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Question: Consider the following statements A and B given below and identify the correct answer: **Statement...

Consider the following statements A and B given below and identify the correct answer:
Statement A: If A\vec{A} is a vector, then the magnitude of the vector is given by A×A\sqrt{\vec{A}\times\vec{A} }
Statement B: If a=mb\vec{a}=m\vec{b}, where ‘m’ is a scalar, the value of ‘m’ is equal to abb2\dfrac{\vec{a}\cdot\vec{b} }{{{b}^{2}}}
A) Both A & B are correct
B) A is correct but B is wrong
C) A is wrong but B is correct
D) Both A and B are wrong

Explanation

Solution

We should have a good knowledge of the Vector analysis. For this question, we should look where the Dot product (Scalar product) is supposed to be used and where the Cross product (Vector product) must be used. Remember the formulae for these products. The formula for magnitude involves Dot product and not the Cross product.

Formula Used:
For Dot product: XY=XYcos(θ)\vec{X}\cdot\vec{Y} =|\vec{X}||\vec{Y}|\cos (\theta ).
For Cross product: X×Y=XYsin(θ)n^\vec{X}\times\vec{Y} =|\vec{X}||\vec{Y}|\sin (\theta )\widehat{n}.
Magnitude of a vector: X=XY|\vec{X}|=\sqrt{\vec{X}\cdot\vec{Y} }.

Complete step by step solution:
We know the formula for the magnitude of a vector is:
X=XY|\vec{X}|=\sqrt{\vec{X}\cdot\vec{Y} }
We know the formula for Dot product is:
XY=XYcos(θ)\vec{X}\cdot\vec{Y} =|\vec{X}||\vec{Y}|\cos (\theta )
And that for Cross product is:
X×Y=XYsin(θ)n^\vec{X}\times\vec{Y} =|\vec{X}||\vec{Y}|\sin (\theta )\widehat{n} ,
where X|\vec{X}|\equiv the magnitude of X\vec{X}, Y|\vec{Y}|\equiv the magnitude for Y\vec{Y}, n^\widehat{n} is the unit normal vector to both X\vec{X} and Y\vec{Y}
From these formulae:
A=AA|\vec{A}|=\sqrt{\vec{A}\cdot\vec{A} }
Therefore, we find that Statement A is wrong.
For Statement B:
a=mb\vec{a}=m\vec{b}
ab=mbb\Rightarrow \vec{a}\cdot\vec{b} =m\vec{b}\cdot\vec{b} (Taking dot product both sides with b\vec{b})
ab=mabcosθ\Rightarrow \vec{a}\cdot\vec{b} =m|\vec{a}||\vec{b}|\cos \theta
ab=mb2\Rightarrow \vec{a}\cdot\vec{b} =m{{b}^{2}} (\because bb=b2cosθ=b\sqrt{\vec{b}\cdot\vec{b}}=\sqrt{|\vec{b}|^{2}}\cos \theta =b )
m=abb2\Rightarrow m=\dfrac{\vec{a}\cdot\vec{b} }{{{b}^{2}}}
Therefore, we find that Statement B is correct.

Hence, Statement A is wrong and Statement B is correct. Therefore, the correct answer is option (C).

Note: We must be very careful about the vector signs () to identify which quantity is a vector and apply vector analysis accordingly. Sometimes, vectors might be identified by bold. Also, Dot product gives a scalar quantity and Cross product gives a vector quantity as a result, and this result must be remembered as sometimes we forget to write that the cross product gives a vector as a result.