Question

vector p=6i+4√2j+4√2k
$6i + 4 \sqrt{2j} + 4 \sqrt{2k}$
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Answer 1

Answer:

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°

Explanation:

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