Search
You can find the results of your search below.
Fulltext results:
- Question 1, Exercise 3.2
- i}+3\hat{j}$, then find $\vec{a}+2\vec{b}$. ====Solution==== \begin{align}\vec{a}+2\vec{b}&=3\hat{i}-5\hat... }+3\hat{j}$, then find $3\vec{a}-2\vec{b}$. ====Solution==== \begin{align}3\vec{a}-2\vec{b}&=3(3\hat{i}-5\... +3\hat{j}$, then find $2(\vec{a}-\vec{b})$. ====Solution==== First we have, \begin{align}\vec{a}-\vec{b}&=... }+3\hat{j}$, then find $|\vec{a}+\vec{b}|$. ====Solution==== We have, \begin{align}\vec{a}+\vec{b}&=3\hat{
- Question 2, Exercise 3.2
- the same direction as the vector $3\hat{i}.$ ====Solution==== Let $$\overset{\scriptscriptstyle\rightharpo... direction as the vector $3\hat{i}-4\hat{j}.$ ====Solution==== Let $$\overset{\scriptscriptstyle\rightharpo... ion as the vector $\hat{i}+\hat{j}-2\hat{k}.$ ====Solution==== Let $$\overset{\scriptscriptstyle\rightharpo... rac{\sqrt{3}}{2}\hat{i}-\dfrac{1}{2}\hat{j}.$ ====Solution==== Let $$\overset{\scriptscriptstyle\rightharpo
- Question 7, Exercise 3.2
- rrightarrow{PQ}$, if $P(-1,2)$, $Q(2,-1)$. ====Solution==== As we have \begin{align}\overrightarrow{PQ}&... \overrightarrow{PQ}.$ If $P(-2,1),Q(2,3).$ ====Solution==== Do yourself. =====Question 7(iii)===== Find ... htarrow{PQ}$. if $P(-1,1,2)$, $Q(2,-1,3)$. ====Solution==== As we have \begin{align}\overrightarrow{PQ}&=... ghtarrow{PQ}$. if $P(2,4,6)$, $Q(1,-2,3)$. ====Solution==== Do yourself. ====Go To==== <text align="lef
- Question 7, Exercise 3.2
- rrightarrow{PQ}$, if $P(-1,2)$, $Q(2,-1)$. ====Solution==== As we have \begin{align}\overrightarrow{PQ}&... \overrightarrow{PQ}.$ If $P(-2,1),Q(2,3).$ ====Solution==== Do yourself. =====Question 7(iii)===== Find ... htarrow{PQ}$. if $P(-1,1,2)$, $Q(2,-1,3)$. ====Solution==== As we have \begin{align}\overrightarrow{PQ}&=... ghtarrow{PQ}$. if $P(2,4,6)$, $Q(1,-2,3)$. ====Solution==== Do yourself. ====Go To==== <text align="lef
- Question 1, Exercise 3.3
- }-5 \hat{k}$ then find $\vec{a}\cdot \vec{b}$ ====Solution==== \begin{align}\vec{a} \cdot \vec{b}&=(3 \hat{i... 5 \hat{k}$ then find $\vec{a} \cdot \vec{c}$. ====Solution==== \begin{align}\vec{a} \cdot \vec{c}&=(3 \hat{i... }$ then find $\vec{a} \cdot(\vec{b}+\vec{c})$ ====Solution==== \begin{align}\vec{b}+\vec{c}&=(\hat{i}-\hat{j... k}$ then find $(\vec{a}-\vec{b})\cdot\vec{c}$ ====Solution==== \begin{align}\vec{a}-\vec{b}&=3 \hat{i}+4 \ha
- Question 2 and 3 Exercise 3.3
- \quad \vec{b}=2 \hat{i}+\hat{j}-7 \hat{k}$$. ====Solution==== We first find the sum \begin{align}\vec{a}+\v... {j}+\hat{k}, \quad-\hat{i}+\hat{j}+2 \hat{k}$ ====Solution==== Let $\vec{a}=\hat{i}-\hat{j}+\hat{k}$ and $\v... \hat{i}+4 \hat{j}, \quad 2 \hat{j}-5 \hat{k}$ ====Solution==== Let $\vec{a}=3 \hat{i}+4 \hat{j}$ and $\vec{b... {i}-3 \hat{k}, \quad \hat{i}+\hat{j}+\hat{k}$ ====Solution==== Let $\vec{a}=2 \hat{i}-3 \hat{k}$ and $\vec{b
- Question 12, 13 & 14, Exercise 3.2
- pha \hat{i}+(\alpha +1)\hat{j}+2\hat{k}|=3$. ====Solution==== We are given \begin{align}|\alpha \hat{i}+(\a... e sides of a triangle. Find the value of $z.$ ====Solution==== {{ :fsc-part1-kpk:sol:unit03:math-11-kpk-3-2-... ow{AB}$ is parallel to $\overrightarrow{CD}.$ ====Solution==== Position vector of $A$ is $\overrightarrow{OA
- Question 7 & 8 Exercise 3.3
- {a}$ on $\vec{b}$ and $\vec{b}$ on $\vec{a}$. ====Solution==== $\vec{a}=-\dfrac{3}{2} \hat{j}+\dfrac{4}{5} \... {a}$ on $\vec{b}$ and $\vec{b}$ on $\vec{a}$. ====Solution==== We compute the dot product\\ \begin{align}\ve... \hat{i}+\hat{j}+\hat{k}$ makes with $y$ axis. ====Solution==== Let $\vec{a}=\sqrt{2} \hat{i}+\hat{j}+\hat{k}
- Question 1 Exercise 3.4
- product $\hat{j} \times(2 \hat{j}+3 \hat{k})$ ====Solution==== Let \begin{align}\vec{a}=\hat{j}&=0 \hat{i}+\... roduct $(2 \hat{i}-3 \hat{j}) \times \hat{k}$ ====Solution==== Let \begin{align}\vec{a}&=2 \hat{i}-3 \hat{j}... {k}) \times(6 \hat{i}+2 \hat{j}-3 \hat{k})$\\ ====Solution==== Let \begin{align}\vec{a}&=2 \hat{i}-3 \hat{j}
- Question 4 Exercise 3.4
- \hat{k},\quad$ find $\vec{a} \times \vec{b}$ ====Solution==== \begin{align}\vec{a} \times \vec{b}&=\left|\b... \hat{k},\quad$ find $\vec{b} \times \vec{c}$ ====Solution==== \begin{align}\vec{b} \times \vec{c}&=\left| \... d $(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})$ ====Solution==== \begin{align}\vec{a}+\vec{b}&=(3 \vec{i}-6 \h
- Question 7 & 8 Exercise 3.4
- c{B} \times \vec{C}=\vec{C} \times \vec{A}.$$ ====Solution==== We are given\\ $$\vec{A}+\vec{B}+\vec{C}=\vec... and $\vec{b}=-2 \hat{i}+\hat{j}-3 \hat{k}$\\ ====Solution==== Let $\hat{n}$ be unit vector perpendicular to... }=4 \hat{i}-2 \hat{j}-4 \hat{k} \text {. }$$ =====Solution===== Let $\hat{n}$ be unil vector perpendicular t
- Question 7 Exercise 3.5
- k} \cdot \vec{w}=3 \hat{i}+\hat{j}+c \hat{k}$ ====Solution==== The given vectors are coplanar, therefore \be... hat{k}, \vec{w}=c \hat{i}+\hat{j}-c \hat{k}$. ====Solution==== The given vectors are coplanar, therefore \be... {k}$. $\vec{n}=c \hat{i}+2 \hat{j}+6 \hat{k}$ ====Solution==== Since the given vectors are coplanar, therefo
- Question 3 & 4, Exercise 3.2
- d $q$ such that $\vec{r}=p\vec{a}+q\vec{b}$. ====Solution==== Given $$\vec{r}=p\vec{a}+q\vec{b}.$$ We put t... value of $x$ such that $|\vec{p}+\vec{q}|=5.$ ====Solution==== We calculate \begin{align}\vec{p}+\vec{q}&=2
- Question 5 & 6, Exercise 3.2
- r in the direction of $\overrightarrow{AB}$. ====Solution==== The position vector of $\vec{A}$ and $\vec{B}... $(0,5).$ find coordinates of the vertex $D.$ ====Solution==== Position vectors of given points are $$\overr
- Question 9 & 10, Exercise 3.2
- }-\overrightarrow{b}+3\overrightarrow{c}.$ ====Solution==== We compute that\\ \begin{align}2\overrightarr... the ratio $2:1$ internally and externally. ====Solution==== We find the position vector of a point $R$ wh