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

Unit vectors Quiz 2

mathematics

 Two position vectors of points Y and Z are given as $\vec{y}=2\hat{i}+4\hat{j}-a\hat{k}$ and $\vec{z}=3\hat{i}-5\hat{j}+4\hat{k}$. If the  resultant is a unit vector, calculate the value of $a$ .

 The resultant of two vectors is given by the vector addition of two vectors.

$\begin{align} &nbsp;&nbsp;& \text{The resultant of the }\vec{y}\text{ and }\vec{z}\text{ is given by;} \\ &nbsp;& \vec{R}=\vec{y}+\vec{z} \\ &nbsp;& \vec{R}=2\hat{i}+4\hat{j}-a\hat{k}+3\hat{i}-5\hat{j}+4\hat{k} \\ &nbsp;& \vec{R}=5\hat{i}-\hat{j}+\left( 4-a \right)\hat{k} \\ &nbsp;& \text{The modulus of the resultant is given by;} \\ &nbsp;& \left| {\vec{R}} \right|=\sqrt{{{5}^{2}}+{{1}^{2}}+{{\left( 4-a \right)}^{2}}} \\ &nbsp;& \Rightarrow \left| {\vec{R}} \right|=\sqrt{26+16+{{a}^{2}}-8a} \\ &nbsp;& \text{Given that;} \\ &nbsp;& \left| {\vec{R}} \right|=1 \\ &nbsp;& \Rightarrow \sqrt{{{a}^{2}}-8a+42}=1 \\ &nbsp;& \Rightarrow {{a}^{2}}-8a+42=1 \\ &nbsp;& \Rightarrow {{a}^{2}}-8a+41=0 \\ &nbsp;& \text{The discriminant of the above equation is given by;} \\ &nbsp;& \Delta ={{b}^{2}}-4ac={{\left( -8 \right)}^{2}}-4\times 1\times 41 \\ &nbsp;& \Rightarrow \Delta =64-164=-100 \\ &nbsp;& \Rightarrow \Delta <0 \\ &nbsp;& \text{Therefore, the resultant cannot be a unit vector}\text{.} \\ &nbsp;& \text{So, option }\left( d \right)\text{ is the correct answer}\text{.} \\ \end{align}$

Imagine two points P and Q in the three-dimensional coordinate system, such that $OP=2\hat{i}-\hat{k}$ and $OQ=3\hat{i}+z\hat{k}$ , find the value of $z$ . <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5fc9ff70-4377-42b0-80d8-7a92040468672731052859242528612.png"/>

 The distance between the two given points P and Q is the vector difference of their position vectors with respect to the origin.

\[\begin{align} &nbsp;&nbsp;& \text{The vector }QP\text{ is given as;} \\ &nbsp;& QP=OQ-OP \\ &nbsp;& \Rightarrow QP=3\hat{i}+z\hat{k}-\left( 2\hat{i}-\hat{k} \right) \\ &nbsp;& \text{Given that;} \\ &nbsp;& \left| QP \right|=1 \\ &nbsp;& \Rightarrow \left| QP \right|=\sqrt{1+{{\left( z+1 \right)}^{2}}}=1 \\ &nbsp;& \Rightarrow 1+{{z}^{2}}+2z+1=1 \\ &nbsp;& \Rightarrow {{z}^{2}}+2z+1=0 \\ &nbsp;& \text{The roots of the above equations are;} \\ &nbsp;& z=-1,-1 \\ &nbsp;& \therefore z=-1 \\ \end{align}\]

 An ellipse on the two-dimensional plane is given as ${{x}^{2}}+3{{y}^{2}}=3$ , two points A and B lie on the ellipse such that the x coordinate of A is $\sqrt{2}$ and y coordinate of B is $-1$ . Determine the unit vector in the direction of $AB$ such that $AB$ lies on the right side of the Y-axis.

 The ordinate of the point B is on the minor axis of the ellipse and the extreme points of the minor axis are $\left( 0,1 \right)$ and $\left( 0,-1 \right)$

$\begin{align} &nbsp;&nbsp;& \text{For the point A;} \\ &nbsp;& {{x}_{a}}=\sqrt{2} \\ &nbsp;& \Rightarrow {{y}_{a}}=\sqrt{\dfrac{3-x_{a}^{2}}{3}}=\pm \dfrac{1}{\sqrt{3}} \\ &nbsp;& \text{Given that the vector is in the first quadrant;} \\ &nbsp;& \text{So, }{{y}_{a}}=\dfrac{1}{\sqrt{3}} \\ &nbsp;& OA=\sqrt{2}\hat{i}+\dfrac{1}{\sqrt{3}}\hat{j} \\ &nbsp;& \text{For point B;} \\ &nbsp;& {{y}_{b}}=-1 \\ &nbsp;& \Rightarrow {{x}_{b}}=\sqrt{3-3\left( y_{b}^{2} \right)}=0 \\ &nbsp;& \text{The required vector is given as;} \\ &nbsp;& AB=OA-OB \\ &nbsp;& \Rightarrow AB=\sqrt{2}\hat{i}+\dfrac{1}{\sqrt{3}}\hat{j}-\left( -\hat{j} \right) \\ &nbsp;& \Rightarrow AB=\sqrt{2}\hat{i}+\left( \dfrac{1+\sqrt{3}}{\sqrt{3}} \right)\hat{j} \\ &nbsp;& \Rightarrow \left| AB \right|=\sqrt{2+{{\left( \dfrac{1+\sqrt{3}}{\sqrt{3}} \right)}^{2}}}=\sqrt{\dfrac{10+2\sqrt{3}}{3}} \\ &nbsp;& \text{Unit vector is given as;} \\ &nbsp;& ab=\dfrac{AB}{\left| AB \right|}=\dfrac{\sqrt{2}\hat{i}+\left( \dfrac{1+\sqrt{3}}{\sqrt{3}} \right)\hat{j}}{\sqrt{\dfrac{10+2\sqrt{3}}{3}}} \\ &nbsp;& \therefore ab=\dfrac{\left[ \sqrt{2}\hat{i}+\left( \dfrac{1+\sqrt{3}}{\sqrt{3}} \right)\hat{j} \right]\sqrt{3}}{\sqrt{10+2\sqrt{3}}} \\ \end{align}$

 A vector $\vec{a}=\dfrac{1}{2}\hat{i}-\dfrac{1}{\sqrt{2}}\hat{j}+x\hat{k}$ is on a three-dimensional coordinate system such that its dot product with a vector along the Z-axis having magnitude as $10\text{ unit}$ is $5$ . Determine which type of vector is $\vec{a}$

 For determining the type of vector, the magnitude must be evaluated first. The magnitude can only be evaluated when the value of $x$ is known.

$\begin{align} &nbsp;&nbsp;& \text{Given that;} \\ &nbsp;& \vec{a}\cdot \left( 10\hat{k} \right)=5 \\ &nbsp;& \text{Substituting the values;} \\ &nbsp;& \Rightarrow \left[ \dfrac{1}{2}i-\dfrac{1}{\sqrt{2}}j+xk \right]\cdot \left( 10k \right)=5 \\ &nbsp;& \Rightarrow 0-0+10x=5 \\ &nbsp;& \Rightarrow x=\dfrac{1}{2} \\ &nbsp;& \text{Modulus of }\vec{a}\text{ is given by;} \\ &nbsp;& \Rightarrow \left| {\vec{a}} \right|=\sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{-1}{\sqrt{2}} \right)}^{2}}+{{\left( \dfrac{1}{2} \right)}^{2}}} \\ &nbsp;& \Rightarrow \left| {\vec{a}} \right|=\sqrt{\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{4}} \\ &nbsp;& \Rightarrow \left| {\vec{a}} \right|=\sqrt{1} \\ &nbsp;& \therefore \left| {\vec{a}} \right|=1 \\ &nbsp;& \text{Therefore, }\vec{a}\text{ is a unit vector}\text{.} \\ \end{align}$

For a certain vector $\vec{R}$ the magnitude of the vector changes with time, determine the times for which the vector is a unit vector. The three components vary as follows: $\begin{align} &nbsp;&nbsp;& x\left( t \right)=t \\ &nbsp;& y\left( t \right)=4+t \\ &nbsp;& z\left( t \right)=5 \\ \end{align}$

 The given vector is assumed to be a unit vector and further solved for the unknown variable.

$\begin{align} &nbsp;&nbsp;& \text{Let the vector }\vec{R}\text{ be given as;} \\ &nbsp;& \vec{R}=x\hat{i}+y\hat{j}+z\hat{k} \\ &nbsp;& \text{According to the given condition;} \\ &nbsp;& \left| {\vec{R}} \right|=1 \\ &nbsp;& \Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=1 \\ &nbsp;& \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1 \\ &nbsp;& \Rightarrow {{t}^{2}}+{{\left( 4+t \right)}^{2}}+{{5}^{2}}=1 \\ &nbsp;& \Rightarrow {{t}^{2}}+16+{{t}^{2}}+8t+25=1 \\ &nbsp;& \Rightarrow 2{{t}^{2}}+8t+40=0 \\ &nbsp;& \Rightarrow {{t}^{2}}+4t+20=0 \\ &nbsp;& \text{The discriminant of the above equation is given as;} \\ &nbsp;& \Rightarrow \Delta ={{b}^{2}}-4ac \\ &nbsp;& \Rightarrow \Delta =16-4\times 1\times 20 \\ &nbsp;& \Rightarrow \Delta =16-80<0 \\ &nbsp;& \text{Therefore, there are no real roots,} \\ &nbsp;& \text{So, there will be no time for which the vector }\vec{R}\text{ will be a unit vector}\text{.} \\ \end{align}$