Find the unit vector perpendicular for the vector i+2j-4k in the xy-plane.
A vector Normal to it is 5i+10j-40k
the unit vector is that divided by sqrt (25+100+1600)
V=|u|cos90 =4.58*0=0
To find the unit vector perpendicular to a given vector in the xy-plane, we first need to determine the normal vector of the plane. The normal vector of the xy-plane is simply the z-axis vector, which in this case is k.
Given the vector i + 2j - 4k, we need to eliminate the z-component by setting it to 0, as we are in the xy-plane. So, we set -4k = 0, which means k = 0.
Therefore, the vector in the xy-plane becomes i + 2j.
To find the unit vector perpendicular to this vector, we can simply divide the vector by its magnitude. The magnitude of the vector can be calculated using the Pythagorean theorem.
Magnitude of the vector = sqrt((i)^2 + (2j)^2) = sqrt(1^2 + 2^2) = sqrt(5)
Now, dividing the vector i + 2j by the magnitude sqrt(5), we get:
Unit vector perpendicular to the vector i + 2j - 4k in the xy-plane = (i + 2j) / sqrt(5)
So, the unit vector perpendicular to the vector i + 2j - 4k in the xy-plane is (1/sqrt(5))i + (2/sqrt(5))j.