Maths · Scalar and vector products
Given that Out of three vectors, two are equal in magnitude and the magnitude of
Given that \( A+B+C=0 . \) Out of three vectors, two are equal in magnitude and the magnitude of third vector is \( \sqrt{2} \) times that of either of the two having equal magnitude. Then, the angles between the vectors are given by.
- A. \( 30^{\circ}, 60^{\circ}, 90^{\circ} \)
- B. \( 45^{\circ}, 45^{\circ} \), \( 90^{\circ} \)
- C. \( 45^{\circ}, 60^{\circ}, 90^{\circ} \)
- D. \( 90^{\circ}, 135^{\circ}, 135^{\circ} \)
Step-by-step solution
Let the two equal magnitude vectors have magnitude 'a', and the third magnitude be √2 a. From A+B+C=0, we have A = -(B+C). Taking magnitude squared: |A|^2 = |B+C|^2 = |B|^2 + |C|^2 + 2B·C = 2a²(1+cosθ). Given |A|² = 2a², we get cosθ=0 → θ=90° between B and C. Then A·B = -(B+C)·B = -a², so cosφ = (A·B)/(|A||B|) = -1/√2 → φ=135°. Similarly, angle between A and C is also 135°. Thus the angles are 90°, 135°, 135°.
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