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Find the equation of a plane passes through the points `(0,0,0)` and `(1,3,5)` and parallel to the line `(x)/(-2) = y/1 = (z+3)/(4)`. |
Answer» Let equation of the plane passing through the point `(0,0,0)` is `a(x-0)+b(y-0)+c(z-0) = 0` `rArr ax+ by+cz = 0 "….."(1)` This plane passes throough the point `(1,3,5)`. `:. a+3b+5+5c=0"……"(2)` This plane is parallel to the line `(x)/(-2) = (y)/(1) = (z+3)/(4)`. `:. -2a+b+4c = 0 "....."(3)` Solving eqs. (2) and (3) by cross multiplication method. `(a)/(12-5) = (b)/(-10-4) = (c)/(1+6)` `rArr a/7 = (b)/(-14) = c/1 = k` `:. a = k, b = - 2 k, c = k` From eq. (1) `kx - 2ky + kz = 0` `rArr x - 2y + z = 0` Which is the requried equation of the plane. |
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