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Let L_(1) : vecr = (veci + vecj + veck) + lambda(veci + vecj), lambda in R, L_(2):hati + (mu + 1)hatj + (mu +1)hatk, mu in R are two lines intersecting at point 'A'. Through point B(3,3,1) a line in drawn making an angle of 60^(@) with L_(2) and intersecting it a point 'C'. Then area of triangleABC is: |
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Answer» `2sqrt(2)` `r =(1,1,1) + MU(0,1,1)` int pt. `A -= (1,1,1)` Let c be `(1, 1+mu, 1+mu)` `vec(AC) = 011` `vec(BC) =2,2- mu, -mu` `AC ^ BC = 60^(@) rArr (0+2-mu - mu)/(sqrt(2) sqrt(4+(2-mu)^(2)+mu^(2)))=1/2`  `(4+4mu)^(2) = 2(4 + 4+mu^(2) - 4mu + mu^(2))` `16(1+mu^(2) -2MU) = (16 + 4mu^(2) - 8mu)` `12mu^(2) - 24mu rArr mu =0, mu=2` `mu=0, mu=2`  C(1,1,1)(1,3,3) `1/2|{:(hati,hatj,hatk),(-2,0,2),(-2,-2,0):}|=1/2{hati(4) - hatj(4) + k(4)}` `|(2hati - 2hatj + 2hatk)|=2sqrt(3)=asqrt(b)` a=2, b=3 |
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