Give P = A + B and Q = A – B. If the magnitudes of vector P and Q are equal, what is the angle between vectors A and B ?

zero

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Solution

The correct option is C
Let $θ$ be the angle between A and B. Then magnitudes of vectors P and Q are given by
$P^{2} = A^{2} + B^{2} + 2 A B cos θ$
and $Q^{2} = A^{2} + B^{2} - 2 A B cos θ$
Since P = Q, it follows that $cos θ = 0 o r θ = \frac{π}{2} .$ Hence correct choice is (c).

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Give P = A + B and Q = A – B. If the magnitudes of vector P and Q are equal, what is the angle between vectors A and B ?

The correct option is C Let θ be the angle between A and B. Then magnitudes of vectors P and Q are given by P2=A2+B2+2ABcosθ and Q2=A2+B2–2ABcosθ Since P = Q, it follows that cosθ=0 or θ=π2. Hence correct choice is (c).

The correct option is C
Let $θ$ be the angle between A and B. Then magnitudes of vectors P and Q are given by
$P^{2} = A^{2} + B^{2} + 2 A B cos θ$
and $Q^{2} = A^{2} + B^{2} - 2 A B cos θ$
Since P = Q, it follows that $cos θ = 0 o r θ = \frac{π}{2} .$ Hence correct choice is (c).