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Question: Which of the following is not a unit vector for all values of \(\theta \) A. \[\cos (\theta )\hat...

Which of the following is not a unit vector for all values of θ\theta
A. cos⁡(θ)i^−sin⁡(θ)j^\cos (\theta )\hat i - \sin (\theta )\hat j
B. sin⁡(θ)i^−cos⁡(θ)j^\sin (\theta )\hat i - \cos (\theta )\hat j
C. sin⁡(2θ)i^−cos⁡(θ)j^\sin (2\theta )\hat i - \cos (\theta )\hat j
D. cos⁡(2θ)i^−sin⁡(2θ)j^\cos (2\theta )\hat i - \sin (2\theta )\hat j

Explanation

Solution

First, we will find the magnitude of all the vectors given to us. The vectors whose magnitude is equal to one are known as unit vectors, Hence we need to find out the vector whose magnitude is not equal to one. Hence, all the vectors whose magnitude is one would be eliminated and the one whose magnitude is not one would be our final answer.

Complete step-by-step answer:
According to the question we need to find the choice which is not a unit vector for any value of θ\theta
Now as we know a unit vector is a vector whose magnitude is unity (equal to 1) so we will check whose magnitude is unity and give the final answer.
Checking Option A
cos⁡(θ)i^−sin⁡(θ)j^\cos (\theta )\hat i - \sin (\theta )\hat j
We know that magnitude of a vector of the form ai^+bj^+ck^a\hat i + b\hat j + c\hat k is a2+b2+c2\sqrt {{a^2} + {b^2} + {c^2}}
⇒cos⁡(θ)i^−sin⁡(θ)j^=cos⁡2(θ)+sin⁡2(θ)\Rightarrow \cos (\theta )\hat i - \sin (\theta )\hat j = \sqrt {{{\cos }^2}(\theta ) + {{\sin }^2}(\theta )}
As we know the identity cos⁡2(θ)+sin⁡2(θ)=1{\cos ^2}(\theta ) + {\sin ^2}(\theta ) = 1 , we get
⇒cos⁡2(θ)+sin⁡2(θ)=1\Rightarrow \sqrt {{{\cos }^2}(\theta ) + {{\sin }^2}(\theta )} = \sqrt 1
On simplification we get,
⇒cos⁡2(θ)+sin⁡2(θ)=1\Rightarrow \sqrt {{{\cos }^2}(\theta ) + {{\sin }^2}(\theta )} = 1
Hence, It is a unit vector
Checking Option B
sin⁡(θ)i^−cos⁡(θ)j^\sin (\theta )\hat i - \cos (\theta )\hat j
We know that magnitude of a vector of the form ai^+bj^+ck^a\hat i + b\hat j + c\hat k is a2+b2+c2\sqrt {{a^2} + {b^2} + {c^2}}
⇒sin⁡(θ)i^−cos⁡(θ)j^=sin2(θ)+cos⁡2(θ)\Rightarrow \sin (\theta )\hat i - \cos (\theta )\hat j = \sqrt {si{n^2}(\theta ) + {{\cos }^2}(\theta )}
As we know the identity cos⁡2(θ)+sin⁡2(θ)=1{\cos ^2}(\theta ) + {\sin ^2}(\theta ) = 1
⇒sin2(θ)+cos⁡2(θ)=1\Rightarrow \sqrt {si{n^2}(\theta ) + {{\cos }^2}(\theta )} = \sqrt 1
On simplification we get,
⇒sin2(θ)+cos⁡2(θ)=1\Rightarrow \sqrt {si{n^2}(\theta ) + {{\cos }^2}(\theta )} = 1
Hence, It is a unit vector
Checking Option C
sin⁡(2θ)i^−cos⁡(θ)j^\sin (2\theta )\hat i - \cos (\theta )\hat j
We know that magnitude of a vector of the form ai^+bj^+ck^a\hat i + b\hat j + c\hat k is a2+b2+c2\sqrt {{a^2} + {b^2} + {c^2}}
⇒sin⁡(2θ)i^−cos⁡(θ)j^=sin⁡2(2θ)+cos⁡2(θ)\Rightarrow \sin (2\theta )\hat i - \cos (\theta )\hat j = \sqrt {{{\sin }^2}(2\theta ) + {{\cos }^2}(\theta )}
Using sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta , we get,
⇒sin⁡(2θ)i^−cos⁡(θ)j^=(2sin⁡(θ)cos⁡(θ))2+cos⁡2(θ)\Rightarrow \sin (2\theta )\hat i - \cos (\theta )\hat j = \sqrt {{{(2\sin (\theta )\cos (\theta ))}^2} + {{\cos }^2}(\theta )}
On simplification we get,
⇒sin⁡(2θ)i^−cos⁡(θ)j^=4sin⁡2(θ)cos⁡2(θ)+cos⁡2(θ)\Rightarrow \sin (2\theta )\hat i - \cos (\theta )\hat j = \sqrt {4{{\sin }^2}(\theta ){{\cos }^2}(\theta ) + {{\cos }^2}(\theta )}
Taking terms common we get,
⇒sin⁡(2θ)i^−cos⁡(θ)j^=cos⁡2(θ)(4sin⁡2(θ)+1)\Rightarrow \sin (2\theta )\hat i - \cos (\theta )\hat j = \sqrt {{{\cos }^2}(\theta )(4{{\sin }^2}(\theta ) + 1)}
We know that, cos⁡(θ)(4sin⁡2(θ)+1)≠1\cos (\theta )\sqrt {(4{{\sin }^2}(\theta ) + 1)} \ne 1 , Hence
sin⁡(2θ)i^−cos⁡(θ)j^≠1\sin (2\theta )\hat i - \cos (\theta )\hat j \ne 1
⇒\Rightarrow It is Not a unit vector.
Checking Option D
cos⁡(2θ)i^−sin⁡(2θ)j^\cos (2\theta )\hat i - \sin (2\theta )\hat j
We know that magnitude of a vector of the form ai^+bj^+ck^a\hat i + b\hat j + c\hat k is a2+b2+c2\sqrt {{a^2} + {b^2} + {c^2}}
⇒cos⁡(2θ)i^−sin⁡(2θ)j^=cos⁡2(2θ)+sin⁡2(2θ)\Rightarrow \cos (2\theta )\hat i - \sin (2\theta )\hat j = \sqrt {{{\cos }^2}(2\theta ) + {{\sin }^2}(2\theta )}
As we know the identity cos⁡2(x)+sin⁡2(x)=1{\cos ^2}(x) + {\sin ^2}(x) = 1
⇒cos⁡2(2θ)+sin⁡2(2θ)=1\Rightarrow \sqrt {{{\cos }^2}(2\theta ) + {{\sin }^2}(2\theta )} = \sqrt 1
On simplification we get,
⇒cos⁡2(2θ)+sin⁡2(2θ)=1\Rightarrow \sqrt {{{\cos }^2}(2\theta ) + {{\sin }^2}(2\theta )} = 1
It is a unit vector
Now, since Option 3 is NOT a unit vector we get OPTION 3 as the final answer.

Note: In the questions which have multiple choice questions, But only one answer is correct, if we find our answer before checking for all the MCQs, it is advisable to stop after getting the correct answer, MCQs are commonly used for competitive exams and in competitive exams, time is the key to success. Hence, we should not waste time checking for all the MCQs if we are able to find our answer before checking for all the MCQs.