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find a vector length of 5m in the xy plane that is perpendicular to A=3i+6j+2k hint use dot product​

Respuesta :

Answer:

Explanation:

vector A =(3i^+4J^-2k^),

 

let the required vector is B = x i^ + y j^

 

sqrt (x2 + y2) = 5,  

 

x2 + y2 = 25  ..............(1)

 

Since vector A and B are perpendicular to each other then,

 

A.B = 0

 

3x + 4 y = 0

 

y = -3x/4 ..............(2)

 

Solving these two equations we get x = 4 and y = -3;

 

hence vector B = 4 i^ -3 j^

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