What angle does vector P=6i+4√2j+4√2k makes with the z-axis?
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Vector P = 6 i + 4√2 j + 4√2 k
Directional cosine of the vector with the Z axis:
Cos Ф = 4√2 / √[6²+ (4√2) + (4√2)² ]
= 4√2 / √[ 36+32+32]
= 4√2/10
Ф = 55.55°
Directional cosine of the vector with the Z axis:
Cos Ф = 4√2 / √[6²+ (4√2) + (4√2)² ]
= 4√2 / √[ 36+32+32]
= 4√2/10
Ф = 55.55°
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