Let u=<4sqrt3,-4>. Find the length and direction of u?

Answer 1

The length of a vector is sqrt(x² + y²)

The direction of the vector is tan^-1(y/x), making sure to add 180 degrees if the vector has a negative x-component.

(Inverse tangent only ranges between -90 and 90, or in Quadrants 1 and 4, if your vector points toward Quadrant 2 or 3, will need to add 180 degrees or pi radians to get the correct direction angle.)

So for this vector, the length is sqrt[ (4√3)² + (-4)² ] = sqrt( 48 + 16 ) = √64 = 8

Direction angle is tan^-1( -4/(4√3) ) = -30 degrees or - pi/6 radians.