Find Angle of Vector h: 3ax+5ay-8az

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The discussion focuses on calculating the angles between the vector h = 3ax + 5ay - 8az and the x, y, and z axes using the dot product method. To find the angle with respect to each axis, the formula h . (unit vector) = (magnitude of h) * cos(theta) is utilized. The magnitude of vector h can be calculated as √(3² + 5² + (-8)²), which simplifies to √(9 + 25 + 64) = √98. The cosine of the angles can then be derived by dividing the component of the vector by its magnitude.

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i have a vector h whose value is 3ax+5ay-8az where a is the unit vector in the direction beside it, how would i find the angle between this vector and the x,y,and z axes
 
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where a is the unit vector in the direction beside it,

Do you mean h = 3x + 5y - 8z ?

In any case, in order to find the angle between vectors, take the dot product.

Eg. Angle between h and x- axis would be given by:

h . (unit vector x) = (magnitued of h) * cos (theta)
 
Or, you can find the sq. root of the sum of the squares (3^2+5^2+(-8)^2)^0.5

and then for each direction divide the magnitude by this value and it will give you the cos for each angle
 

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