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- Question 9 Exercise 3.4
- ec{b}\\ &=-\hat{i}+\dfrac{3}{2} \hat{j}+2 \hat{k}\end{align} From $\triangle A E B$, we have\\ \begin{a... -\hat{j}-3 \hat{k} \ldots \ldots \ldots \ldots(1)\end{align} From $\triangle A E D$. we have\\ \begin{a... d}&=\hat{i}+2 \hat{j}+ \hat{k} \text {. }....(2) \end{align} By using (1) and (2), we have,\\ \begin{al... & \hat{j} & \hat{k} \\ 3 & -1 & -3 \\ 1 & 2 & 1 \end{array}\right| \\ & =(-1+6) \hat{i}-(3+3) \hat{j}+
- Question 2 Exercise 3.4
- & \hat{j} & \hat{k} \\ -1 & 2 & -3 \\ 2 & -4 & 6 \end{array}\right| \\ & =(12-12) \hat{i}-(-6+6) \hat{j... vec{b}&=0 . \\ & \Rightarrow \vec{a} \| \vec{b} .\end{align} Second Way \begin{align}\vec{a} \cdot \ve... \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-28 .\end{align} Also \begin{align}|\vec{a}|&=\sqrt{(-1)^2+... circ} . \\ \Rightarrow \quad \vec{a} \| \vec{b} .\end{align} =====Question 2(ii)===== Show in two diff
- Question 7 & 8 Exercise 3.4
- c{A} \times \vec{B}&=\vec{C} \times \vec{A}...(2)\end{align} $\because \quad$ cross product is anti-com... {B} \times \vec{C}&=\vec{A} \times \vec{B}....(3)\end{align} $\because$ crass product is anti-commuativ... } & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & 1 & -3 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{... c{a} \times \vec{b}&=-5 \hat{i}+7 \hat{j}-\hat{k}\end{align} \begin{align}\vec{a} \times \vec{b} \mid&=
- Question 3 & 4 Exercise 3.5
- {array}{ccc} 3 & 0 & 2 \\ 1 & 2 & 1 \\ 0 & -1 & 4\end{array}\right|\\ \vec{a} \cdot \vec{b} \times \vec... array}{ccc} 1 & 2 & 1 \\ 0 & -1 & 4 \\ 3 & 0 & 2 \end{array}\right|\\ \vec{b} \cdot \vec{c} \times \vec... array}{ccc} 0 & -1 & 4 \\ 3 & 0 & 2 \\ 1 & 2 & 1 \end{array}\right|\\ \vec{c}\cdot\vec{a}\times\vec{b}&... vec{a} \times \vec{b}&=1+24=25 \ldots \ldots (3) \end{align} From (1), (2) and (3), we get that \begin
- Question 2 and 3 Exercise 3.3
- 4} \\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{169}=13\end{align} Now let say $\hat{c}$ be the unit vector $... rrow &=\dfrac{1}{13}(4\hat{i}+3\hat{j}-12\hat{k})\end{align} =====Question 3(i)===== Find the angles b... \Rightarrow \quad \vec{a} \cdot \vec{b}&=-1-1+2=0\end{align} $ \vec{a}$ and $\vec{b}$ are orthogonal. ... t{(2)^2+(-5)^2}\\ \Rightarrow|\vec{b}|&=\sqrt{29}\end{align} Now putting all these in (1) \begin{align}
- Question 12, 13 & 14, Exercise 3.2
- }|\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, \beg... \alpha -4=0\\ \implies & {\alpha ^2}+\alpha -2=0.\end{align} This is quadratic equation in $\alpha $. $... 1\quad \text{or} \quad \alpha =\dfrac{-1-3}{2}=-2\end{align} $\alpha=1$ or $\alpha =-2$, no real value
- Question 5(iii) & 5(iv) Exercise 3.5
- \vec{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{ali... 3) . \\ |\vec{a}|&=\sqrt{(a_1)^2-(a_2)^2+(a_3)^2}\end{align} Taking square of the both sides \begin{al... } \\ |\vec{b} |&=\sqrt{(b_1)^2+(b_2)^2-(b_3)^2} .\end{align} Taking square of the both sides \begin{al... _2)^2+(b_3)^2 \\ |\vec{b}|^2&=b_1^2+b_2^2+b_3^2 .\end{align} =====Question 5(iv)===== Let $\vec{a}=a_1
- Question 1, Exercise 3.3
- times 1)+(4 \times-1)+(-1 \times 3)\\ & =3-4-3=-4\end{align}. =====Question(ii)===== If $\vec{a}=3 \ha... times 2)+(4 \times 1)+(-1 \times-5)\\ & =6+4+5=15\end{align}. =====Question(iii)===== If $\vec{a}=3 \h... \Rightarrow \vec{b}+\vec{c}&=3 \hat{i}-2 \hat{k}\end{align} Taking dot product with $\vec{a}$ \begin{a... \\ \Rightarrow \vec{a} \cdot(\vec{b}+\vec{c})&=11\end{align}. =====Question(iii)===== If $\vec{a}=3 \h
- Question 12 & 13, Exercise 3.3
- w{O A}-\overrightarrow{O B}=\vec{a}-\vec{b}...(1)\end{align} Also from $\triangle A C O$, we have \begi... errightarrow{O A}=\vec{c}-\vec{a} \text {...(2) }\end{align} Now \begin{align}\overrightarrow{B A} \cd... w{A C}&=0 \quad \because \quad|\vec{a}|=|\vec{c}|\end{align} Triangle formed looking in semicircle is r... overrightarrow{O E} \cdot \overrightarrow{C A}&=0\end{align} \begin{align} \Rightarrow \dfrac{\vec{c}+\
- Question 1 Exercise 3.4
- {i} & \hat{j} & \hat{k} \\ 0 & 1 & 0 \\ 0 & 2 & 3\end{array}\right|\\ \Rightarrow \vec{a} \times \vec{b}&=(3-0) \hat{i}=3 \hat{i}.\end{align} =====Question 1(ii)===== Find the cross p... } & \hat{j} & \hat{k} \\ 2 & -3 & 0 \\ 0 & 0 & 1 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{... ow \vec{a} \times \vec{b}&=-3 \hat{i}-2 \hat{j} .\end{align} =====Question 1(iii)===== Find the cross
- Question 4 Exercise 3.4
- & \hat{j} & \hat{k} \\ 3 & -6 & 5\\ 2 & -1 & 4\\ \end{array}\right| \\ & =(-24+5) \hat{i}-(12-10) \hat{... \times \vec{b}&=-19 \hat{i}-2 \hat{j}+9 \hat{k} .\end{align} =====Question 4(ii)===== If $\vec{a}=3 \h... & \hat{j} & \hat{k} \\ 2 & -1 & 4 \\ 1 & 1 & -1 \end{array}\right| \\ &=(1-4) \hat{i}-(-2-4) \hat{j}+(... \times \vec{c}&=-3 \hat{i}+6 \hat{j}+3 \hat{k} .\end{align} =====Question 4(iii)===== If $\vec{a}=3 \
- Question 5(i) & 5(ii) Exercise 3.5
- a_2 & a_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array}\right|\\ &=0\quad \because \text{two rows ... row \quad \vec{a} \cdot \vec{a} \times \vec{b}&=0\end{align}\\ Which implies that $\vec{a} \times \vec{... b_2 & b_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array}\right|\\ &=0\quad\because \quad\text{ two ... htarrow \vec{a} \cdot(\vec{a} \times \vec{b})&=0.\end{align} Which implies that $\vec{a} \times \vec{b}
- Question 7 Exercise 3.5
- n{array}{ccc}1 & 2 & 3 \\ 2 & -3 & 4 \\ 3 & 1 & c\end{array}\right|&=0\\ 1(-3 c-4)-2(2 c-12)+3(2+9)&=0\... w \quad-7 c+53&=0\\ \Rightarrow c&=\dfrac{53}{7}.\end{align} which is required value of $c$ for which t... ray}{ccc} 1 & 1 & -1 \\ 1 & -2 & 1 \\ c & 1 & -c \end{array}\right|&=0\\ (2 c-1)-(-c-c)-(1+2 c)&=0 \\ \... 2 c&=0 \\ \Rightarrow 2 c-2&=0\\ \Rightarrow c&=1\end{align} which is the required value of $c$ for whi
- Question 8 Exercise 3.5
- hat{j}+6 \hat{k}, \\ \vec{c}&=7 \hat{j}+8 \hat{k}\end{align} ====Solution==== The volume of tetrahedron... in{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 0 & 7 & 8\end{array}\right|\\ V&=\dfrac{1}{6} \cdot 1(40-42)-4(... htarrow V&=\dfrac{1}{6}(-2+20)=3 \text { units. }\end{align} =====Question 8(ii)===== Find the volume ... ightarrow \vec{c}&=-2 \hat{i}-2 \hat{j}-3 \hat{k}\end{align} The volume of tetiahedron is: \begin{align
- Question 1, Exercise 3.2
- }-5\hat{j}-4\hat{i}+6\hat{j}\\ &=-\hat{i}+\hat{j}\end{align} =====Question.1(ii)===== If $\vec{a}=3\ha... \hat{j}+4\hat{i}-6\hat{j}\\ &=13\hat{i}-21\hat{j}\end{align} =====Question.1(iii)===== If $\vec{a}=3\h... -5\hat{j}+2\hat{i}-3\hat{j}\\ &=5\hat{i}-8\hat{j}\end{align} Multiply both sides by $2$. We have, $$2(... }-5\hat{j}-2\hat{i}+3\hat{j}\\ &=\hat{i}-2\hat{j}\end{align} Taking modulus of both sides. We have, $$|