Vector Algebra Practice Questions

From given figure, we can write all the forces in vector notation as {\text{OP}}=(20 \sin 30° {{i}}^{\^}+20 \cos 30° {{j}}^{\^}) {\text{N}} =(10 {{i}}^{\^}+17.32 {{j}}^{\^}) {\text{N}} {\text{OQ}}=(10 \cos 30° {{i}}^{\^}+10 \sin 30° {{j}}^{\^}) {\text{N}} =(8.66 {{i}}^{\^}+5 {{j}}^{\^}) {\text{N}} {\text{OR}}=(20 \cos 45° {{i}}^{\^}-20 \sin 45° {{j}}^{\^}) {\text{N}} =(14.14 {{i}}^{\^}-14.14 {{j}}^{\^}) {\text{N}} {\text{OS}}=(-15 \cos 45° {{i}}^{\^}-15 \sin 45° {{j}}^{\^}) {\text{N}} =(-10.60 {{i}}^{\^}-10.60 {{j}}^{\^}) {\text{N}} {\text{OT}}=(-15 \sin 60°\hat{i}+15 \cos 60°{{j}}^{\^}) {\text{N}} =(-12.99 {{i}}^{\^}+7.5 {{j}}^{\^}) {\text{N}} The resultant of these forces is given as F = OP + OQ + OR + OS + OT {\text{F}}=[(10 {{i}}^{\^}+17.32 {{j}}^{\^})+(8.66 {{i}}^{\^}+5 {{j}}^{\^}) +(14.14 {{i}}^{\^}-14.14 {{i}}^{\^})+(-10.60 {{i}}^{\^}-10.60 {{j}}^{\^}) +(-12.99 {{i}}^{\^}+7.5 {{j}}^{\^})] {\text{F}}=9.21 {{i}}^{\^}+5.08 {{j}}^{\^} {\text{N}} {\text{F}} \approx 9.25 {{i}}^{\^}+5 {{j}}^{\^} {\text{N}} Thus, the resultant force is {\text{F}}\approx 9.25 {{i}}^{\^}+5 {{j}}^{\^} {\text{N}}

Given, $P \times Q=Q \times P$ $\Rightarrow \;\; P \times Q=-P \times Q \Rightarrow 2(P \times Q)=0$ or $P \times Q=0$ It is possible only, if $P=0$ or $Q=0$, angle between them is $180^{\circ}$. $\therefore \;\; \theta =180^{\circ}$

(A \cos \theta ) {\hat{B}}=A(\frac{\vec{A} \.\vec{B}}{A B}) \hat{B}=\frac{{A} ^{\to }\.{B} ^{\to }}{B} \hat{B} $=\frac{2}{\\sqrt{2}}(\frac{\hat{i}+\hat{j}}{\\sqrt{2}})=\hat{i}+\hat{j}$

Given ${X}={Y}$ $\\sqrt{X^{2}+Y^{2}-2 \\times Y \cos \theta }$ $=n \\sqrt{X^{2}+Y^{2}+2 \\times Y \cos \theta }$ Square both sides $2 X^{2}(1-\cos \theta )=n^{2} \.2 X^{2}(1+\cos \theta )$ $1-\cos \theta =n^{2}+n^{2} \cos \theta$ $\cos \theta =\frac{1-{n}^{2}}{1+{n}^{2}}$ $\theta =\cos ^{-1}[\frac{n^{2}-1}{-n^{2}-1}]$

Given, force vectors ${\text{P}} \perp {\text{Q}}$, i.e. $\theta =90°$ Let resultant of P + Q = x and resultant of P − Q = y $\therefore x^{2}=\text{P}^{2}+\text{Q}^{2}+2 \text{PQ} \cos 90°=\text{P}^{2}+\text{Q}^{2}$ ...(i) $y^{2}=\text{P}^{2}+\text{Q}^{2}-2 \text{PQ} \cos 90°=\text{P}^{2}+\text{Q}^{2}$ ...(ii) When $\theta _1$ is the angle between (P + Q) and (P − Q), then, their resultant is given by $|{x}+{y}|=\\sqrt{3(\text{P}^{2}+\text{Q}^{2})}$ $\therefore \\sqrt{x^{2}+y^{2}+2 x y \cos \theta _{1}}=\\sqrt{3 \text{P}^{2}+3 \text{Q}^{2}}$ ...(iii) Substituting the values of $x^{2}$ and $y^{2}$ in Eq. (iii), we get $\text{P}^{2}+\text{Q}^{2}+\text{P}^{2}+\text{Q}^{2}+2 \\sqrt{\text{P}^{2}+\text{Q}^{2}} \\sqrt{\text{P}^{2}+\text{Q}^{2}} \cos \theta _{1}=3 \text{P}^{2}+3 \text{Q}^{2}$ $2 \text{P}^{2}+2 \text{Q}^{2}+(2 \text{P}^{2}+2 \text{Q}^{2}) \cos \theta _{1}=3 \text{P}^{2}+3 \text{Q}^{2}$ $\Rightarrow \cos \theta _{1}=\frac{\text{P}^{2}+\text{Q}^{2}}{2(\text{P}^{2}+\text{Q}^{2})}=\frac{1}{2}$ $\therefore \theta _{1}=60°$ When $\theta _2$ is the angle between (P + Q) and (P − Q), then the magnitude of their resultant is given by $|{x}+{y}|=\\sqrt{2(\text{P}^{2}+\text{Q}^{2})}$ $x^{2}+y^{2}+2 x y \cos \theta _{2}=2 \text{P}^{2}+2 \text{Q}^{2}$ ...(iv) Substituting the values, we get $\text{P}^{2}+\text{Q}^{2}+\text{P}^{2}+\text{Q}^{2}+2(\text{P}^{2}+\text{Q}^{2}) \cos \theta _{2}$ $=2 \text{P}^{2}+2 \text{Q}^{2}$ cos $\theta _2 = 0$ $\Rightarrow \theta _2 = 90°$ So, $\theta _{1}=60°$ and $\theta _{2}=90°$ and $\theta _{1}<\theta _{2}$ Hence, both statement I and statement II are true.

Direction of $P\;\;\; \hat{v}_{1}=\±\frac{\vec{A} \\times \vec{B}}{|{A} ^{\to }\\times \vec{B}|}=\±\frac{\hat{i}-\hat{j}+\hat{k}}{\\sqrt{3}}$ Direction of ${Q}\;\;\; \hat{v}_{2}=\±{{{\vec{A}}} \\times {\vec{C}}}/{|{\vec{A}} \\times {\vec{C}}|}=\±\frac{2 \hat{k}}{2}=\±\hat{k}$ Angle between $\hat{v}_{1}$ and $\hat{v}_{2}$ {\hat{v}_{1} \.{{v}}^{\^}_{2}}/{|{{v}}^{\^}_{1}||\hat{v}_{2}|}=\frac{\±1 / \\sqrt{3}}{(1)(1)}=\±\frac{1}{\\sqrt{3}} $\Rightarrow{x}=3$

Polygon law is applicable in both but the equation given in the reason is not useful in explaining the assertion.

$|{\vec{P}}|=|{\vec{Q}}|={x}$ .......(i) $|{\vec{P}}+{\vec{Q}}|={n}|{\vec{P}}-{{Q}}^{\to }|$ ${P}^{2}+{Q}^{2}+2 {PQ} \cos \theta$ $={n}^{2}({P}^{2}+{Q}^{2}-2 {P} {Q} \cos \theta )$ Using (i) in above equation $\cos \theta =\frac{{n}^{2}-1}{1+{n}^{2}}$ $\theta =\cos ^{-1}(\frac{{n}^{2}-1}{{n}^{2}+1})$

${\vec{A}} \.{{B}}^{\to }=|{{A}}^{\to } \\times {\vec{B}}|$ ${AB} \cos \theta ={AB} \sin \theta \Rightarrow \theta =45°$ $|{{A}}^{\to }-{\vec{B}}|=\\sqrt{{A}^{2}+{B}^{2}-2 {AB} \cos 45°}$ $=\\sqrt{{A}^{2}+{B}^{2}-\\sqrt{2} {AB}}$ Hence option (4).

In the given figure, the magnitude of vectors OA, OB and OC areequal. Let it be m. From figure, we can write {\text{OA}}=m(\cos 30° {{i}}^{\^}+\sin 30° {{j}}^{\^}) =m(\frac{\\sqrt{3}}{2} {{i}}^{\^}+\frac{1}{2} {{j}}^{\^}) {\text{OB}}=m(\cos 60° {{i}}^{\^}-\sin 60°{{j}}^{\^}) =m(\frac{1}{2} {{i}}^{\^}-\frac{\\sqrt{3}}{2} {{j}}^{\^}) {\text{OC}}=m(\cos 45°(-{{i}}^{\^})+\sin 45°{{j}}^{\^}) =m\{\frac{-1}{\\sqrt{2}} {{i}}^{\^}+\frac{1}{\\sqrt{2}} {{j}}^{\^}\} According to the question,{\text{OA}}+{\text{OB}}-{\text{OC}}=m(\frac{\\sqrt{3}}{2} {{i}}^{\^}+\frac{1}{2} {{j}}^{\^})+m(\frac{1}{2} {{i}}^{\^}-\frac{\\sqrt{3}}{2} {{j}}^{\^}) -m(-\frac{1}{\\sqrt{2}} {{i}}^{\^}+\frac{1}{\\sqrt{2}} {{j}}^{\^}) =m {{i}}^{\^}(\frac{\\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{\\sqrt{2}})+m {{j}}^{\^}(\frac{1}{2}-\frac{\\sqrt{3}}{2}-\frac{1}{\\sqrt{2}}) ∴ Angle with X - axis is given by $\tan \theta =\frac{y}{x}={m(\frac{1}{2}-\frac{\\sqrt{3}}{2}-\frac{1}{\\sqrt{2}})}/{m(\frac{\\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{\\sqrt{2}})}$ $\Rightarrow \tan \theta ={\frac{1}{2}-\frac{\\sqrt{3}}{2}-\frac{\\sqrt{2}}{2}}/{\frac{\\sqrt{3}}{2}+\frac{1}{2}+\frac{\\sqrt{2}}{2}}$ $\Rightarrow \tan \theta =\frac{1-\\sqrt{3}-\sqrt{2}}{\sqrt{3}+1+\\sqrt{2}}$ $\Rightarrow \theta =\tan ^{-1}(\frac{1-\\sqrt{3}-\\sqrt{2}}{1+\\sqrt{3}+\\sqrt{2}})$

$|\vec{A}+\vec{B}|$ = 2acosθ/2 ... (1) $|\vec{A}-\vec{B}|$ = 2acos $(\pi -\theta )/2$ = 2a sin θ/2 ... (2) \Rightarrow n $(2acos\theta /2)$ = 2a ${sin\theta }/2$ \Rightarrow tan $\theta /2$ = n

$( {{\vec{A}}_{1}}+ {{\vec{A}}_{2}} ) .( {{\vec{A}}_{1}}+ {{\vec{A}}_{2}} )=25$ $\Rightarrow 9+25+2 ( {{\vec{A}}_{1}} .{{\vec{A}}_{2}} )=25$ $\Rightarrow {{\vec{A}}_{1}} .{{\vec{A}}_{2}}=-9/2$ $=6 A_{1}^{2}-6 A_{2}^{2}+5 {{\vec{A}}_{1}} .{{\vec{A}}_{2}}$ $\Rightarrow 6 \times 9-6 \times 25+5 \times \frac{-9}{2}$ $= - 69 + (-22.5) = - 118.5$

r_\vec{g} = $a/2\hat{i} + a/2 \hat{k}$ r_\vec{H} = $a/2 \hat{j}+a/2 \hat{k}$ r_\vec{H}-r_\vec{g} = $1/2 a(\hat{j}-\hat{i})$

$\;\vec{A} \times \vec{B}=\vec{B} \times \vec{A}$ $\;\vec{A} \times \vec{B}-\vec{B} \times \vec{A}=0$ $\;\Rightarrow \vec{A} \times \vec{B}+\vec{A} \times \vec{B}=0$ $\; \therefore \vec{A} \times \vec{B}=0$ $\;\Rightarrow A B \sin \;\theta =0$ $\;\theta =0, \pi , 2 \pi . . . . . . .$from the given options, $\theta =\pi$