vector problem help please?

I converted both to spherical, and found that vector 1 is magnitude 12.083 @ 65.556 degrees, and vector 2 is 2.236 @ -26.565 degrees, The bisecting angle between these 2 vectors would be 19.496 degrees. So you have a vector of magnitude 1.00 @ 19.496 degrees. Converting back to rectangular coordinates gives a vector: 0.943i + 0.334j.

Magnitude 1 means it's a unit vector, so to check: (0.943^2 + 0.334^2)^(1/2) = 1.00

You can calculate the angle between the given vectors by using the scalar product.

"I converted both to spherical, and found that vector 1 is magnitude 12.083 @ 65.556 degrees, and vector 2 is 2.236 @ -26.565 degrees, The bisecting angle between these 2 vectors would be 19.496 degrees" No, about 46 degrees.

My "46 degrees": I have only used the angles from Leitey and made a mentual calculation with his angles. - Make a sketch, you'll see that the angle between the given vectors is about 90 degr. For a correct calculation use the scalar product and the arccosine. - About 90 deg is the angle BETWEEN the vectors, but not the absolute angle in the coordinates system. Therefore must "0.69 , 0.72 " be wrong.

Yes, Arctan is very much easier than using the scalar product and arcsine, I agree.

I have explained, why ( 0.69 , 0.72 ) must be wrong.

Control it youself by making a little sketch drawing.

This problem or one very like it, looks familiar to me. Let's see if I can remember how I did it.

Draw a rough sketch of the vectors which will make things easier to follow.

The first vector points up and to the right i.e it is in the first quadrant

The second vector points down and to the right i.e it is in the fourth quadrant.

What I'll do is to work out the angle that each vector makes with the positive x-axis.

For vector 1, angle = arctan (11/5) = 65.65 degrees to two decimal places.

For vector 2, angle = - arctan (1/2) = - 26.57 degrees.

So the angle between them is 65.65 + 26.57 = 92.22 degrees

Half of this is 46.11 degrees. So the unit vector we are looking for poinst into the first quadrant.

Since its magnitude = 1, using basic trig we can work out the other two sides

Horizontal component = 1 cos (65.65 - 46.11) = 1 cos 19.54 = 0.942

Vertical component = 1 sin 19.54 = 0.334

So, using i-j notation for our unit vector we have

0.942i + 0.334j

The magnitude of this is √(0.942^2 + 0.334^2) = 0.999 which is close enough to 1 as makes no difference!! So that answer IS the right one. What 'program' are you using?