Question

Show that the lines $\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}$ and 3x − 2y + z + 5 = 0 = 2x + 3y + 4z − 4 intersect. Find the equation of the plane in which they lie and also their point of intersection.

Answer 1

$$\begin{gathered} \text{The equation of the given line is} \\ \frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2} \\ \text{The coordinates of any point on this line are of the form} \\ \frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}=\lambda \\ \Rightarrow x=3\lambda -4;y=5\lambda -6;z=-2\lambda +1 \\ \text{So, the coordinates of the point on the given line are}\left(3\lambda -4,5\lambda -6,-2\lambda +1\right)\text{. Since this point lies on the plane} \\ 3x-2y+z+5=0, \\ 3\left(3\lambda -4\right)-2\left(5\lambda -6\right)+\left(-2\lambda +1\right)+5=0 \\ \Rightarrow 9\lambda -12-10\lambda +12-2\lambda +1+5=0 \\ \Rightarrow -3\lambda +6=0 \\ \Rightarrow \lambda =2 \\ \mathrm{So,\; the\; coordinates\; of\; the\; point}\text{are} \\ \left(3\lambda -4,5\lambda -6,-2\lambda +1\right) \\ =\left(3\left(2\right)-4,5\left(2\right)-6,-2\left(2\right)+1\right) \\ =\left(2,4,-3\right) \\ \text{Substituting this point in another plane equation}2x+3y+4z-4=0,\text{we get} \\ 2\left(2\right)+3\left(4\right)+4\left(-3\right)-4=0 \\ \Rightarrow 4+12-12-4=0 \\ \Rightarrow 0=0 \\ \text{So, the point (2, 4, -3) lies on another plane too. So, this is the point of intersection of both the lines.} \\ \mathbf{\text{Finding the plane equation}} \\ \text{Let the direction ratios be proportional to}a,b,c. \\ \text{Since the plane contains the line}\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2},\text{it must pass through the point (-4, -6, 1) and is parallel to this line.} \\ \text{So, the equation of plane is} \\ a\left(x+4\right)+b\left(y+6\right)+c\left(z-1\right)=0...\left(1\right) \\ \text{and} \\ 3a+5b-2c=0...\left(2\right) \\ \text{Since the given plane contains the planes}3x-2y+z+5=0=2x+3y+4z-4, \\ 3a-2b+c=0...\left(3\right) \\ 2a+3b+4z=0...\left(4\right) \\ \text{Solving (3) and (4) using cross-multiplication, we get} \\ \frac{a}{-11}=\frac{b}{-10}=\frac{c}{13}...\left(5\right) \\ \text{Using (1), (2) and (5), the equation of plane is} \\ \left|\begin{array}{ccc}x+4& y+6& z-1\\ 3& 5& -2\\ 11& 10& -13\end{array}\right|=0 \\ \Rightarrow -45\left(x+4\right)+17\left(y+6\right)-25\left(z-1\right)=0 \\ \Rightarrow 45\left(x+4\right)-17\left(y+6\right)+25\left(z-1\right)=0 \\ \Rightarrow 45x-17y+25z+53=0 \end{gathered}$$