The magnitude of vectors A - B and C are respectively 12-5 and 13 units and A - B - C- then the angle between A and B isA- 0B- TC- - - 2D- - - 4

The correct option is C π/2Given, C=A+B C.C=(A+B).(A+B) C^2=A^2+B^2+2AB cosθ 13^2=12^2+5^2+2(12)(5)cosθ cosθ=0 ⇒θ=π2 ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Unit Vectors

The magnitude...

Question

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 $\to B$ 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 $π / 2$
Given, $\to C = \to A + \to B$
$\to C . \to C = (\to A + \to B) . (\to A + \to B)$
$C^{2} = A^{2} + B^{2} + 2 A B cos θ$
$13^{2} = 12^{2} + 5^{2} + 2 (12) (5) cos θ$
$cos θ = 0 \Rightarrow θ = \frac{π}{2}$

Suggest Corrections

117

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. Given that

\to C = \to A + \to B

. The magnitude of

\to A, \to B

and

\to C

are 12, 5 and 13 units respectively. The angle between

\to A

and

\to B

The magnitude of vector $\to A, \to B$ and $\to C$ are respectively 12, 5 and 13 units and $\to A + \to B = \to C$ then the angle between $\to A$ and $\to B$ is

$\to a, \to b$ 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

Q. If a,b,c are non-coplanar unit vectors such that

a \times (b \times c) = \frac{b + c}{\sqrt{2}}

, then the angle between a and b is

Join BYJU'S Learning Program

The magnitude of vectors →A,→B and →C are respectively 12,5 and 13 units and →A+→B=→C, then the angle between →A and →B is

The correct option is C π/2Given, →C=→A+→B →C.→C=(→A+→B).(→A+→B) C2=A2+B2+2ABcosθ 132=122+52+2(12)(5)cosθ cosθ=0⇒θ=π2

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 $\to B$ is

The correct option is C $π / 2$
Given, $\to C = \to A + \to B$
$\to C . \to C = (\to A + \to B) . (\to A + \to B)$
$C^{2} = A^{2} + B^{2} + 2 A B cos θ$
$13^{2} = 12^{2} + 5^{2} + 2 (12) (5) cos θ$
$cos θ = 0 \Rightarrow θ = \frac{π}{2}$