Question

(1) If \[\left| {\bar A + \bar B} \right| = \left| {\bar A - \bar B} \right|\], then angle between the $\bar A$ and $\bar B$is
A. ${0^ \circ }$
B. ${45^ \circ }$
C. ${90^ \circ }$
D. ${60^ \circ }$

(2) The resultant of two forces $3N$and $4N$ is $5N$. The angle between the forces is
A. ${30^ \circ }$
B. ${60^ \circ }$
C. ${90^ \circ }$
D. ${120^ \circ }$

Answer 1

Hint: To find the angle between the two vectors we make use of the concept of addition of two vectors and subtraction of two vectors. The expression for the resultant of the two vectors gives us the angle between the two vectors.

Formulae used:
Magnitude of the resultant vector for addition of two vectors is given by
$R = \left| {\bar A + \bar B} \right| = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } $
And the Magnitude of the resultant vector for subtraction of two vectors is given by
$R = \left| {\bar A - \bar B} \right| = \sqrt {{A^2} + {B^2} - 2AB\cos \theta } $
Where, $A,B$ - magnitudes of the vectors $\bar A\& \bar B$, respectively and $\theta $ - angle between vectors$\bar A\& \bar B$.

Complete step by step answer:
(1) Given that , \[\left| {\bar A + \bar B} \right| = \left| {\bar A - \bar B} \right|\]
So, writing the formula of resultant vector for addition and subtraction of two vectors, we get
$\sqrt {{A^2} + {B^2} + 2AB\cos \theta } = \sqrt {{A^2} + {B^2} - 2AB\cos \theta } $
Squaring both sides, then
\[{A^2} + {B^2} + 2AB\cos \theta = {A^2} + {B^2} - 2AB\cos \theta \]
\[\Rightarrow 2AB\cos \theta = - 2AB\cos \theta \]
\[\Rightarrow 4AB\cos \theta = 0\]
Then, \[\therefore \cos \theta = 0\] which gives,
\[\therefore \theta = {90^ \circ }\].
The angle between the $\bar A$ and $\bar B$is \[\theta = {90^ \circ }\].

Hence, option C is correct.

(2) Let us assume the given force vectors as
$P = 3N$, $Q = 4N$ and the resultant of these vectors is given as $R = 5N$.
We have to find the angle between these two vectors $(\theta )$
By the magnitude of resultant vector of two vectors,
$R = \left| {\bar P + \bar Q} \right| = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } $
Substituting the given values, we get
$5 = \sqrt {{3^2} + {4^2} + 2 \times 3 \times 4\cos \theta } $
$\Rightarrow 5 = \sqrt {9 + 16 + 24\cos \theta } $
Squaring on both sides,
$25 = 9 + 16 + 24\cos \theta $
$\Rightarrow 0 = 24\cos \theta $
$\therefore \cos \theta = 0$
This gives, \[\therefore \theta = {90^ \circ }\].
Thus, the angle between two force vectors is \[\theta = {90^ \circ }\].

Hence, option C is correct.

Note:If two vectors are such that their sum and their difference have equal magnitude, then the angle between the given vectors \[\theta = {90^ \circ }\] . Also, If the magnitudes of the two vectors and their resultant vector’s magnitude are such that they are forming an Pythagorean triplets $(3,4,5),(5,12,13),(6,8,10),$ etc , then the angle between the vectors is also \[\theta = {90^ \circ }\].