Question

The angle between the Z - axis and th
vector î + 9 + 2k is​

Answer 1

Step-by-step explanation:

AB= i+9j+2k

unit vector along z axis is k

the angle b/w z axis AB vector is given as:

cos = AB.k/|AB|

cos = (i+9j+2k)k/√1^2+9^2+2^2)

= 2/√86

so, theeta = cos^-1(2/√86)

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Answer 2

Answer:

z axis is rep as x=0,y=0 hence unit ector is k

using cos product

cos theta = 2/ ( root ( 1^2+9^2+2^2). 1

cos theta = 2/root 86

theta is cos inverse 2/ root 86