CBSE Class 12-science Answered
find the vector having a length 5m in the XY plane that is perpendicular to vector A =(3i^+4J^-2k^) ?
Asked by pooja dev | 09 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 | 13 Jun, 2011, 02:49: PM
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