Question

Angle made between vector a=2i-3j+4k and the z axis

Answer 1

Answer:

Let,©be the angle made by vector a with z axis respectively.

Given, vector a =2i-3j+4k

|a|=√2^2+(-3)^2+4^2

=√4+9+16

=√29

Now,

cos©=az/a

i.e. cos©=4/√29

i.e. ©=cos^-1 4/√29

Explanation:

hey mate...the angle has taken in the answer for my help ...so u can take theta..alpha.. whatever u want.... hope it helps you

Answer 2

The angle between A and the z axis is 42.01 degrees.

Explanation:

Vector A, $A=2i-3j+4k$

We need to find the angle between vector A and the z axis. It is given by :

$\cos\theta=\dfrac{z}{|A|}$

$|A|=\sqrt{2^2+(-3)^2+4^2} =5.38$

$\cos\theta=\dfrac{4}{5.38}$

$\theta=42.01^{\circ}$

So, the angle between A and the z axis is 42.01 degrees. Hence, this is the required solution.

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