MHT CET · Physics · Mathematics in Physics
The resultant of two vectors \(\vec{A}\) and \(\vec{B}\) is \(\vec{C}\). If the magnitude of \(\vec{B}\) is doubled, the new resultant vector becomes perpendicular to \(\overrightarrow{\mathrm{A}}\). Then the magnitude of \(\overrightarrow{\mathrm{C}}\) is
- A 3B
- B 2B
- C B
- D 4B
Answer & Solution
Correct Answer
(B) 2B
Step-by-step Solution
Detailed explanation
According to the question, \(A+B=C \ldots . .(0)\)
Also, when \(B\) is doubled, \(C\) becomes
perpendicular to \(A\). Let the new magnitude of \(C\)
be \(C^{\prime}\)
That means \(A . C^{\prime}=0\left(A C^{\prime} \cos 90=0\right)\)
\(>(A)(A+2 B)=0 \ldots \ldots .(1)\)
\(>\left(A^{\wedge} 2+2 A B\right)=0\)
\(>|A|=-2 A B\)
To find the magnitude of \(C\), square eq. (0)
\(>(A+B)^{\wedge} 2=|C|\)
\(>|A|+|B|+2 A B\)
but \(2 A B=-|A| .\) from \((1)\)
which means \(|C|=|B|\)
Also, when \(B\) is doubled, \(C\) becomes
perpendicular to \(A\). Let the new magnitude of \(C\)
be \(C^{\prime}\)
That means \(A . C^{\prime}=0\left(A C^{\prime} \cos 90=0\right)\)
\(>(A)(A+2 B)=0 \ldots \ldots .(1)\)
\(>\left(A^{\wedge} 2+2 A B\right)=0\)
\(>|A|=-2 A B\)
To find the magnitude of \(C\), square eq. (0)
\(>(A+B)^{\wedge} 2=|C|\)
\(>|A|+|B|+2 A B\)
but \(2 A B=-|A| .\) from \((1)\)
which means \(|C|=|B|\)
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