Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$in degrees for $\sqrt{3}\overrightarrow{a}-\overrightarrow{b}$ to be unit vector is:
A. 60
B. 45
C. 30
D. 90

Answer 1

Hint: First will understand what unit vector is and then we will use the information that is given in the question and we will express the two vectors in form of $\widehat{i}$ and $\widehat{j}$ , after doing that we put the modulus of $\sqrt{3}\overrightarrow{a}-\overrightarrow{b}$= 1, as it is a unit vector.

Complete step-by-step answer:
Now we will state what the unit vector is:
Unit vector: A unit vector in a normed vector space is a vector of length 1.
Now we will express the two vectors in form of $\widehat{i}$ and $\widehat{j}$ ,
$\begin{aligned} & \overrightarrow{a}=(\cos \alpha )\widehat{i}+(\sin \alpha )\widehat{j} \\\ & \overrightarrow{b}=(\cos \beta )\widehat{i}+(\sin \beta )\widehat{j} \\\ \end{aligned}$
Here $\beta$ and $\alpha$ are the angles made by the two vectors with the x – axis.
Now we will put the values of $\overrightarrow{a}$ and $\overrightarrow{b}$ in $\sqrt{3}\overrightarrow{a}-\overrightarrow{b}$,
$\begin{aligned} & \sqrt{3}\left( (\cos \alpha )\widehat{i}+(\sin \alpha )\widehat{j} \right)-\left( (\cos \beta )\widehat{i}+(\sin \beta )\widehat{j} \right) \\\ & \left( \sqrt{3}\cos \alpha -\cos \beta \right)\widehat{i}+\left( \sqrt{3}\sin \alpha -\sin \beta \right)\widehat{j} \\\ \end{aligned}$
As it is given that this is a unit vector, hence it’s magnitude must be equal to 1.
$\begin{aligned} & \sqrt{{{\left( \sqrt{3}\cos \alpha -\cos \beta \right)}^{2}}+{{\left( \sqrt{3}\sin \alpha -\sin \beta \right)}^{2}}}=1 \\\ & {{\left( \sqrt{3}\cos \alpha -\cos \beta \right)}^{2}}+{{\left( \sqrt{3}\sin \alpha -\sin \beta \right)}^{2}}=1 \\\ & 3{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta -2\sqrt{3}\cos \alpha \cos \beta +3{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta -2\sqrt{3}\sin \alpha \sin \beta =1 \\\ \end{aligned}$
Now we are going to use these formula,
$\begin{aligned} & {{\cos }^{2}}x+{{\sin }^{2}}x=1 \\\ & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\\ & \cos \left( x-y \right)=\cos x\cos y+\sin x\sin y \\\ \end{aligned}$
After using these formula in the above equation we get,
$\begin{aligned} & 3({{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha )+({{\sin }^{2}}\beta +{{\cos }^{2}}\beta )-2\sqrt{3}(\cos \alpha \cos \beta +\sin \alpha \sin \beta )=1 \\\ & 3+1-2\sqrt{3}\cos \left( \alpha -\beta \right)=1 \\\ & 3=2\sqrt{3}\cos \left( \alpha -\beta \right) \\\ & \cos \left( \alpha -\beta \right)=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$
We have been asked to find the value of angle between the two lines which is $\alpha -\beta$ .
And $\cos {{60}^{0}}=\dfrac{\sqrt{3}}{2}$ ,
Hence, from the above equation we can conclude that $\alpha -\beta$= ${{60}^{0}}$ .
Hence, option (a) is the correct answer.

Note: We can also solve this question by taking two unit vectors instead of taking it as a variable, that will be easy to solve and will also take less time. Representing vectors into angles made with the x and y axis is important. Since its asked angle between the two vectors and no difference of alpha and beta is required.