the value of n so that vector (2i-3j+k) may be perpendicular to the vector (3i+4j+nk) is.
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a.b = abcos(theta)
When theta equal to pi/2 then cos(theta) is become zero then
a.b=0
(2i-3j+k).(3i+4j+nk) = 0
6-12+n=0
n=6
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