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- Question 7 Exercise 3.5
- 3 \hat{i}+\hat{j}+c \hat{k}$ ====Solution==== The given vectors are coplanar, therefore \begin{align}\vec... ign} which is required value of $c$ for which the given vectors become coplanar. =====Question 7(ii)====... \hat{i}+\hat{j}-c \hat{k}$. ====Solution==== The given vectors are coplanar, therefore \begin{align}\vec... which is the required value of $c$ for which the given vectors become coplanar. =====Question 7(iii)===
- Question 3 & 4, Exercise 3.2
- hat $\vec{r}=p\vec{a}+q\vec{b}$. ====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 $$\hat{i}-9\hat{j}=p(\hat{i}+2\h... 4x+4+4}\\ &=\sqrt{x^2+4x+8}\end{align} But we are given that \begin{align}&|\vec{p}+\vec{q}|=5 \\ \impli
- Question 7 & 8 Exercise 3.3
- TBB) Peshawar, Pakistan. =====Question 7(i)===== Given the vectors $\vec{a}$ and $\vec{b}$ as $\vec{a}=-... \vec{a}=\dfrac{14}{17}$ =====Question 7(ii)===== Given the vectors $\vec{a}$ and $\vec{b}$ as $\vec{a}=-... ng $y-a x i s$.\\ The cosine of angie between the given vector and $y-a x i s$ is now actually cosine of
- Question 6 Exercise 3.5
- ie in a plane? ====Solution==== Let we denote the given points with $A(4,-2,1), B(5,1,6)$. $C(2,2,-5)$ an... {c}$ are the vectors formed by the combination of given points such that in which all the points are used... } \cdot \vec{b} \times \vec{c}$ is zero. Thus the given points lie in same plane. ====Go To==== <text
- Question 6 & 7 Review Exercise 3
- \hat{k}$ are coplanar. ====Solution==== Since the given vectors are coplanar, therefore, \begin{align}\v... $\theta$ be the angle between two vectors. We are given\\ $$|\vec{a} \times \vec{b}|=1,|\vec{a}|=\sqrt{3}... b}|=|\vec{a}||\vec{b}| \sin \theta$$. Putting the given in above, we get \begin{align}1&=\sqrt{3} \cdot \
- Question 5 & 6, Exercise 3.2
- nates of the vertices $A, B$ and $C$ respectively given by $(-2,-3),(1,4)$ and $(0,5).$ find coordinates ... vertex $D.$ ====Solution==== Position vectors of given points are $$\overrightarrow{OA}=-2\hat{i}-3\hat{
- Question 3 Exercise 3.4
- ==== Find a unit vector that is orthogonal to the given vector $\vec{a}=\hat{i}- 2 \hat{j}+3 \hat{k}, \qu... ==== Find a unit vector that is orthogonal to the given vector $\vec{a}=3 \hat{i}-\hat{j}+6 \hat{k}, \qua
- Question 5 Exercise 3.4
- duct to compute the area of the triangle with the given vertices $P(-2,-3), \quad Q(3,2)\quad$ and $\quad... duct to compute the area of the triangle with the given vertices $P(-2,-1,3), Q(1,2,-1)$ and $R(4.3,-3)$
- Question 9 & 10, Exercise 3.2
- 4\hat{k}\end{align} This is a desired vector with given conditions. =====Question 10===== Find the posit
- Question 12, 13 & 14, Exercise 3.2
- +1)\hat{j}+2\hat{k}|=3$. ====Solution==== We are given \begin{align}|\alpha \hat{i}+(\alpha +1)\hat{j}+2
- Question 11, Exercise 3.3
- esenting the sides of triangle from the ver-tices given. \begin{align}\overrightarrow{P Q}&=\overrightarr
- Question 7 & 8 Exercise 3.4
- \vec{C} \times \vec{A}.$$ ====Solution==== We are given\\ $$\vec{A}+\vec{B}+\vec{C}=\vec{O} \text {. }$$\
- Question 9 Exercise 3.4
- t{i}-3 \hat{j}+4 \hat{k}$ ====Solution==== We are given the diagonals as shown in figure instead of Iwo :
- Question 3 & 4 Exercise 3.5
- end{align} Hence the scalar triple product of the given vectors is zero. ====Go To==== <text align="l
- Question 2 & 3 Review Exercise 3
- htarrow{0} \text {. }$$\\ ====Solution==== We are given\\ \begin{align}(\hat{i}+3 \hat{j}+9 \hat{k}) \tim