Question

If vectors $P,Q{\text{ and R}}$ have magnitude $5,12{\text{ and 13}}$units$\vec P + \vec Q = \vec R$, the angle between $Q{\text{ and R}}$ is:
$\left( a \right){\text{ co}}{{\text{s}}^{ - 1}}\dfrac{5}{{12}}$
$\left( b \right){\text{ co}}{{\text{s}}^{ - 1}}\dfrac{5}{{13}}$
$\left( c \right){\text{ co}}{{\text{s}}^{ - 1}}\dfrac{{12}}{{13}}$
$\left( d \right){\text{ co}}{{\text{s}}^{ - 1}}\dfrac{7}{{13}}$

Answer 1

Hint We already have the magnitude of the vectors given to us. And also there is a relation between the vectors given to us. On squaring the relation, we will get the angle between $Q{\text{ and R}}$ from there. If the work magnitude is lesser than the product of the force and the displacement, then it means that the force has been applied in an inclined direction to the body. And in this way, we can get the answer.
Formula used:
$ \Rightarrow {\left( {R - Q} \right)^2} = {Q^2} + {R^2} - 2QR\cos \theta $
Here,
$R{\text{ }}and{\text{ }}Q$, will be the vectors.
$\theta $, will be the angle between them.

Complete Step by Step Solution we have values given as-
$\left| P \right| = 5$
$\left| Q \right| = 12$
$\left| R \right| = 13$
According to the question, there is a relation between these vectors, given by
$\vec P + \vec Q = \vec R$
It can also be written as
$\vec P = \vec R - \vec Q$
Now on squaring both the sides, we get
$ \Rightarrow {P^2} = {Q^2} + {R^2} - 2QR\cos \theta $
On substituting the values, we get
$ \Rightarrow {5^2} = {13^2} + {12^2} - 2 \times 12 \times 13\cos \theta $
Now, on solving the above equation, we will get
$ \Rightarrow 25 = 169 + 144 - 312\cos \theta $
Now again solving the above equation,
$ \Rightarrow 25 - 313 = - 312\cos \theta $
Solving the LHS, we get
$ \Rightarrow - 288 = - 312\cos \theta $
Here, the minus sign will be canceled out from both sides.
Therefore,
$ \Rightarrow \cos \theta = \dfrac{{288}}{{312}}$
Now we will reduce the number to the lowest and we get
$ \Rightarrow \cos \theta = \dfrac{{12}}{{13}}$
And from here, the angle will be
$ \Rightarrow \theta = {\cos ^{ - 1}}\dfrac{{12}}{{13}}$

Hence, the option $\left( c \right)$ will be correct.

Note One simple way to find the angle between two vectors is to use the concept of Dot (Or Scalar) product. The dot product between two vectors. Magnitude is the value of the parameter, like distance, speed, etc. Vector gives the Magnitude as well as direction, like Displacement, Velocity, etc.
A vector is an imaginary measurement component that is used to mark certain variables which have a magnitude and an effective direction. In other words, magnitude means the effective intensity of the variable.