Vector Algebra Practice Questions

\;\overline{ {\text{N}}}={|\overline{ {\text{A}}}|}\ { {\text{B}}^{\^}}=\;{5(4 { {\text{i}}^{\^}}+3 { {\text{j}}^{\^}})}/{5}=4 { {\text{i}}^{\^}}+3 { {\text{j}}^{\^}} $\; \therefore {\text{x}}=4$

The magnitude of resultant vector$R^{'}={\sqrt{a^{2+}b^{2+}2 a b \cos \theta }}$Here $a=b=R$Then $R^{'}={\sqrt{R^{2+}R^{2+}2 R^2 \cos \theta }}$ $\;=R {\sqrt{2}} {\sqrt{1+\cos \theta }}$ $\;={\sqrt{2}} R \sqrt{2 \cos ^2 \;\frac{\theta }{2}}$ $\;=2 R \cos \;\frac{\theta }{2}$

$\;\vec{R}= (A+\;{A}/{{\sqrt{2}}}) \hat{i}+ (A-\;{A}/{{\sqrt{2}}}) \hat{j}$ $\;|\vec{R}|=\sqrt{ (A+\;{A}/{{\sqrt{2}}}) ^{2+} (A-\;{A}/{{\sqrt{2}}}) ^2}={\sqrt{3}} A$

$\;B \cos \theta =\;\frac{B}{2}$ $\;\Rightarrow \theta =60^{\circ}$So, angle between ${ {\text{A}}}^{\to }$ & ${ {\text{B}}}^{\to }$ is $90^{\circ}+60^{\circ}=150^{\circ}$

$\;\vec{B} \times \vec{C}={\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 3 \\ -8 & -1 & 3\end{vmatrix}}=15 \hat{i}-21 \hat{j}+33 \hat{k}$ $\;\vec{A} \cdot (\vec{B} \times \vec{C})=(-x \hat{i}-6 \hat{j}-2 \hat{k}) \cdot (15 \hat{i}-21 \hat{j}+33 \hat{k})$ $\;0=-15 x+126-66$ $\;15 x=60$ $\;x=4$

$\;{ {\text{R}}}^{\to }=(2 { {\text{Q}}}^{\to }+2 { {\text{P}}}^{\to })+(2 { {\text{Q}}}^{\to }-2 { {\text{P}}}^{\to })$ $\;{ {\text{R}}}^{\to }=4 { {\text{Q}}}^{\to }$Angle between $\vec{Q}$ and $\vec{R}$ is zeroOption (1) is correct

$\;\cos \theta =\;\frac{5}{9}$ $\;\;\frac{\vec{a} \cdot \vec{b}}{a b}=\;\frac{5}{9} \;\; . . . . . . .(1)$ $\;|\vec{a}+\vec{b}|={\sqrt{2}}|\vec{a}-\vec{b}|$ $\;a^{2+}b^{2+}2 \vec{a} \cdot \vec{b}=2 a^{2+}2 b^{2-}4 \vec{a} \cdot \vec{b}$ $\;6 \vec{a} \cdot \vec{b}=a^{2+}b^2$ $\;6 \times \;\frac{5}{9} a b=a^{2+}b^2$ \;\;\frac{10}{3} a b=a^{2+}b^2 \;\; & \;\; a=n b $\;\;\frac{10}{3} n b^2=n^2 b^{2+}b^2$ $\;3 n^{2-}10 n+3=0$ $\;n=\;\frac{1}{3} \;\text{ and }\; n=3$ $\;\;\text{ integer value }\; n=3$

$\;{| { {\text{F}}}^{\to }_1 |}= {\text{A}},{| { {\text{F}}}^{\to }_2 |}=\;{ {\text{A}}}/{2} \;\; \theta =\;\frac{\pi }{2}$ $\;{| { {\text{F}}}^{\to }_{\;\text{net }\;} |}=\sqrt{ {\text{F}}_1^{2+} {\text{F}}_2^2}$ $\;=\sqrt{ {\text{A}}^{2+} (\;{ {\text{A}}}/{2}) ^2}$ $\;{| { {\text{F}}}^{\to }_{\;\text{net }\;} |}=\;{{\sqrt{5}} {\text{A}}}/{2}$

\;{ {\text{A}}}^{\to }=2 { {\text{i}}^{\^}}+3 { {\text{j}}^{\^}}+2 { {\text{k}}^{\^}} \;{ {\text{B}}}^{\to }-{ {\text{A}}}^{\to }=2 { {\text{j}}^{\^}} \;{ {\text{B}}}^{\to }=2 { {\text{j}}^{\^}}+{ {\text{A}}}^{\to }=2 { {\text{i}}^{\^}}+5 { {\text{j}}^{\^}}+2 { {\text{k}}^{\^}} $\;|{ {\text{B}}}^{\to }|={\sqrt{2^{2+}5^{2+}2^2}}={\sqrt{4+25+4}}={\sqrt{8+25}}={\sqrt{33}}$

\vec{\text{P}} \times \vec{\text{Q}}={\begin{vmatrix}{\text{i}}^{\^} & {\text{j}}^{\^} & {\text{k}}^{\^} \\ 3 & {\sqrt{3}} & 2 \\ 4 & {\sqrt{3}} & 2.5\end{vmatrix}} ={\sqrt{3}} \;{{ {\text{i}}^{\^}}}/{2}+\;{{ {\text{j}}^{\^}}}/{2}-{\sqrt{3}} { {\text{k}}^{\^}} \Rightarrow \;{{ {\text{P}}}^{\to } \times { {\text{Q}}}^{\to }}/{|{ {\text{P}}}^{\to } \times { {\text{Q}}}^{\to }|}=\;\frac{1}{2} ({\sqrt{3}} \;{{ {\text{i}}^{\^}}}/{2}+\;{{ {\text{j}}^{\^}}}/{2}-{\sqrt{3}} { {\text{k}}^{\^}}) \;=\;\frac{1}{4}({\sqrt{3}} { {\text{i}}^{\^}+{ {\text{j}}^{\^}-2 {\sqrt{3}} { {\text{k}}^{\^}) \;\; {\text{x}}=4

\;{ \vec{\text{P}} \cdot { \vec{\text{Q}}=0 \;(\;{ {\text{i}}}^{\^}+2 {\text{m}} \;{ {\text{j}}}^{\^}+ {\text{m}} \;{ {\text{k}}}^{\^}) \cdot (4 \;{ {\text{i}}}^{\^}-2 \;{ {\text{j}}}^{\^}+ {\text{m}} \;{ {\text{k}}}^{\^})=0 $\;\Rightarrow 4-4 {\text{m}}+ {\text{m}}^2=0$ $\;\Rightarrow ( {\text{m}}-2)^2=0 \Rightarrow {\text{m}}=2$

For two perpendicular vectors$\;(a {i}^{\wedge }+b {j}^{\wedge }+{k}^{\wedge }) \cdot (2 {i}^{\wedge }-3 {j}^{\wedge }+4 {k}^{\wedge })=0$ $\;2 a-3 b+4=0$On solving, $2 a-3 b=-4$Also given$3 a+2 b=7$We get ${\text{a}}=1, {\text{b}}=2$ $\;\;{ {\text{a}}}/{ {\text{b}}}=\;{ {\text{x}}}/{2} \Rightarrow {\text{x}}=\;{2 {\text{a}}}/{ {\text{b}}}=\;\frac{2 \times 1}{2}$ $\Rightarrow {\text{x}}=1$

$\; {\text{A}}_{ {\text{y}}}= {\text{A}} \cos 30^{\circ}=2 {\sqrt{3}}$ $\; {\text{A}}=4$ $\; {\text{A}}_{ {\text{x}}}= {\text{A}} sin \;30^{\circ}=2$

{ \vec{\text{a}} \cdot { \vec{\text{b}}=0 \therefore { \vec{\text{a}} \cdot { \vec{\text{b}}=0 $\therefore 2 \times 1+4 \times 2-2 \times \alpha =0$ $\therefore \alpha =5$

{ \vec{\text{A}}=(2 { {\text{i}}^{\^}+3 { {\text{j}}^{\^}-{ {\text{k}}^{\^}) {\text{m}} and { \vec{\text{B}}=({ {\text{i}}^{\^}+2 { {\text{j}}^{\^}+2 { {\text{k}}^{\^}) {\text{m}}Component of $\vec{A}$ along $\vec{B}=\vec{A} \cdot \hat{B}$ =\;{{ \vec{\text{A}} \cdot { \vec{\text{B}}}/{|{ \vec{\text{B}}|}=\;\frac{2+6-2}{\sqrt{1^{2+}2^{2+}2^2}} $=\;\frac{6}{3}=2$

$\vec{A} \times \vec{A}=A \times A \times \sin 0^{\circ}$ $=0$

$\because |\hat{A}-\hat{B}|=2 \sin (\;\frac{\theta }{2})$and, $|\hat{A}+\hat{B}|=2 \cos (\;\frac{\theta }{2})$ $\Rightarrow \;\frac{|\hat{A}-\hat{B}|}{|\hat{A}+\hat{B}|}=\tan (\;\frac{\theta }{2})$

${\sqrt{A^{2+}A^{2+}2 A^2 \cos \theta }}=2 {\sqrt{A^{2+}A^{2+}2 A^2(-\cos \theta )}}$ $\Rightarrow 2 A^{2+}2 A^2 \cos \theta =8 A^{2+}8 A^{2-}(-\cos \theta )$ $\Rightarrow 5 \cos \theta =3$ $\Rightarrow \theta =\cos ^{-1} (\;\frac{3}{5})$

Given, angle between A and B, θ = 120° The angle between A and - B, α = 180° − θ = 180° − 120° = 60° Now, from diagram the resultant vector of A and −B will be A − B and the angle between A − B and A is denoted by β. So, angle between A and −B is calculated as, ϕ = 180°− α = 180°− 60° = 120° Now, from parallelogram law of vector addition angle β can becalculated as follows $\tan \beta =\frac{-\text{B} \sin \phi }{-\text{A}+(-\text{B} \cos \phi )}$ $=\frac{-\text{B} \sin 120°}{-\text{A}-\text{B} \cos 120°}={-\text{B} \frac{\\sqrt{3}}{2}}/{-\text{A}+\frac{\text{B}}{2}}$ $=\frac{\\sqrt{3} \text{B}}{2 \text{A}\-\text{B}}$ $\beta =\tan ^{-1}(\frac{\\sqrt{3} \text{B}}{2 \text{A} - \text{B}})$ Hence, the angle between vector A and (A - B) is $\tan ^{-1}(\frac{\\sqrt{3} \text{B}}{2 \text{A} - \text{B}})$.