Question: Find the equation of the plane containing two lines (x-1)/2=(y-2)/3=(z-3)/4 and (x-2)/3=(y-3)/4=(z-4)/5 .
Solution :
First we shall find direction vector of the lines \[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\] and \[\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\] are \[\vec{d_1}=2\hat{i}+3\hat{j}+4\hat{k},\ \vec{d_2}=3\hat{i}+4\hat{j}+5\hat{k}\] Respectively. Therefore direction vector of the perpendicular line to the given two lines is given as \[\hat{d}=\left|\begin{matrix}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\3&4&5\\\end{matrix}\right|=-\hat{i}+2\hat{j}-\hat{k}\] Therefore direction ratios of the perpendicular line are a =-1, b=2, c=-1 Since the point (1, 2, 3) will lie on plane and direction ratios of the perpendicular line (-1, 2, -1). Therefore (x1,y1,z1) = (1, 2, 3) , (a,b,c)= (-1, 2, -1) Now we know that the equation of a plane passing through (x1,y1,z1) and perpendicular to a line with direction ratios a, b, c is given as \[a(x-x_1)\ +\ b(y-y_1)\ +\ c(z-z_1)\ =\ 0\] Substitute the values we get =-1(x-1) +2(y-2) -1(z-3) = 0 => -x + 1 +2y – 4 -z+3 = 0 => -x +2y -z = 0 =>x-2y+z=0 Is the required equation of the plane.