If √(A² + B²) represents the magnitude of resultant of two vectors (A + B) and (A - B), then the angle between two vectors is (a) cos ⁻¹[-(2(A² - B²))/((A² + B²))] (b) cos ⁻¹[-(A² - B²)/(A² B²)] (c) cos ⁻¹[-((A² + B²))/(2(A² - B²))] (d) cos ⁻¹[-((A² - B²))/(A² + B²)]

Answer: (c). Treat the two vectors as having magnitudes (A + B) and (A - B) with angle θ between them. (A + B)² + (A - B)² + 2(A + B)(A - B)cos θ = A² + B² 2(A² + B²) + 2(A² - B²)cos θ = A² + B² cos θ = -(A² + B²)/(2(A² - B²)) So the answer is (c).

If √(A² + B²) represents the magnitude of resultant of two vectors (A… | MHT CET 2019 (2 May, Shift 1)

Q28MHT CET 2019 · 2 MAY · SHIFT 1Medium

Addition & Subtraction of Vectors

If $\sqrt{A^2 + B^2}$ represents the magnitude of resultant of two vectors $(\mathbf{A} + \mathbf{B})$ and $(\mathbf{A} - \mathbf{B})$ , then the angle between two vectors is

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Answer: (c)

Treat the two vectors as having magnitudes $(A + B)$ and $(A - B)$ with angle $\theta$ between them.
$(A + B)^2 + (A - B)^2 + 2(A + B)(A - B)\cos\theta = A^2 + B^2$
$2(A^2 + B^2) + 2(A^2 - B^2)\cos\theta = A^2 + B^2$
$\cos\theta = -\frac{A^2 + B^2}{2(A^2 - B^2)}$
So the answer is (c).

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Addition & Subtraction of Vectors question from MHT CET 2019 (2 May, Shift 1)

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