Question

If $\left| \overrightarrow{A+B} \right|=\left| \overrightarrow{A} \right|+\left| \overrightarrow{B} \right|$ then angle between $\overrightarrow{A}\text{ and }\overrightarrow{B}$ will be

& a){{90}^{\circ }} \\\ & b){{120}^{\circ }} \\\ & c){{0}^{\circ }} \\\ & d){{60}^{\circ }} \\\ \end{aligned}$$

Answer 1

It is given to us in the question that the resultant of two vectors has a magnitude equal to the sum of the magnitude of the individual vectors. First we will take the square of the resultant i.e. basically the square of the sum of the two vectors. As a result we will obtain an expression consisting of the Magnitude of the vector and the and the angle between them. Hence using this expression, we will obtain the angle between the vectors.
Formula used:
${{\left| \overrightarrow{A+B} \right|}^{2}}={{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|Cos\theta$

Complete answer:
Let us consider two vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ . The resultant of the two vectors is given by $\overrightarrow{A+B}$ . If we multiply the resultant to itself, we will obtain the square of the magnitude of the resultant i.e. ${{\left| \overrightarrow{A+B} \right|}^{2}}$
The above result obtained is also equal to,
${{\left| \overrightarrow{A+B} \right|}^{2}}={{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|Cos\theta$ where $\theta$ is the angle between vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$
It is given to us that, $\left| \overrightarrow{A+B} \right|=\left| \overrightarrow{A} \right|+\left| \overrightarrow{B} \right|$ . hence substituting this in the above equation we get,
$\begin{aligned} & {{\left| \overrightarrow{A+B} \right|}^{2}}={{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|Cos\theta \\\ & {{\left( \left| \overrightarrow{A} \right|+\left| \overrightarrow{B} \right| \right)}^{2}}={{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|Cos\theta \\\ \end{aligned}$
$\left| \overrightarrow{A} \right|\text{ and }\left| \overrightarrow{B} \right|$ are scalars. Therefore we can use the identity i.e. ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\begin{aligned} & {{\left( \left| \overrightarrow{A} \right|+\left| \overrightarrow{B} \right| \right)}^{2}}={{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|Cos\theta \\\ & \Rightarrow {{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|={{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|Cos\theta \\\ & \Rightarrow 2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|=2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|Cos\theta \\\ & \Rightarrow 1=Cos\theta \text{, since }Cos{{0}^{\circ }}=1, \\\ & \Rightarrow \theta ={{0}^{\circ }} \\\ & \\\ \end{aligned}$

Hence the correct answer of the above question is option is c.

Note:
In the above case we wrote the magnitude of the resultant as the sum of the two vectors. Similarly the magnitude of the difference of the two vectors is given by ${{\left| \overrightarrow{A-B} \right|}^{2}}={{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}-2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|Cos\theta$ .One of the key points to be noted is that if you multiply the vector to itself then we obtain the square of the magnitude of that vector. For any two vectors it is also to be noted that the result $\left| \overrightarrow{A+B} \right|\le \left| \overrightarrow{A} \right|+\left| \overrightarrow{B} \right|$ is always true.