Search
You can find the results of your search below.
Fulltext results:
- Question 9 Exercise 3.4
- }+2 \hat{k}\end{align} From $\triangle A E B$, we have\\ \begin{align}\vec{c}&=\overrightarrow{A E}+\ove... s \ldots(1)\end{align} From $\triangle A E D$. we have\\ \begin{align}\vec{d}&=\overrightarrow{A E}+\ove... {. }....(2) \end{align} By using (1) and (2), we have,\\ \begin{align}\vec{c} \times \vec{d}&=\left| \b... at{k}\end{align} Which from $\triangle A E B$, we have\\ \begin{align}\vec{c} & =\overrightarrow{A E}+\v
- Question 1, Exercise 3.2
- $2(\vec{a}-\vec{b})$. ====Solution==== First we have, \begin{align}\vec{a}-\vec{b}&=3\hat{i}-5\hat{j}-... \hat{j}\end{align} Multiply both sides by $2$. We have, $$2(\vec{a}-\vec{b})=10\hat{i}-16\hat{j}$$ ===... n find $|\vec{a}+\vec{b}|$. ====Solution==== We have, \begin{align}\vec{a}+\vec{b}&=3\hat{i}-5\hat{j}+... at{j}\end{align} Taking modulus of both sides. We have, $$|\vec{a}+\vec{b}|=\sqrt{(1)^2+(-2)^2}=\sqrt{5}
- Question 3 & 4, Exercise 3.2
- ing the coeffients of $\hat{i}$ and $\hat{j}$, we have, $$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+\lim... \implies q=1$$ \\ Put the value of $q$ in (i). We have, $$p+5(1)=1 \quad \implies p=-4$$ Hence we have $p=-4$ and $q=1$. =====Question 4===== If $\vec{p}=2\h
- Question 12, 13 & 14, Exercise 3.2
- 2}&=3.\end{align} Taking square on both sides, we have, \begin{align}&{\alpha ^2+(\alpha +1)^2}+4=9\\ \i... 13}} By head to tail rule of vectors addition, we have\\ \begin{align}\vec{u}+\vec{v}&=\vec{w}\\ (2\hat... By comparison $\hat{i},\hat{j}$ and $\hat{k}.$ we have,\\ $$3=-z$$ $$-z=3$$ $$\Rightarrow \,\,\,z=-3$$ ... ghtarrow{OD}=-\hat{i}-2\hat{j}+\hat{k}$. Now, we have \begin{align}\overrightarrow{AB}&=\overrightarrow
- Question 12 & 13, Exercise 3.3
- ering a triangle inside a semicircle as shown. We have to show $\overrightarrow{B A} \cdot \overrightarr... opposile in direction. From $\triangle A B O$, we have \begin{align}\overrightarrow{O B}+\overrightarrow... ...(1)\end{align} Also from $\triangle A C O$, we have \begin{align}\overrightarrow{O A}+\overrightarrow... $ intersect at point $O$ as shown in figure.\\ We have to show that side bisector of $A B$ also passes t
- Question 5 & 6, Exercise 3.2
- $\overrightarrow{OB}=7\hat{i}+9\hat{j}.$$ Thus we have \begin{align}\overrightarrow{AB}&=\overrightarrow... t{i}+y\hat{j}$. Since $ABCD$ is parallelogram, we have that $$\overrightarrow{AB}=\overrightarrow{DC}.$$... \hat{i}+(5-y)\hat{j}\end{align} By comparison, we have $$3=-x \text{ and } 7=5-y$$ This gives $$x=-3 \te
- Question 7, Exercise 3.2
- f $P(-1,2)$, $Q(2,-1)$. ====Solution==== As we have \begin{align}\overrightarrow{PQ}&=\overrightarro... (-1,1,2)$, $Q(2,-1,3)$. ====Solution==== As we have \begin{align}\overrightarrow{PQ}&=\overrightarrow
- Question 7, Exercise 3.2
- f $P(-1,2)$, $Q(2,-1)$. ====Solution==== As we have \begin{align}\overrightarrow{PQ}&=\overrightarro... (-1,1,2)$, $Q(2,-1,3)$. ====Solution==== As we have \begin{align}\overrightarrow{PQ}&=\overrightarrow
- Question 9 & 10, Exercise 3.2
- the points internally. By using ratio theorem, we have \begin{align}\overrightarrow{OR}&=\dfrac{1\cdot \... the points externally. By using ratio theorem, we have \begin{align}\overrightarrow{OR}&=\dfrac{1\cdot \
- Question 7 & 8 Exercise 3.4
- ec{b}|&=\sqrt{75} .\end{align} Putting in (1), we have \\ \begin{align}\hat{n}&=\dfrac{\vec{a} \times \v... {b}|&=2 \sqrt{65} .\end{align} Putting in (1), we have\\ \begin{align}\hat{n}&=\dfrac{\vec{a} \times \ve
- Question 4 & 5 Review Exercise 3
- (\bar{r} \times \hat{j})+x y$ ====Solution==== We have to find\\ $$(\vec{r} \times \hat{i}) \cdot(\vec{r... on of $\vec{a}$ on $\vec{b}$. ====Solution==== We have to compute Projection of $\vec{a}$ on $\vec{b}=\
- Question 11, Exercise 3.2
- ratio $2:5$internally, then by ratio theorem, we have position vector $H$ is: \begin{align}\overrightar
- Question 4 and 5 Exercise 3.3
- d{array}\right| \end{align} expanding by $R_1$ we have \begin{align} & \vec{c}=(-2-1) \hat{i}+(-1-2) \ha
- Question 7 & 8 Exercise 3.3
- qrt{(1)^2}=1.\end{align} Now from dot product, we have\\ $$\cos \theta=\dfrac{\vec{a} \cdot \vec{b}}{|\v
- Question 3 Exercise 3.4
- $\vec{a}$ and $\vec{b}$. then by cross product we have\\ \begin{align}\hat{n}&=\dfrac{\vec{a} \times \ba