find the vector having a length 5m in the XY plane that is perpendicular to vector A =(3i^+4J^-2k^) ?

Asked by pooja dev | 9th Jun, 2011, 12:00: AM

Expert Answer:

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^

Answered by  | 13th Jun, 2011, 02:49: PM

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