Find the magnitude and direction angle (in degrees) of the following vector.v=12(cos(68°)i+sin(68°)j)

The magnitude of a vector v = ai + bj is given by √(a² + b²). In this case, a = 12cos(68°) and b = 12sin(68°).

Step 1: Calculate a² and b² a² = (12cos(68°))² b² = (12sin(68°))²

Step 2: Add a² and b² a² + b² = (12cos(68°))² + (12sin(68°))²

Step 3: Take the square root of the sum to find the magnitude |v| = √((12cos(68°))² + (12sin(68°))²)

However, we know that (cos(θ))² + (sin(θ))² = 1 for any angle θ. So, the magnitude of the vector is √(12²) = 12.

The direction angle of the vector is the same as the angle given in the vector, which is 68°.

So, the magnitude of the vector v is 12 and its direction angle is 68°.