Question

The magnitudes of vectors A,B and C are 3,4 and 5 units respectively. If A+B=C, the angle between A and B is____?

(a) π/2
(b) cos=3/5
(c) tan=7/5
(d) π/4​

Answer 1

Answer:

(a) π/2

Step-by-step explanation:

Step 1:  

Squaring the given equation on both sides

Let θ be the angle between A and B

A+B =C

Squaring on both sides

$(A+B)^2=(C)^2$  

⇒$|A|^2+|B|^2+2A.B=|C|^2$

⇒$|A|^2+|B|^2+2|A||B|cos(\theta) = |C|^2$         ....(1)

Step 2: Calculations

Put the values of all the variables in equation (1) to get θ

$(3)^2+(4)^2+2*3*4cos(\theta)=(5)^2$

⇒ cosθ=0

⇒ θ=90∘ =  π/2

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Answer 2

Answer:

The angle between A and B is π/2.

Step-by-step explanation:

Given

• Magnitude of vector A = |A| = 3
• Magnitude of vector B = |B| = 4
• Magnitude of vector C = |C| = 5

            and A+B = C

         |A+B| = |C|

             |C|²= |A+B|²

             |C|²= |A|²+B|²+2|A||B|cosθ

           substitute the values of |A|,|B| and |C| in above equation

               5²= 3²+4²+2(3)(4)cosθ

               25 = 9+16+2(12)cosθ

               25 = 25+24cosθ

            cosθ = 0

                  θ = cos⁻¹(0)

                  θ = π/2

  Hence angle between the vectors A and B is π/2.

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