v20
v20
A
`1/(lambda^(2))=1/(lambda_(1)^(2))+1/(lambda_(2)^(2))`
B
`lambda=sqrt(lambda_(1)lambda_(2))`
C
`2/lambda=1/(lambda_(1))+1/(lambda_(2))`
D
`lambda=(lambda_(1)+lambda_(2))/2`
Text Solution
Verified by Experts
The correct Answer is:
A
`P hat(i) + P hat(j) = vec(P)`
`h/(lambda_(1))hat(i)+h/(lambda_(2))hat(j)=h/lambda`
`1/lambda =sqrt(1/(lambda_(1)^(2))+1/(lambda_(2)^(2)))`
`lambda = (lambda_(1)lambda_(2))/(sqrt(lambda_(1)^(2)+lambda_(2)^(2)))`
`h/(lambda_(1))hat(i)+h/(lambda_(2))hat(j)=h/lambda`
`1/lambda =sqrt(1/(lambda_(1)^(2))+1/(lambda_(2)^(2)))`
`lambda = (lambda_(1)lambda_(2))/(sqrt(lambda_(1)^(2)+lambda_(2)^(2)))`
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