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Find the equations of the line passing through the point `(3,0,1)` parallel to the planes `x+2y=0` and `3y-z=0`. |
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Answer» Let the direction rations of the required line be a,b,c Then its equation is given by `(x-3)/(a)=(y-0)/(b) =(z-1)/(c )` Since the line (i) is parallel to ech of the planes xd+2y+oz=0 and 0x+3y-z=0 so it must be perpendicular to the normal of each of these planes `therefore axx1+bxx2+cxx0=0 rarr a+2b+0c =0` ltbvrgt `and `axx0 +bxx3+cxx(-1)=0 rarrr 0a + 3b -c =0` on solving (ii) and (iii) by cross multiplication we get `(a)/(-2-0)=(b)/(0+1)=(c )/(3-0)` `rarr (a)/(-2)=(b)/(1)=(c )/(3) =lambda(say) rarr a= 3 lambda , b = lambda` and ` c= 3lambda` putting these values of a,b,c in (i) we get `(x-3)/(-2lambda)=(y-0)/(lambda)=(z-1)/(3 lambda) rarr (x-3)/(-2)=(y)/(1)=(z-1)/(3)` Hence the required equation of the line is `(x-3)/(-2)=(y)/(1)=(z-1)/(3)` |
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