Find the magnitude and direction (in degrees) of the vector. (Assume 0° ≤ 𝜃 < 360°.) 𝘃 =〈-root2/5,-root2/5.

Question

Find the magnitude and direction (in degrees) of the vector. (Assume 0° ≤ 𝜃 < 360°.) v=〈-(√2)/5,-(√2)/5)

i was getting 2/25 and 45° but my homework program said it was wrong

Answer 1

The correct answer is 2/5 and 225 degrees. First, for the magnitude: (√(2/)/5)² = 2/25 (it doesn't matter that the actual vector component is negative, because (-n)² =n²). Since we have both components the same, x² + y² = 4/25, and the square root of this gives the magnitude of the vector as 2/5. Then, for the angle: Since both components are negative, the vector is in Quadrant III; and, since y/x = 1, the related angle to the actual angle is tan^-1 (1) = 45°. The related angle to any angle in Quadrant III is 180° less than the actual angle, so the actual angle must be 45° + 180° = 225°.

Answer 2

square root of sum of sqr2/5 squared + sqr2/5 squared = sqr(2/25 + 2/25) = sqr(4/25) = 2/5

magnitude = 2/5 = hypotenuse of a right triangle with both legs = -sqr2/5

direction = 180+45 = 225 degrees or pi+pi/4 = 5pi/4 radians

the angle is in quadrant III

the reference angle is 45 degrees, but it's in quadrant III, so add 180 degrees