Question

Vector P=6i+4√2j+4√2k makes angle from z-axis
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Answer 1

Answer:

Here the coefficients of i, j, k represents the lengths of projections of vector P on X,Y,Z axes respectively.

Length of P on X-axis is 6

Length of P on Y-axis is 42(4root2)

Length of P on Z-axis is 42(4root2)

Length of P in space is square root of (6*6)+(42*42)+(42*42) =square root of 100 = 10.

Now to find angle of P with Z-axis,

Cos  = length of P on Z-axis/ length of P in space = 42/10 =22/5

Angle of P with Z-axis = cos^-1(22/5).