The magnitudes of vectors A - B - C are 12-5-13 units respectively and A - B - C- the angle between vectors A and B is-

The correct option is C π /2 We know the resultant of A⃗ and #160; #160; #160; B⃗ #160; has resultant of magnitude R= √(A^2 + B^2 + ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Definition of a Projectile

The magnitude...

Question

The magnitudes of vectors $\to A, \to B, \to C$ are $12, 5, 13$ units respectively and $\to A + \to B = \to C$ , the angle between vectors $\to A$ and $\to B$ is:

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

π

No worries! We‘ve got your back. Try BYJU‘S free classes today!

π / 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

π / 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $π / 2$
We know the resultant of $\to A$ and $\to B$ has resultant of magnitude $R = \sqrt[2]{A^{2} + B^{2} + 2 A B C o s θ} . . . . . . . . . . . . .1$
where $θ$ is angle between $\to A$ and $\to B$ .
but resultant is $\to C$ so $R = C = 13$
putting above values in equation-1 we get $θ = \frac{π}{2}$
Npte- We have used $A = 12, B = 5$
Option C is correct.

Suggest Corrections

Similar questions

Given $\to A + \to B = \to C$ and the magnitudes of vectors $\to A$ , $\to B$ and $\to C$ are respectively 12, 5 and 13 units. The angle between $\to A$ and $\to B$ is

Q. The magnitude of vectors

\to A, \to B

and

\to C

are respectively

12, 5

and

13

units and

\to A + \to B = \to C

, then the angle between

\to A

and

\to B

$\to a, \to b$ and $\to c$ be three vectors having magnitudes 1, 1 and 2 respectively. If $\to a \times (\to a \times \to c) + \to b = 0$ , then the acute angle between $\to a$ and $\to c$ is