The distance between a (cos theta + b theta ,0) and (0,a single theta - b cos theta )
Answers
Answered by
0
√[acos α + bsinα-0]^2+(-asinα+bcosα)^2
=√ a^2cos^2α +b^2sin^2α+2abcosαsinα+a^2sin^2α+b^2cos^2α-2abcosαsinα
√4=+-2
=√ a^2cos^2α +b^2sin^2α+2abcosαsinα+a^2sin^2α+b^2cos^2α-2abcosαsinα
√4=+-2
saurabhsemalti:
Yup also... it will be equal to 2..as -2 distance is not possible
Answered by
1
use A instead of theta
distance=√{-(acosA+bsinA)}^2+(asinA-bcosA)^2
=√a^2cos^2A+b^2sin^2A+2abcosAsinA+a^2sin^2A+b^2cos^2A-2abcosAsinA
=√a^2(sin^2A+cos^2A) + b^2(sin^2A+cos^2A)
=√a^2+b^2
distance=√{-(acosA+bsinA)}^2+(asinA-bcosA)^2
=√a^2cos^2A+b^2sin^2A+2abcosAsinA+a^2sin^2A+b^2cos^2A-2abcosAsinA
=√a^2(sin^2A+cos^2A) + b^2(sin^2A+cos^2A)
=√a^2+b^2
Similar questions