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3-1.Vectors
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If $| A + B |=| A |+| B |$ the angle between $\overrightarrow A $and $\overrightarrow B $ is ....... $^o$
A
$0$
B
$60$
C
$120$
D
$90$
Solution
For two vectors $A \square A \rightarrow$ and $B \square B \rightarrow$, the angle between them being $\theta \theta$, the magnitude of the resultant vector $A + B \rightarrow A + B \rightarrow$ is given by,
$| A + B |=\sqrt{| A |^2+| B |^2+2 AB \cos \theta}$
Now in the problem we've,
$| A + B |=| A |+| B |$
Squaring on both sides,
$\begin{array}{l}| A + B |^2=(| A |+| B |)^2 \\| A |^2+| B |^2+2| A || B | \cos \theta=| A |^2+| B |^2+2| A || B | \\\Rightarrow \cos \theta=1\end{array}$
assuming neither of the vectors are $zero$ vectors.
Standard 11
Physics
