The resultant of two vectors $\vec{A}$ and $\vec{B}$ is $\vec{C}$. If the magnitude of $\vec{B}$ is doubled,…
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
3B
2B
B
4B
Solution
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 $|C|=|B|$