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Question

Question: Write all the unit vectors in XY- plane....

Write all the unit vectors in XY- plane.

Explanation

Solution

For solving this question you should know about the unit vectors in planes. A unit vector in the XY- plane indicates that there is no Z-axis, we have to find the value of a=xi^+yj^\overrightarrow{a}=x\widehat{i}+y\widehat{j} and for that we will let a unit vector a\overrightarrow{a} and using the formula of dot product a.b=abcosθ\overrightarrow{a}.\overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \theta we will find the value of X and Y.

Complete step-by-step solution:
According to our question it is asked to write all the unit vectors in the XY- plane. Let the unit vector be a\overrightarrow{a}, we know that a=xi^+yj^+zk^\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}, where x-axis is i^\widehat{i}, y-axis is j^\widehat{j} and z-axis isk^\widehat{k}. Since the vector is in XY-plane, that means there is no z-coordinate (z = 0). Then,
a=xi^+yj^+0k^ a=xi^+yj^ \begin{aligned} & \overrightarrow{a}=x\widehat{i}+y\widehat{j}+0\widehat{k} \\\ & \Rightarrow \overrightarrow{a}=x\widehat{i}+y\widehat{j} \\\ \end{aligned}
The unit vector in the direction of x-axis is i^\widehat{i} and y-axis is j^\widehat{j}.

Angle with x-axis:
Since a\overrightarrow{a} makes an angle of θ\theta with x-axis, so angle between a\overrightarrow{a} and i^\widehat{i} is θ\theta .
We know that: a.b=abcosθ\overrightarrow{a}.\overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \theta
So, putting a=a,b=i,^θ=θ\overrightarrow{a}=\overrightarrow{a,}\overrightarrow{b}=\widehat{i,}\theta =\theta
a.i^=ai^cosθ\overrightarrow{a}.\widehat{i}=\left| \overrightarrow{a} \right|\left| \widehat{i} \right|\cos \theta
As a\overrightarrow{a} is a unit vector, a=1\left| \overrightarrow{a} \right|=1 and i^\widehat{i} is a unit vector, i^=1\left| \widehat{i} \right|=1. So,
a.i^=1×1×cosθ a.i^=cosθ (xi^+yj^+0k^).i^=cosθ (xi^+yj^+0k^).(1i^+0j^+0k^)=cosθ x.1+y.0+0.0=cosθ x=cosθ \begin{aligned} & \overrightarrow{a}.\widehat{i}=1\times 1\times \cos \theta \\\ &\Rightarrow \overrightarrow{a}.\widehat{i}=\cos \theta \\\ &\Rightarrow \left( x\widehat{i}+y\widehat{j}+0\widehat{k} \right).\widehat{i}=\cos \theta \\\ &\Rightarrow \left( x\widehat{i}+y\widehat{j}+0\widehat{k} \right).\left( 1\widehat{i}+0\widehat{j}+0\widehat{k} \right)=\cos \theta \\\ &\Rightarrow x.1+y.0+0.0=\cos \theta \\\ &\Rightarrow x=\cos \theta \\\ \end{aligned}

Angle with y-axis:
Since a\overrightarrow{a} makes an angle of (90θ)\left( {{90}^{\circ }}-\theta \right) with y-axis, so angle between a\overrightarrow{a} and j^\widehat{j} is (90θ)\left( {{90}^{\circ }}-\theta \right).
We know that: a.b=abcosθ\overrightarrow{a}.\overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \theta
So, putting a=a,b=j,^θ=(90θ)\overrightarrow{a}=\overrightarrow{a,}\overrightarrow{b}=\widehat{j,}\theta =\left( {{90}^{\circ }}-\theta \right)
a.j^=aj^cos(90θ) a.j^=1×1×cos(90θ) a.j^=cos(90θ) a.j^=sinθ (xi^+yj^+0k^).j^=sinθ (xi^+yj^+0k^).(0i^+1j^+0k^)=sinθ x.0+y.1+0.0=sinθ y=sinθ \begin{aligned} & \overrightarrow{a}.\widehat{j}=\left| \overrightarrow{a} \right|\left| \widehat{j} \right|\cos \left( {{90}^{\circ }}-\theta \right) \\\ &\Rightarrow \overrightarrow{a}.\widehat{j}=1\times 1\times \cos \left( {{90}^{\circ }}-\theta \right) \\\ &\Rightarrow \overrightarrow{a}.\widehat{j}=\cos \left( {{90}^{\circ }}-\theta \right) \\\ &\Rightarrow \overrightarrow{a}.\widehat{j}=\sin \theta \\\ &\Rightarrow \left( x\widehat{i}+y\widehat{j}+0\widehat{k} \right).\widehat{j}=\sin \theta \\\ &\Rightarrow \left( x\widehat{i}+y\widehat{j}+0\widehat{k} \right).\left( 0\widehat{i}+1\widehat{j}+0\widehat{k} \right)=\sin \theta \\\ &\Rightarrow x.0+y.1+0.0=\sin \theta \\\ &\Rightarrow y=\sin \theta \\\ \end{aligned}
Thus,
a=xi^+yj^ a=cosθi^+sinθj^ \begin{aligned} & \overrightarrow{a}=x\widehat{i}+y\widehat{j} \\\ &\Rightarrow \overrightarrow{a}=\cos \theta \widehat{i}+\sin \theta \widehat{j} \\\ \end{aligned}
This value will be true in all quadrants. So, 0θ2π0\le \theta \le 2\pi .
Therefore a=cosθi^+sinθj^\overrightarrow{a}=\cos \theta \widehat{i}+\sin \theta \widehat{j}; for 0θ2π0\le \theta \le 2\pi .

Note: While solving this type of questions you should be careful of the angles from the x-axis and y-axis and always, we have to make a unit vector. And this will be mandatory to be in the XY-plane. And if it is in the XY-plane, then the rest Z-plane coordinate must be always zero.