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This is the second of our many quizzes on Vector basics. Click on this to access more such competitive quizzes on Vectors.

Vector basics Quiz 2

mathematics

 A student rotates the co – ordinate axis to some angle so that the X – axis is along the vector $8\hat{i}+15\hat{j}$. Now, he wants to find the unit vector along the new X – axis. Help him to find it.

 Use the condition that the unit vector along $\vec{A}$ is given as $\hat{A}=\dfrac{{\vec{A}}}{\left| {\vec{A}} \right|}$ and then substitute the given vector to get the required answer.

 $\text{Assume }\vec{A}=8\hat{i}+15\hat{j} \\  \text{The required unit vector is,} \\  \Rightarrow \hat{A}=\dfrac{{\vec{A}}}{\left| {\vec{A}} \right|} \\  \Rightarrow \hat{A}=\dfrac{8\hat{i}+15\hat{j}}{\sqrt{{{8}^{2}}+{{15}^{2}}}} \\  \Rightarrow \hat{A}=\dfrac{1}{17}\left( 8\hat{i}+15\hat{j} \right)$

 Two points A and B are located at $\left( 5,10 \right)$ and $\left( 6,13 \right)$ find the vector form of $\overset{\to }{\mathop{AB}}\,$

 Find the position vectors using the co – ordinates of given points. Then obtain the required vector using the concept of vectors that form a triangle.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/da0e402d-3618-49e2-be77-3dafe9eb6a855636921817493043997.png"/>  $ \\ \text{The coordinates of points are }A=\left( 5,10 \right);B=\left( 6,13 \right) \\  \text{The position vectors of given points are,} \\  \Rightarrow \overset{\to }{\mathop{OA}}\,=5\hat{i}+10\hat{j} \\  \Rightarrow \overset{\to }{\mathop{OB}}\,=6\hat{i}+13\hat{j} \\  \text{The required vector is,} \\  \Rightarrow \overset{\to }{\mathop{AB}}\,=\overset{\to }{\mathop{OB}}\,-\overset{\to }{\mathop{OA}}\, \\  \Rightarrow \overset{\to }{\mathop{AB}}\,=\left( 5\hat{i}+10\hat{j} \right)-\left( 6\hat{i}+13\hat{j} \right) \\  \Rightarrow \overset{\to }{\mathop{AB}}\,=-\hat{i}-3\hat{j}$

 A bird is resting on a top of a tree of height 10m. Suddenly it flew to a distance of 20m making an angle ${{45}^{\circ }}$ with the horizontal and landed on ground at a distance of 50m from the foot of the tree. Find the displacement from initial position to final position of bird.

 Draw the graph in a co – ordinate plane and check for vectors that needed to find the required displacement.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0dd206d1-e7fd-46a5-82a9-36664ae4783a3942805816564423140.png"/> $ \\   \text{From figure,} \\  \Rightarrow \overset{\to }{\mathop{OA}}\,=10\hat{j} \\  \Rightarrow \overset{\to }{\mathop{AC}}\,=50\hat{i} \\  \text{The resultant of two vectors from }\Delta OAC\text{ is,} \\  \Rightarrow \overset{\to }{\mathop{OA}}\,+\overset{\to }{\mathop{AC}}\,=\overset{\to }{\mathop{OC}}\, \\  \Rightarrow 10\hat{j}+\overset{\to }{\mathop{AC}}\,=50\hat{i} \\  \Rightarrow \overset{\to }{\mathop{AC}}\,=50\hat{i}-10\hat{j} \\  \Rightarrow \left| \overset{\to }{\mathop{AC}}\, \right|=\sqrt{{{\left( 50 \right)}^{2}}+{{\left( 10 \right)}^{2}}} \\  \Rightarrow \left| \overset{\to }{\mathop{AC}}\, \right|=\sqrt{2600}=10\sqrt{26}$

 If there are two vectors $\vec{a}=\hat{i}+\hat{j}$ and $\vec{b}=\hat{j}+\hat{k}$ then find the value of $\left| \vec{a}\cdot \vec{b} \right|+\left| \vec{a}\times \vec{b} \right|$

 Find the dot product and cross product of given vectors. Then find the magnitude and then add them up to get the required answer.

 $ \text{Dot product,} \\  \Rightarrow \vec{a}\cdot \vec{b}=\left( \hat{i}+\hat{j} \right)\cdot \left( \hat{j}+\hat{k} \right) \\  \Rightarrow \vec{a}\cdot \vec{b}=1 \\  \Rightarrow \left| \vec{a}\cdot \vec{b} \right|=1 \\  \text{Now, cross product,} \\  \Rightarrow \vec{a}\times \vec{b}=\left| \begin{matrix}   {\hat{i}} {\hat{j}} {\hat{k}}  \\   1 1 0  \\   0 1 1  \\\end{matrix} \right| \\  \Rightarrow \vec{a}\times \vec{b}=\hat{i}+\hat{k} \\  \Rightarrow \left| \vec{a}\times \vec{b} \right|=\sqrt{{{1}^{2}}+{{1}^{2}}}=\sqrt{2} \\  \text{The required value is,} \\  \Rightarrow \left| \vec{a}\cdot \vec{b} \right|+\left| \vec{a}\times \vec{b} \right|=1+\sqrt{2}$

 There are two vectors $\bar{A}=6\hat{i}+8\hat{j}$ and $\vec{B}=12\hat{i}+5\hat{j}$. Arun and Sandeep calculated a vector parallel to $\vec{A}$ with magnitude of $\vec{B}$ and gave the answers like this: $ \\ $ Arun: The required vector is $\dfrac{13}{5}\left( 2\hat{i}+4\hat{j} \right)$ $ \\ $ Sandeep: The required vector is $\dfrac{10}{13}\left( 12\hat{i}+5\hat{j} \right)$ $ \\ $ Who is correct?

 Find the unit vector along which the required vector is parallel and then multiply the magnitude of the required vector to the unit vector to get the required answer.

 $\text{The required vector is parallel }\vec{A} \\ \text{So, the unit vector along }\vec{A}\text{ is,} \\  \Rightarrow \hat{A}=\dfrac{{\vec{A}}}{\left| {\vec{A}} \right|} \\  \Rightarrow \hat{A}=\dfrac{6\hat{i}+8\hat{j}}{\sqrt{{{6}^{2}}+{{8}^{2}}}} \\  \Rightarrow \hat{A}=\dfrac{1}{5}\left( 2\hat{i}+4\hat{j} \right) \\  \text{Now, the required vector is,} \\  \Rightarrow \vec{C}=\left| {\vec{B}} \right|.\hat{A} \\  \Rightarrow \vec{C}=\sqrt{{{12}^{2}}+{{5}^{2}}}\left( \dfrac{1}{5}\left( 2\hat{i}+4\hat{j} \right) \right) \\  \Rightarrow \vec{C}=\dfrac{13}{5}\left( 2\hat{i}+4\hat{j} \right) \\  \text{So, Arun is correct}\text{.}$