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- Question 9 Exercise 3.4
- is the midpoint of both diagonals. Thus\\ \begin{align}\overrightarrow{A E}&=\overrightarrow{E C}\\ &=\d... }\\ &=-\hat{i}+\dfrac{3}{2} \hat{j}+2 \hat{k}\end{align} From $\triangle A E B$, we have\\ \begin{align}\vec{c}&=\overrightarrow{A E}+\overrightarrow{E B} \\ \R... t{j}-3 \hat{k} \ldots \ldots \ldots \ldots(1)\end{align} From $\triangle A E D$. we have\\ \begin{align}\
- Question 2 and 3 Exercise 3.3
- $$. ====Solution==== We first find the sum \begin{align}\vec{a}+\vec{b}&=(2 \hat{i}+2 \hat{j}-5 \hat{k})+... \ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{169}=13\end{align} Now let say $\hat{c}$ be the unit vector $x$ the sum of $\vec{a}$ and $\vec{b}$ then \begin{align}\hat{c}&=\dfrac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}... &=\dfrac{1}{13}(4\hat{i}+3\hat{j}-12\hat{k})\end{align} =====Question 3(i)===== Find the angles between
- Question 2 Exercise 3.4
- }+$ $6 \hat{k}$ ====Solution==== First Way \begin{align}\vec{a} \times \vec{b}&=\left|\begin{array}{ccc} ... b}&=0 . \\ & \Rightarrow \vec{a} \| \vec{b} .\end{align} Second Way \begin{align}\vec{a} \cdot \vec{b}&=(-\hat{i}+2 \hat{j}-3 \hat{k}) \cdot(2 \hat{i}-4 \hat{j... Rightarrow \quad \vec{a} \cdot \vec{b}&=-28 .\end{align} Also \begin{align}|\vec{a}|&=\sqrt{(-1)^2+(2)^2+
- Question 7 & 8 Exercise 3.4
- c{A} \times(\vec{A}+\vec{B}+\vec{C})=0$$\\ \begin{align}\Rightarrow \vec{A} \times \vec{A}+\vec{A} \times... \times \vec{B}&=\vec{C} \times \vec{A}...(2)\end{align} $\because \quad$ cross product is anti-commutati... ss product of $\vec{B}$ with (1), we get\\ \begin{align}\vec{B} \times(\vec{A}+\vec{B}+\vec{C})&=0 \\ \Ri... \times \vec{C}&=\vec{A} \times \vec{B}....(3)\end{align} $\because$ crass product is anti-commuative\\ $$
- Question 5(iii) & 5(iv) Exercise 3.5
- ^2,\quad|a|^2,\quad|b|^2$ ====Solution==== \begin{align}\vec{a} \cdot \vec{b}&=(a_1 \hat{i}+a_2 \hat{j} +... c{a} \cdot \vec{b}&=a_1 b_1+ a_2 b_2+a_3 b_3 \end{align} Taking square of the both sides \begin{align}(\vec{a} \cdot \vec{b})^2&=(a_1 b_1 + a_2 b_2+a_3 b_3)^2 .... \\ |\vec{a}|&=\sqrt{(a_1)^2-(a_2)^2+(a_3)^2}\end{align} Taking square of the both sides \begin{align}|\
- Question 12, 13 & 14, Exercise 3.2
- hat{k}|=3$. ====Solution==== We are given \begin{align}|\alpha \hat{i}+(\alpha +1)\hat{j}+2\hat{k}|&=3.\end{align} This gives \begin{align}\sqrt{(\alpha )^2+(\alpha +1)^2+(2)^2}&=3.\end{align} Taking square on both sides, we have, \begin{align
- Question 1, Exercise 3.3
- nd $\vec{a}\cdot \vec{b}$ ====Solution==== \begin{align}\vec{a} \cdot \vec{b}&=(3 \hat{i}+4 \hat{j}-\hat{... s 1)+(4 \times-1)+(-1 \times 3)\\ & =3-4-3=-4\end{align}. =====Question(ii)===== If $\vec{a}=3 \hat{i}+4... $\vec{a} \cdot \vec{c}$. ====Solution==== \begin{align}\vec{a} \cdot \vec{c}&=(3 \hat{i}+4 \hat{j}-\hat{... s 2)+(4 \times 1)+(-1 \times-5)\\ & =6+4+5=15\end{align}. =====Question(iii)===== If $\vec{a}=3 \hat{i}+
- Question 12 & 13, Exercise 3.3
- direction. From $\triangle A B O$, we have \begin{align}\overrightarrow{O B}+\overrightarrow{A B}&=\overr... A}-\overrightarrow{O B}=\vec{a}-\vec{b}...(1)\end{align} Also from $\triangle A C O$, we have \begin{align}\overrightarrow{O A}+\overrightarrow{A C}&=\overright... ghtarrow{O A}=\vec{c}-\vec{a} \text {...(2) }\end{align} Now \begin{align}\overrightarrow{B A} \cdot \ov
- Question 9 & 10, Exercise 3.2
- c}.$ ====Solution==== We compute that\\ \begin{align}2\overrightarrow{a}-\overrightarrow{b}+3\overrigh... c}|&=\sqrt{(1)^2+(-2)^2+(2)^2}\\ &=\sqrt{9}=3\end{align} Let sat $\hat{a}$ be unit vector in direction of... }-\overrightarrow{b}+3\overrightarrow{c}.$ \begin{align}\hat{a}&=\dfrac{2\overrightarrow{a}-\overrightarr... c}|}\\ &=\dfrac{\hat{i}-2\hat{j}+2\hat{k}}{3}\end{align} Now vector of magnitude of $6$ unit which is pa
- Question 1, Exercise 3.2
- ind $\vec{a}+2\vec{b}$. ====Solution==== \begin{align}\vec{a}+2\vec{b}&=3\hat{i}-5\hat{j}+2(-2\hat{i}+3... hat{j}-4\hat{i}+6\hat{j}\\ &=-\hat{i}+\hat{j}\end{align} =====Question.1(ii)===== If $\vec{a}=3\hat{i}-5... nd $3\vec{a}-2\vec{b}$. ====Solution==== \begin{align}3\vec{a}-2\vec{b}&=3(3\hat{i}-5\hat{j})-2(-2\hat{... {j}+4\hat{i}-6\hat{j}\\ &=13\hat{i}-21\hat{j}\end{align} =====Question.1(iii)===== If $\vec{a}=3\hat{i}-
- Question 3 & 4, Exercise 3.2
- vec{q}|=5.$ ====Solution==== We calculate \begin{align}\vec{p}+\vec{q}&=2\hat{i}-\hat{j}+x\hat{i}+3\hat{j}\\ &=(2+x)\hat{i}+2\hat{j}\end{align} Thus \begin{align}|\vec{p}+\vec{q}|&=\sqrt{(2+x)^2+2^2}\\ &=\sqrt{x^2+4x+4+4}\\ &=\sqrt{x^2+4x+8}\end{align} But we are given that \begin{align}&|\vec{p}+\v
- Question 5 & 6, Exercise 3.2
- rrow{OB}=7\hat{i}+9\hat{j}.$$ Thus we have \begin{align}\overrightarrow{AB}&=\overrightarrow{OB}-\overrig... (-3\hat{i}+5\hat{j}) \\ &=10\hat{i}+4\hat{j}.\end{align} This gives \begin{align}|\overrightarrow{AB}|&=\sqrt{(10)^2+(4)^2}\\ &=\sqrt{116} = 2\sqrt{29}.\end{align} Let $\hat{r}$ be unit vector in the direction of
- Question 7, Exercise 3.2
- $Q(2,-1)$. ====Solution==== As we have \begin{align}\overrightarrow{PQ}&=\overrightarrow{OQ}-\overrig... })-(-\hat{i}+2\hat{j})\\ &=3\hat{i}-3\hat{j}\end{align} Also \begin{align}|\overrightarrow{PQ}|&=\sqrt{(3)^2+(3)^2}\\ &=\sqrt{18}=3\sqrt{2}.\end{align} Hence components of $\overrightarrow{PQ}$ are $3
- Question 7, Exercise 3.2
- $Q(2,-1)$. ====Solution==== As we have \begin{align}\overrightarrow{PQ}&=\overrightarrow{OQ}-\overrig... })-(-\hat{i}+2\hat{j})\\ &=3\hat{i}-3\hat{j}\end{align} Also \begin{align}|\overrightarrow{PQ}|&=\sqrt{(3)^2+(3)^2}\\ &=\sqrt{18}=3\sqrt{2}.\end{align} Hence components of $\overrightarrow{PQ}$ are $3
- Question 11, Exercise 3.3
- vec{c}=2 \hat{i}+\hat{j}-4 \hat{k}$. Then \begin{align}|\vec{a}|&=\sqrt{(3)^2+(-2)^2+(1)^2}\\ \Rightarro... |\vec{c}|^2&=|\vec{b}|^2\\ 14+21&=35\\ 35&=35\end{align} Thus by Pytagorous theorem, the vectors $\vec{a}... form right angle triangle. Also if we see \begin{align}\vec{a} \cdot \vec{c}&=-(3 \hat{i}-2 \hat{j}+\hat... Rightarrow \vec{a} \cdot \vec{c}&=6 - 2-4=0 .\end{align} $\therefore \quad \vec{a} \cdot \vec{c}$. or sid