Search
You can find the results of your search below.
Fulltext results:
- Question 2 and 3 Exercise 3.3
- t{j}-5 \hat{k})+(2 \hat{i}+\hat{j}-7 \hat{k}) \\ \Rightarrow &=4 \hat{i}+3 \hat{j}-12 \hat{k}\\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{(4)^2+(3)^2+(-12)^2} \\ \Rightarrow &=\sqrt{16+9+144} \\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{169}=13\end{align} Now let say $\hat{c}$ be the unit
- Question 2 Exercise 3.4
- =(12-12) \hat{i}-(-6+6) \hat{j}+(4-4) \hat{k} \\ \Rightarrow \vec{a} \times \vec{b}&=0 . \\ & \Rightarrow \vec{a} \| \vec{b} .\end{align} Second Way \begin{align}\vec{a}... \hat{k}) \cdot(2 \hat{i}-4 \hat{j}+6 \hat{k}) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-1(2)+2(-4)-3(6) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-28 .\end{align} Als
- Question 7 & 8 Exercise 3.4
- imes(\vec{A}+\vec{B}+\vec{C})=0$$\\ \begin{align}\Rightarrow \vec{A} \times \vec{A}+\vec{A} \times \vec{B}+\vec{A} \times \vec{C}&=\vec{O}...(1) \\ \Rightarrow \vec{A} \times \vec{B}+\vec{A} \times \vec{C} &= \vec{O} \quad \because \vec{A} \| \vec{A} \\ \Rightarrow \vec{A} \times \vec{B}&=-\vec{A} \times \vec{C} \\ \Rightarrow \vec{A} \times \vec{B}&=\vec{C} \times \vec{A}...
- Question 12 & 13, Exercise 3.3
- B}+\overrightarrow{A B}&=\overrightarrow{O A}\\ \Rightarrow \overrightarrow{B A}&=\overrightarrow{O A}-\overr... A}+\overrightarrow{A C}&=\overrightarrow{O C}\\ \Rightarrow \overrightarrow{A C}&=\overrightarrow{O C}-\overr... \vec{a}=|\vec{a}|^2 and\quad \vec{b}=-\vec{c}\\ \Rightarrow \overrightarrow{B A} \cdot \overrightarrow{A C}&=... rightarrow{O D} \cdot \overrightarrow{B C}&=0 \\ \Rightarrow \dfrac{\vec{b}-\vec{c}}{2} \cdot(\vec{c}-\vec{b})
- Question 6 Exercise 3.3
- in{align}(\vec{a}+m \vec{b}) \cdot \vec{a}&=0 \\ \Rightarrow \quad[(1+2 m) \hat{j}+(3-3 m) \hat{j}+(5 m- 4) \h... at{k}] \cdot [\hat{i}+3 \hat{j}-4 \hat{k}]&=0 \\ \Rightarrow \quad 1(1+2 m)+3(3-3 m)-4(5 m-4)&=0 \\ \Rightarrow \quad 2 m-9 m-20 m+1+9+16&=0 \\ \Rightarrow \quad-27 m&=-26 \\ \Rightarrow \quad m&=\dfrac{26}{27}\end{align}
- Question 7 & 8 Exercise 3.3
- {k}\right) \cdot(\hat{i}-2 \hat{j}-2 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{b}&=0(1)+\left(-\dfrac{3}{2}\right)(-2)+\dfrac{4}{5}(-2) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=3-\dfrac{8}{5}=\dfra... c{3}{2}\right)^2+\left(\dfrac{4}{5}\right)^2} \\ \Rightarrow|\vec{a}|&=\sqrt{\dfrac{9}{4}+\dfrac{16}{25}} \\ \Rightarrow|\vec{a}|&=\sqrt{\dfrac{225+64}{100}}=\dfrac{\sqrt
- Question 7 Exercise 3.5
- }&=0\\ \vec{u} \cdot \vec{v} \times \vec{w}&=0\\ \Rightarrow\left|\begin{array}{ccc}1 & 2 & 3 \\ 2 & -3 & 4 \\... ray}\right|&=0\\ 1(-3 c-4)-2(2 c-12)+3(2+9)&=0\\ \Rightarrow-3 c-4-4 c+24+33&=0\\ \Rightarrow \quad-7 c+53&=0\\ \Rightarrow c&=\dfrac{53}{7}.\end{align} which is required value of $c$ for which the given
- Question 5 Exercise 3.4
- Since } \overrightarrow{P Q}&=(3,2)-(-2 ,-3) \\ \Rightarrow \overrightarrow{P Q}&=(5,5) \\ \overrightarrow{P R}&=(-1,-8)-(-2 ,-3) \\ \Rightarrow \overrightarrow{P R}&=(1,-5) \\ \overrightarrow{P... } \\ 5 & 5 & 0\\ 1 & -5 & 0 \end{array}\right|\\ \Rightarrow \overrightarrow{P Q} \times \overrightarrow{P R}&=(-25-5) \hat{k}=-30 \hat{k} \\ \Rightarrow|\overrightarrow{P Q} \times \overrightarrow{P R}|
- Question 3 Exercise 3.4
- \ 1 & -2 & 3 \\ 2 & 1 & -1 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{b}&=(2+3) \hat{i}-(-1-6) \hat{j}+(1-4) \hat{h} \\ \rightarrow \vec{a} \times \vec{b}&=-\hat{i}+7 \hat{j}+5 \hat{k} \\ \Rightarrow | \vec{a} \times \vec{b}|&=\sqrt{(-1)^2+(7)^2+(5)^2} \\ \Rightarrow \quad|\vec{a} \times \vec{b}|&=\sqrt{75}=5\sqrt{3
- Question 8 & 9 Review Exercise 3
- $\vec{a}=\overrightarrow{A B}=(-1,3,2)-(0,0,2)$ $\Rightarrow \vec{a}=(-1,3,0)$ $\vec{b}=\overrightarrow{B C}=(1,0,4)-(-1,3,2)$ $\Rightarrow \vec{b}=(2,-3,2)$. We know that area of triangle... \vec{b} &=(\hat{i}+2 \hat{j} \cdot 3 \hat{k} \\ \Rightarrow | \vec{a} \times \vec{b} |&=\sqrt{(6)^2+(2)^2+(-3)^2} \\ \Rightarrow | \vec{a} \times \vec{b}|&=\sqrt{49}= 7 .\end{ali
- Question 1, Exercise 3.3
- t{j}-\hat{k}) \cdot(\hat{i}-\hat{j}+3 \hat{k})\\ \Rightarrow &=(3 \times 1)+(4 \times-1)+(-1 \times 3)\\ & =3-... j}-\hat{k}) \cdot(2 \hat{i}+\hat{j}-5 \hat{k})\\ \Rightarrow \vec{a} \cdot \vec{c}&=(3 \times 2)+(4 \times 1)+... }-\hat{j}+3 \hat{k})+ (2 i+\hat{j}-5 \hat{k}) \\ \Rightarrow \vec{b}+\vec{c}&=3 \hat{i}-2 \hat{k}\end{align} T... 4 \hat{j}-\hat{k}) \cdot(3 \hat{i}-2 \hat{k}) \\ \Rightarrow \vec{a} \cdot(\vec{b}+\vec{c})&=3.3+4.0+-1 .-2 \\
- Question 9 Exercise 3.4
- c}&=\overrightarrow{A E}+\overrightarrow{E B} \\ \Rightarrow \vec{c}&=2 \hat{i}+\dfrac{1}{2} \hat{j}-\hat{k}-(... se \overrightarrow{B E}=-\overrightarrow{E B} \\ \Rightarrow \vec{c}&=3 \hat{i}-\hat{j}-3 \hat{k} \ldots \ldot... {d}&=\overrightarrow{A E}+\overrightarrow{E D}\\ \Rightarrow \bar{d}&=2 \hat{i}+\dfrac{1}{2} \hat{j}-\hat{k}+(-\hat{i}+\dfrac{3}{2} \hat{j}+2 \hat{k}) \\ \Rightarrow \vec{d}&=\hat{i}+2 \hat{j}+ \hat{k} \text {. }...
- Question 6 & 7 Review Exercise 3
- align}\vec{a} \cdot \vec{b} \times \vec{c}&=0 \\ \Rightarrow\left|\begin{array}{ccc} 1 & 3 & 1 \\ 2 & -1 & -1 \\ 0 & \lambda & 3 \end{array}\right|&=0 \\ \Rightarrow \quad 1(-3+\lambda)-3(6+0)+1(2 \lambda-0)&=0\\ \Rightarrow \quad-3+\lambda-18+2 \lambda&=0 \\ \Rightarrow \quad 3 \lambda - 21&=0 \\ \Rightarrow \quad \lambda&=\dfrac{2
- Question 4 and 5 Exercise 3.3
- 3 \hat{k}) \cdot(\hat{i}-\hat{j}+2 \hat{k}) . \\ \Rightarrow \vec{a} \cdot \vec{b}&=1(1)+7(-1)+3(2) \\ \Rightarrow \vec{a} \cdot \vec{b}&=1-7+6=0 \\ \Rightarrow \vec{a} \perp \vec{b} \cdot \text { Now } \\ \vec{a} \cdot \vec... 3 \hat{k}) \cdot(2 \hat{i}+\hat{j}-3 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{c}&=1(2)+7(1)+3(-3) \\ \Righta
- Question 6 Exercise 3.5
- A B}=\overrightarrow{O B}-\overrightarrow{O A}\\ \Rightarrow \vec{a}&=(5 \hat{i}+\hat{j}+6 \hat{k})-(4 \hat{i}-2 \hat{j}+\hat{k}) \\ \Rightarrow \vec{a}&=\hat{i}+3 \hat{j}+5 \hat{k} \ldots \ldot... C}=\overrightarrow{O C}-\overrightarrow{O A} \\ \Rightarrow \vec{b}&=2 \hat{i}+2 \hat{j} \quad 5 \hat{k}-(4 \hat{i}-2 \hat{j}-\hat{k}) \\ \Rightarrow \vec{b}&=-2 \hat{i}+4 \hat{j}-6 \hat{k} ....(2)\\