Question

If north is the direction of the positive y-axis and east is the direction of the positive x-axis, give the unit vector pointing northwest.

Answer 1

North is the direction of positive y-axis. East is the direction of positive x-axis. So West will be the direction of negative x-axis.

Northwest will mean, in between north and west i.e. in between y-axis and the negative x-axis which is the mid of the 2nd quadrant. Thus the vector pointing northwest will form an angle of 135 degrees with positive x-axis.

The magnitude of unit vector is 1 and is forming an angle of 135 degrees. In terms of its components, we can write:

x-component = 1 cos (135) = $- \frac{ \sqrt{2} }{2}$
y-component = 1 sin (135) = $\frac{ \sqrt{2} }{2}$

Thus the unit vector will be = $- \frac{ \sqrt{2} }{2}x+ \frac{ \sqrt{2} }{2}y$

In vector form, component form the vector can be written as:

$(- \frac{ \sqrt{2} }{2}, \frac{ \sqrt{2} }{2})$

Answer 2

A vector pointing northwest passes through point (-1, 1).

Thus an example of a unit vector pointing northwest is $-i+j$.

Recall that a vector is made a unit vector by dividing each component of the vector by the magnitude of the vector.

The magnitude of vector $-i+j$ is given by $|-i+j|=\sqrt{(-1)^2+1^2}=\sqrt{1+1}}=\sqrt{2}$.

Thus, a unit vector pointing northwest is $- \frac{1}{\sqrt{2}} i+ \frac{1}{\sqrt{2}} j$ which when we rationalize we have $- \frac{\sqrt{2}}{2} i+ \frac{\sqrt{2}}{2} j$.