Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Vectors. Click on this to access more such competitive quizzes on Vectors.

Vectors Quiz 2

mathematics

  Two vectors on a 2D cartesian plane are given as $\left( 16,13 \right)$ and $\left( 52,14 \right)$. Using the aforementioned, calculate the sum of the two vectors.

   Name the two vectors as $\overrightarrow{u}$ and $\overrightarrow{v}$. You should next identify the value of ${{u}_{1}},{{u}_{2}},{{v}_{1}}$ and ${{v}_{2}}$ to add the two vectors using the formula $\overrightarrow{u}+\overrightarrow{v}=\left( {{u}_{1}}+{{v}_{1}},{{u}_{2}}+{{v}_{2}} \right)$ to calculate the sum. 

$ \overrightarrow{u}=\left( 16,13 \right) \\ \overrightarrow{v}=\left( 52,14 \right) \\ \overrightarrow{u}+\overrightarrow{v}=\left( {{u}_{1}}+{{v}_{1}},{{u}_{2}}+{{v}_{2}} \right) \\ \Rightarrow \overrightarrow{u}+\overrightarrow{v}=\left( 16+52,13+14 \right) \\ \Rightarrow \overrightarrow{u}+\overrightarrow{v}=\left( 68,27 \right) $

 A student is trying to calculate the value of a vector given the difference between them. If the vectors are $\left( 10,12 \right)$ and $\left( x,y \right)$ with their difference being $\left( 16,21 \right)$, calculate the value of vector $\left( x,y \right)$. 

 Name the two vectors as $\overrightarrow{u}$ and $\overrightarrow{v}$. You should next identify the value of ${{u}_{1}},{{u}_{2}},x$ and $y$ to subtract the two vectors using the formula $\overrightarrow{u}-\overrightarrow{v}=\left( {{u}_{1}}-x,{{u}_{2}}-y \right)$ to calculate the value of the unknown vector.

 $ \overrightarrow{u}=\left( 10,12 \right) \\ \overrightarrow{v}=\left( x,y \right) \\ \overrightarrow{u}-\overrightarrow{v}=\left( {{u}_{1}}-x,{{u}_{2}}-y \right) \\ \Rightarrow \left( 16,21 \right)=\left( 10-x,12-y \right) \\ \Rightarrow x=-6,y=-9 \\ \Rightarrow \left( x,y \right)=\left( -6,-9 \right)$

  A student is trying to calculate the cross product of the vector $x=\left( 5,7,5 \right)$ and $y=\left( 7,5,7 \right)$. Determine the cross product of the aforementioned vectors. 

 Put the values of the vectors $x$ and $y$ into the formula of a cross product $\overrightarrow{x}\times \overrightarrow{y}=\left| \begin{matrix}   i & j & k  \\   {{x}_{1}} & {{x}_{2}} & {{x}_{3}}  \\   {{y}_{1}} & {{y}_{2}} & {{y}_{3}}  \\\end{matrix} \right|$ and use it to solve the determinant of the matrix. 

$ \overrightarrow{x}\times \overrightarrow{y}=\left| \begin{matrix} &nbsp;&nbsp;&nbsp;i & j & k \\ &nbsp;&nbsp;&nbsp;{{x}_{1}} & {{x}_{2}} & {{x}_{3}} \\ &nbsp;&nbsp;&nbsp;{{y}_{1}} & {{y}_{2}} & {{y}_{3}} \\ \end{matrix} \right| \\ \Rightarrow \overrightarrow{x}\times \overrightarrow{y}=\left| \begin{matrix} &nbsp;&nbsp;&nbsp;i & j & k \\ &nbsp;&nbsp;&nbsp;5 & 7 & 5 \\ &nbsp;&nbsp;&nbsp;7 & 5 & 7 \\ \end{matrix} \right| \\ \Rightarrow \overrightarrow{x}\times \overrightarrow{y}=i\left( 7\times 7-5\times 5 \right)-j\left( 5\times 7-7\times 5 \right)+k\left( 5\times 5-7\times 7 \right) \\ \Rightarrow \overrightarrow{x}\times \overrightarrow{y}=i\left( 49-25 \right)-j\left( 35-35 \right)+k\left( 25-49 \right) \\ \Rightarrow \overrightarrow{x}\times \overrightarrow{y}=24i-24k $

 A professor has given a question to his students based on the concept of dot product. If $\left\| \overrightarrow{a} \right\|=5$ and $\left\| \overrightarrow{b} \right\|=\dfrac{\sqrt{2}}{10}$. Calculate the dot product of the two vectors if the angle formed between them is $\dfrac{\pi }{4}$.

 Identify the magnitude from the question and put them in the formula for the geometrical representation of the dot product of two vectors $\overrightarrow{a}\cdot \overrightarrow{b}=\left\| \overrightarrow{a} \right\|\cdot \left\| \overrightarrow{b} \right\|\cos \theta $. Then use the given angle between the two vectors and solve the equation to calculate the dot product. 

 $\overrightarrow{a}\cdot \overrightarrow{b}=\left\| \overrightarrow{a} \right\|\cdot \left\| \overrightarrow{b} \right\|\cos \theta  \\ \Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=5\cdot \dfrac{\sqrt{2}}{10}\cdot \cos \dfrac{\pi }{4} \\ \Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=5\cdot \dfrac{\sqrt{2}}{10}\cdot \cos \dfrac{\pi }{4} \\ \Rightarrow \dfrac{\sqrt{2}}{2}\cdot \dfrac{\sqrt{2}}{2}=\dfrac{1}{2} $

 A student is trying to find a vector which has the same direction of a vector $20\widehat{i}-\sqrt{200}\widehat{j}+5\widehat{k}$. If the vector the student is trying to find has a magnitude of $32$ units, then find the vector. 

  Assign a variable to the given vector and calculate its magnitude. Using the magnitude, you can calculate the value of $\widehat{u}$ by using the formula $\widehat{u}=\dfrac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}$. Next, you can again use $\widehat{u}=\dfrac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}$ again to calculate the vector with the magnitude of $32$ units. 

 $ \text{Let }\overrightarrow{u}=20\widehat{i}-\sqrt{200}\widehat{j}+5\widehat{k}\\ \left\| \overrightarrow{u} \right\|=\sqrt{{{\left( 20 \right)}^{2}}+{{\left( -\sqrt{200} \right)}^{2}}+{{\left( 5 \right)}^{2}}} \\ \Rightarrow \left\| \overrightarrow{u} \right\|=\sqrt{400+200+25}=\sqrt{625}=25 \\ \Rightarrow \widehat{u}=\dfrac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}=\dfrac{20\widehat{i}-\sqrt{200}\widehat{j}+5\widehat{k}}{25} \\ \text{For another vector to be in the same direction as }\\ \overrightarrow{u}=20\widehat{i}-\sqrt{200}\widehat{j}+5\widehat{k}\\ \text{and have a magnitude of 32 units can be calculated by}\\ 32\widehat{u}=32\left( \dfrac{20\widehat{i}-\sqrt{200}\widehat{j}+5\widehat{k}}{25} \right)=\dfrac{640}{25}\widehat{i}-\dfrac{32\sqrt{200}}{25}\widehat{j}+\dfrac{160}{25}\widehat{k} \\ \Rightarrow \dfrac{128}{5}\widehat{i}-\dfrac{64\sqrt{2}}{5}\widehat{j}+\dfrac{32}{5}\widehat{k} $