=5$$ Now we know that $$\hat{a}=\dfrac{{\overset{\scriptscrip... }{a}|=\sqrt{(1)^2+(1)^2+(-2)^2}=\sqrt{6}$$ \\ Now we know that\\ $$\hat{a}=\dfrac{{\overset{\scriptscr... {3}}{2})}^{2}}+{{(\dfrac{1}{2})}^{2}}}=1$$ \\ Now we know that\\ $$\hat{a}=\dfrac{{\overset{\scriptscr
- Question 5(iii) & 5(iv) Exercise 3.5
- {b})-(\vec{a} \cdot \vec{b})^2$. ====Solution==== We have already calculated L.H.S in (ii) that\\ \be... 1^2, a_2^2 b_2^2$ and $ a_3^2 b_3^2$ in the above we get\\ \begin{align}| \vec{a} \times \vec{b}|^2&=a... _2 b_3+2 a_1 a_3 b_1 b_3)....(1)\end{align}\\ Now we know that\\ \begin{align}\vec{a} . \vec{a}&=|\vec... b_1 b_3\end{align}\\ Putting (3) and (2) in (1), we get the desired that is: $$| \vec{a} \times \vec{
- Question 5 & 6, Exercise 3.2
- $$\overrightarrow{OB}=7\hat{i}+9\hat{j}.$$ Thus we have \begin{align}\overrightarrow{AB}&=\overright... \hat{i}+y\hat{j}$. Since $ABCD$ is parallelogram, we have that $$\overrightarrow{AB}=\overrightarrow{D... =-x\hat{i}+(5-y)\hat{j}\end{align} By comparison, we have $$3=-x \text{ and } 7=5-y$$ This gives $$x=-
- Question 7 & 8 Exercise 3.3
- at{k}\quad$ $\vec{b}=\hat{i}-2 \hat{j}-2 \hat{k}$ We compute the dot product \begin{align}\vec{a} \cdo... {b}$ and $\vec{b}$ on $\vec{a}$. ====Solution==== We compute the dot product\\ \begin{align}\vec{a} \c... =\sqrt{(1)^2}=1.\end{align} Now from dot product, we have\\ $$\cos \theta=\dfrac{\vec{a} \cdot \vec{b}
- Question 11, Exercise 3.3
- angle and they form right angle triangle. Also if we see \begin{align}\vec{a} \cdot \vec{c}&=-(3 \hat{... orms a right isosceles triangle. ====Solution==== We find vectors representing the sides of triangle f... {P R}|&=\sqrt{(1)^2+(-1)^2}=\sqrt{2} .\end{align} We observe that\\ \begin{align}|\overrightarrow{P Q}
- Question 3 Exercise 3.4
- =5\sqrt{3}\end{align} Putting (2) and (3) in (1), we get\\ $$\hat{n}=\dfrac{\vec{a} \times \vec{b}}{\v... th $\vec{a}$ and $\vec{b}$. then by cross product we have\\ \begin{align}\hat{n}&=\dfrac{\vec{a} \time... \ldots(3)\end{align} Pulting (2) and (3) in (1), we get\\ \begin{align}\hat{n}&=\dfrac{\vec{a} \times
- Question 3 & 4 Exercise 3.5
- ots \ldots (3) \end{align} From (1), (2) and (3), we get that \begin{align}\vec{a}\cdot \vec{b} \time... {k}\end{align} Taking dut product with $\vec{a}$. we have \begin{align}\vec{c} \times \vec{b} \cdot \... ec{b} \cdot \vec{a}=25....(4)$$ From (1) and (4), we get that\\ $$\vec{a} \cdot \vec{b} \times \vec{c}
- Question 6 & 7 Review Exercise 3
- == Let $\theta$ be the angle between two vectors. We are given\\ $$|\vec{a} \times \vec{b}|=1,|\vec{a}... \text { and }|\vec{c}|=\dfrac{2}{3} \text {. }$$ We know that $$|\vec{a} \times \vec{b}|=|\vec{a}||\v... ec{b}| \sin \theta$$. Putting the given in above, we get \begin{align}1&=\sqrt{3} \cdot \dfrac{2}{3} \