How do you find the magnitude and positive direction angle of the vector
v = 6i − 6j?

Step-by-step explanation:

Given the vector v = 6i-6j, the magnitude is expressed as shown;

|v| = √6²+(-6)²

|v| = √36+36

|v| = √72

|v| = √2×36

|v| = √36×√2

|v| = 6√2

The modulus of the vector is 6√2

To get the direction, we will use the formula;

(theta) = tan^-1(y/x)

(theta) = tan^-1(-6/6)

(theta) = tan^-1(-1)

(theta) = -45°

Since the angle is negative and tan theta is negative the in the second quadrant, the positive direction angle of the vector will be 180° - 45° = 135°

3.76%

step-by-step explanation:

the answer is 31.27mph