what is the angle betwean x axis and vector 3i-2j+4k
You can use the dot product and/or the cross product
dot product:
a dot b = |a| |b| cos theta
a = 1 i + 0 j + 0 k
b = 3 i - 2 j + 4 k
|a| = 1
|b| = sqrt(9+4+16) = sqrt 29
a dot b = 3
so
sqrt 29 cos theta = 3
cos theta = 3/sqrt 29
theta = 56.1 deg
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using cross or vector product
a cross b:
i j k
1 0 0
3 -2 4
= 0i -4j -2k
magnitude = sqrt(20)
sqrt 29 sin theta = sqrt 20
theta = 56.1 deg again