Question

If vector $a + b{\text{ }} =$ $c$ and$a + b{\text{ }} = {\text{ }}c$. What is the angle between $a$ and $b$ ?
(A) $90$
(B) $45$
(C) $0$
(D) $60$

Answer 1

Set an equation using the given relations between the vectors by simplifying them. Use the formula dot product of two vectors. Find the cosine of the angle from the dot product and then find the angle between two vectors.

Formula used:
$\overrightarrow a \overrightarrow b = ab\cos \theta$
Where the angle between two vectors $\overrightarrow a$ and $\overrightarrow b$ is $\theta$ .

Complete step by step answer:
The resultant vector of the two vectors $\overrightarrow a$ and $\overrightarrow b$ is $\overrightarrow c$ i.e $\overrightarrow a + \overrightarrow b = \overrightarrow c$………$(1)$
The modulus of the two vectors $\overrightarrow a$ and $\overrightarrow b$is $a$ and $b$ .
The sum of $a$ and $b$is $c$ i.e $a + b = c$…………$(2)$

By squaring both sides of eq. $(1)$ we get,
$\overrightarrow {(a} + \overrightarrow b {)^2} = {c^2}$
$\Rightarrow {c^2} = \overrightarrow {(a} + \overrightarrow b )\overrightarrow {(a} + \overrightarrow b )$
$\Rightarrow {c^2} = {a^2} + {b^2} + 2\overrightarrow a .\overrightarrow b$

Since $\overrightarrow a .\overrightarrow b = ab\cos \theta$
$\Rightarrow {c^2} = {a^2} + {b^2} + 2ab\cos \theta$
$\Rightarrow {(a + b)^2} = {a^2} + {b^2} + 2ab\cos \theta$ [ from the eq. $(2)$ ]
$\Rightarrow {a^2} + {b^2} + 2ab = {a^2} + {b^2} + 2ab\cos \theta$
$\Rightarrow \cos \theta = 1$
$\Rightarrow \theta = 0^\circ$
So, the angle between the vectors $\overrightarrow a$ and $\overrightarrow b$ is $\theta = 0^\circ$.

Hence, the correct answer is option (C).

Additional information:
If the resultant vector of the two vectors $\overrightarrow a$ and $\overrightarrow b$ is $\overrightarrow c$ and the angle between the vectors $\overrightarrow a$ and $\overrightarrow b$ is $\theta$then the value of $\overrightarrow c$,
$c = \sqrt {{a^2} + {b^2} + 2ab\cos \theta }$

And if the angle between the vectors $\overrightarrow a$ and $\overrightarrow c$ is $\alpha$
Then, $\tan \theta = \dfrac{{b\sin \alpha }}{{a + b\cos \alpha }}$ this the relation between two angles.
When, $\alpha = 0$
$c = a + b = {c_{\max }}$ (the possible highest value of the resultant) and also $\theta = 0^\circ$.

Note: Here we use one of the properties of scalar products. This is $\overrightarrow A .\overrightarrow A = {A^2}$ . means if we take a vector two times, we get the square of the value of the vector.
We get in the answer that the angle between the vectors is zero. This implies that the vectors are either parallel or stay along one single line.
The angle between the two vectors is presented as, $\theta = {\cos ^{ - 1}}\left( {\dfrac{{{\text{dot product of two vectors}}}}{{{\text{product of values of two vectors}}}}} \right)$