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This is second of our many quizzes on Vectors and scalars. Click on this to access more such competitive quizzes on Mechanics.

Vectors and scalars Quiz 2

physics

If $\sqrt{3}\left( \vec{A}\times \vec{B} \right)=\vec{A}.\vec{B}$ then angle between $\vec{A}$ and $\vec{B}$ is  

Use $\vec{A}\times \vec{B}=AB\sin \theta $ and $\vec{A}.\vec{B}=AB\cos \theta $ to solve. 

  $\sqrt{3}\left( \vec{A}\times \vec{B} \right)=\vec{A}.\vec{B} \\ \Rightarrow \sqrt{3}AB\sin \theta =AB\cos \theta  \\ \Rightarrow \tan \theta =\dfrac{1}{\sqrt{3}} \\ \therefore \theta =30{}^\circ$

Prachi is sitting in a car moving in the north direction with velocity $20\,m/s$. She saw Rashmi moving eastwards with a speed of $15\,m/s$ and said “Hi” to her. A person watches both of them resting on the ground. The actual direction of motion of Rashmi will be 

 Find actual velocity of Rashmi with respect to ground by using the concept of relative motion.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/96f8f864-450f-4de5-8c42-dfb031c5213c7019094584436532087.png"/>$ \\ $ Here, prachi observes Rashmi moving eastwards relatively to her. $ \\ $ For a stationary observer on ground Rashmi may move in a different direction. $ \\ $ Let velocity of Prachi with respect to ground$={{{\vec{V}}}_{PG}}$ $ \\ $ Velocity of Rashmi with respect to ground$={{{\vec{V}}}_{RG}}$ $ \\ $ Velocity of Rashmi with respect to Prachi $={{{\vec{V}}}_{RP}} $ $ \\ $ Given, ${{{\vec{V}}}_{PG}}$ is along North $ \\ \therefore {{{\vec{V}}}_{PG}}=20\hat{j} \\As {{\vec{V}}}_{\text{RP}}\text{ is along East} \\{{\vec{V}}}_{\text{RP}}=15\hat{i} \\ As {{\vec{V}}}_{\text{RP}}={{{\vec{V}}}_{RG}}-{{{\vec{V}}}_{PG}} \\ 15\hat{i}={{{\vec{V}}}_{RG}}-20\hat{j} \\{{{\vec{V}}}_{RG}}=15\hat{i}+20\hat{j} \\ $ Thus, Rashmi is seem to be moving North East by an observer on ground.

 If $\vec{A}$, $\vec{B}$ and $\vec{C}$ are mutually perpendicular to each other then.  

 If three vectors are mutually perpendicular, that is, if the first two vectors have an angle of $90^\circ $ with each other then a dot product generates a scalar number, while a cross-product produces a vector perpendicular to the vectors used in the cross-product.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e8d2b57c-cf27-48f2-a46c-09749c6b13787165811675544084062.png"/>$ \\ $ Let vector $\vec{A}\text{, }\vec{B}$ and $\vec{C}$ are in $\hat{i}\text{, }\hat{j}$ and $\hat{k}$ respectively. $ \\ \text{(i) }\vec{A}\times \vec{B}\times \vec{C}=\left( \vec{A}\times \vec{B} \right)\times \vec{C}=\left( \hat{i}\times \hat{j} \right)\times \hat{k}=\hat{k}\times \hat{k}=0 \\ \text{(ii) }\vec{A}.\vec{B}.\vec{C}=\left( \vec{A}.\vec{B} \right).\vec{C}=\left( AB \right).\vec{C}= $ not possible as AB is a scalar quantity.

$\vec{a}=\hat{i}+2\hat{j}$, $\vec{b}=3\hat{i}+4\hat{j}$, $\vec{c}=5\hat{i}+6\hat{j}$ then value of $\left( \vec{a}+\vec{c} \right).\hat{b}$. 

 Find unit vector of $\vec{b}$ first by given expression $\hat{b}=\dfrac{{\vec{b}}}{\left| {\vec{b}} \right|}$ 

 $\vec{a}+\vec{c}=\left( 1+5 \right)\hat{i}+8\hat{j}=6\hat{i}+8\hat{j} \\ \Rightarrow \hat{b}=\dfrac{{\vec{b}}}{\left| {\vec{b}} \right|}\\ \Rightarrow \hat{b}=\dfrac{3\hat{i}+4\hat{j}}{\sqrt{{{3}^{2}}+{{4}^{2}}}}\\ \Rightarrow \hat{b}=\dfrac{3\hat{i}+4\hat{j}}{5} \\ \Rightarrow \left( \vec{a}+\vec{c} \right).\hat{b}=\left( 6\hat{i}+8\hat{j} \right).\dfrac{\left( 3\hat{i}+4\hat{j} \right)}{5} \\ \Rightarrow \left( \vec{a}+\vec{c} \right).\hat{b}=\dfrac{18+32}{5}=\dfrac{50}{5} \\ \therefore \left( \vec{a}+\vec{c} \right).\hat{b}=10$

 A ball is dropped from a height $''H''$ from a building. At the same time a boy moves towards the building to catch the ball. The motion of boy with respect to the ball will be $(\text{Assuming boy is moving with constant velocity})$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a7478789-aa10-402f-b2e2-ba2da657700a125050083705731884.png"/> 

Use concept of relative motion and find relative velocity of boy with respect to ball. 

 Ball at height $H$ dropped from rest, initial velocity of ball will be zero $ \\{{V}_{ball}}=u+gt\left( \therefore {{V}_{ball}}=\text{ Velocity of ball at instant t} \right) \\ \Rightarrow {{V}_{ball}}=u+gt \\ \Rightarrow {{V}_{ball}}=gt\left( -\hat{j} \right)m/s \\ $ As the boy is moving towards positive x axis with velocity V, $ \\ \therefore \,\,{{V}_{boy}}=V\hat{i}\text{ m/s, } \\ $ Relative velocity of boy with respect to ball $={{V}_{boy}}-{{V}_{ball}}  \\ \Rightarrow V\hat{i}+gt\hat{j}\text{ m/s} \\ $ Presence of gravitational acceleration $g$ and time $t$ makes the boy seem to have accelerated motion w.r.t ball.