Thread: Vector with Trig Function Math Question

I am reading a book on electromagnetic fields and in one of the examples describing vector analysis, the text reads, "express the vector velocity v = a_v 55 (ms^-1) at the point P(0,2,5), shown in Fig. 1-4 to be in the y-z plane, in component form, using unit vectors."




So the text goes on, "The vector velocity v can be expressed as v = xv_x + yv_y + zv_z. From Fig. 1-4, v_x = 0, v_y = 55 sin β, v_z = -55 cos β, cos β = 4/5, and sin β = 3/5."

I guess I have a couple of questions, all related:

First, how did the authors derive a vector originating from point P(0, 2, 5) to 4 units down the z-axis and 3 units right on the y-axis? How do they derive this graphic depiction from the original information presented? What is the meaning of (ms^-1)?

Second, how is the vector component, v_y, equal to 55 sin β?

I understand sin is a function of the ratio of the opposite side of the angle to the hypotenuse, and so the sin of β will render 3/5 from the depiction, but how did they get that in the first place? Also, why is it 55 multiplied by this ratio?

NOTE: I am new to this forum and I have not expressed these formulas with the corresponding "hats" or "carets" or vector bars in their notation.

Hey KClose1983.

It is basically as you have pointed out to do with Pythagoras' theorem.

We know that sin(theta) = opposite/hypotenuse and hypotenuse = 55 so opposite = 55*sin(theta) = vy

Also 55*cos(theta) = vx using a similar argument.

It's just a matter of identifying the right angles and sides of the triangle.