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- Question 11, Exercise 3.3
- \hat{k}$ and $2 \hat{i}+\hat{j}-4 \hat{k}$ form a right angle triangle. ====Solution==== Let $\vec{a}=3 \... c}$ represent the sides of triangle and they form right angle triangle. Also if we see \begin{align}\vec{... sides represented by $\vec{a}$ and $\vec{c}$ form right angle with each other. =====Question 11 (ii)====... that $P(1,0,1), Q(1,1,1)$ and $R(1,1.0)$ forms a right isosceles triangle. ====Solution==== We find vect
- Question 2 Exercise 3.4
- \hat{k} \\ -1 & 2 & -3 \\ 2 & -4 & 6 \end{array}\right| \\ & =(12-12) \hat{i}-(-6+6) \hat{j}+(4-4) \hat{... \cos ^{-1}\left(\frac{-28}{2 \sqrt{14} \sqrt{14}}\right) \\ \Rightarrow \quad \theta&=\cos ^{-1}(-1)=180^... & \hat{k} \\ 3 & 6 & -9 \\ 1 & 2 & -3 \end{array}\right| \\ & =(-18+18) \hat{i}+(-9+9) \hat{j}+(6-6) \hat... }\left(\sqrt{\dfrac{42 \times 42}{14 \times 126}}\right) \\ \Rightarrow \theta&=\cos^{-1}\left(\sqrt{\dfr
- Question 9 Exercise 3.4
- & \hat{k} \\ 3 & -1 & -3 \\ 1 & 2 & 1 \end{array}\right| \\ & =(-1+6) \hat{i}-(3+3) \hat{j}+(6+1) \hat{k}... frac{1}{2} \hat{i}-\dfrac{3}{2} \hat{j}+2 \hat{k}\right) \\ \Rightarrow \vec{d}&=2 \hat{i}-\dfrac{1}{2} \... c{5}{2} & -3 \\ 2 & -\dfrac{1}{2} & 1 \end{array}\right|\\ \Rightarrow \vec{c} \times \vec{d}&=\left(\dfrac{5}{2}-\dfrac{3}{2}\right) \hat{i}-(1+6) \hat{j}+\left(-\dfrac{1}{2}-\dfrac
- Question 3 & 4 Exercise 3.5
- } 3 & 0 & 2 \\ 1 & 2 & 1 \\ 0 & -1 & 4\end{array}\right|\\ \vec{a} \cdot \vec{b} \times \vec{c}&=3(8+1)+2... 1 & 2 & 1 \\ 0 & -1 & 4 \\ 3 & 0 & 2 \end{array}\right|\\ \vec{b} \cdot \vec{c} \times \vec{a}&=1(-2-0)+... 0 & -1 & 4 \\ 3 & 0 & 2 \\ 1 & 2 & 1 \end{array}\right|\\ \vec{c}\cdot\vec{a}\times\vec{b}&=1(3-2)+4(6-0... & \hat{k} \\ 0& -1 & 4 \\ 1 & 2 & 1 \end{array}\right|\\ \vec{c} \times \vec{b}&=(-1-8) \hat{i}-(0-4) \
- Question 7 & 8 Exercise 3.3
- =\left(-\dfrac{3}{2} \hat{j}+\dfrac{4}{5} \hat{k}\right) \cdot(\hat{i}-2 \hat{j}-2 \hat{k}) \\ \Rightarro... w \vec{a} \cdot \vec{b}&=0(1)+\left(-\dfrac{3}{2}\right)(-2)+\dfrac{4}{5}(-2) \\ \Rightarrow \quad \vec{a... text { Also }|\vec{a}|&=\sqrt{\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{4}{5}\right)^2} \\ \Rightarrow|\vec{a}|&=\sqrt{\dfrac{9}{4}+\dfrac{16}{25}} \\ \Rightarr
- Question 3 Exercise 3.4
- & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{b}&=(2+3) \ha... & \hat{k} \\ 3 & -1 & 6 \\ 1 & 4 & 1 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{b}&=(-1-24) \... ow|\vec{a} \times \vec{b}|&=\sqrt{\left(-\left.25\right|^2+(3)^2+(13)^2\right.} \\ \Rightarrow|\vec{a} \times \vec{b}|&=\sqrt{803} \ldots \ldots \ldots(3)\end{a
- Question 2 and 3 Exercise 3.3
- eta&=\cos ^{-1}\left(\dfrac{8}{5 \cdot \sqrt{29}}\right) \\ \Rightarrow \theta&=\cos ^1(0.2971)=72.72^{\c... { and }|\vec{b}|&=\sqrt{(1)^2+\left(1^{12}+(1)^2\right.} \\ \Rightarrow|\vec{b}|&=\sqrt{3}\end{align} No... s ^{-1}\left(\dfrac{-1}{\sqrt{3} \cdot \sqrt{13}}\right) \\ \Rightarrow \theta&=\cos ^1(-0.1601)=99^{\cir... x3-3-p1 |< Question 1]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3
- Question 1 Exercise 3.4
- j} & \hat{k} \\ 0 & 1 & 0 \\ 0 & 2 & 3\end{array}\right|\\ \Rightarrow \vec{a} \times \vec{b}&=(3-0) \hat... & \hat{k} \\ 2 & -3 & 0 \\ 0 & 0 & 1 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{b}&=(-3-0) \h... & \hat{k} \\ 2 & -3 & 5 \\ 6 & 2 & -3 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{b}&=(9-10) \h... \hat{k} .\end{align} ====Go To==== <text align="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3
- Question 4 Exercise 3.4
- \hat{k} \\ 3 & -6 & 5\\ 2 & -1 & 4\\ \end{array}\right| \\ & =(-24+5) \hat{i}-(12-10) \hat{j}+(-3+12) \h... & \hat{k} \\ 2 & -1 & 4 \\ 1 & 1 & -1 \end{array}\right| \\ &=(1-4) \hat{i}-(-2-4) \hat{j}+(2+1) \hat{k} ... j} & \hat{k} \\ 5 & -7 & 9 1 & -5 & 1 \end{array}\right| \\ \Rightarrow(\vec{a}+\vec{b}) \times(\vec{a}-\... x3-4-p3 |< Question 3]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3
- Question 7 & 8 Exercise 3.4
- & \hat{k} \\ 1 & 1 & 2 \\ 2 & 1 & -3 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{b}&=(-3-2) \h... \hat{k} \\ 2 & -3 & 4 \\ 4 & -2 & -4 \end{array}\right| \\ \vec{a} \times \vec{b}&=(12+8) \hat{i}-(-8-16... dfrac{10 \hat{i}+12 \hat{j}+4 \hat{k}}{\sqrt{65}}\right)$, is the required vector.\\ ====Go To==== <tex... x3-4-p6 |< Question 6]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3
- Question 5(i) & 5(ii) Exercise 3.5
- \ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array}\right|\\ &=0\quad \because \text{two rows are identical... \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array}\right|\\ &=0\quad\because \quad\text{ two rows are iden... \ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array}\right|\\ \vec{a} \times \vec{b}&=(a_2 b_3-a_3 b_2) \hat... -p2 |< Question 3 & 4]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3
- Question 7 Exercise 3.5
- c}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\\ \Rightarrow... & 1 & -1 \\ 1 & -2 & 1 \\ c & 1 & -c \end{array}\right|&=0\\ (2 c-1)-(-c-c)-(1+2 c)&=0 \\ \Rightarrow 2 ... } 1 & 1 & 2 \\ 2 & 3 & 1 \\ c & 2 & 6 \end{array}\right|&=0\\ 1(18-2)-1(12-c)+2(4-3 c)&=0 \\ 16-12+c+8-6 ... x3-5-p5 |< Question 6]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3
- Question 1 Review Exercise 3
- %%(a)%% Equilateral * (b) Isosceles but not right angle * %%(c)%% Right angled but not isosceles * (d) right angled ind isosceles \\ <btn type="link" collapse="a2">See An... lapse> </panel> ====Go To==== <text align="right"><btn type="success">[[math-11-kpk:sol:unit03:Rev
- Question 5 Exercise 3.4
- } & \hat{k} \\ 5 & 5 & 0\\ 1 & -5 & 0 \end{array}\right|\\ \Rightarrow \overrightarrow{P Q} \times \overr... & \hat{k} \\ 3 & 3 & -4 \\ 3 & 1 & -2 \end{array}\right| \\ \Rightarrow \overrightarrow{P Q} \times \over... x3-4-p4 |< Question 4]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3
- Question 6 Exercise 3.4
- & j & k \\ 1 & -2 & 2 \\ 3 & -2 & 5 \end{array} \right| \\ \Rightarrow \vec{M}&=(-10+4) \hat{i}-(5-6) \h... & \hat{k} \\ 0 & -4 & 1 \\ 3 & -2 & 5 \end{array}\right| \\ \Rightarrow \vec{M}&=(-20+2) \hat{i}-(0-3) \h... x3-4-p5 |< Question 5]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3