Question

Show that the lines $\frac{x+4}{3}=\frac{y+6}{5}=\frac{x-1}{-2}$ 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

We have,

Given that,

$\frac{x+4}{3}=\frac{y+6}{5}=\frac{x-1}{-2}$ …… (A)

$3x-2y+z+5=0......\left(1\right)$

$2x+3y+4z=4......\left(2\right)$

Then, the coordinates of any point on this line are of the form

$\frac{x+4}{3}=\frac{y+6}{5}=\frac{x-1}{-2}=\lambda$

So, the coordinates of the point on the given line are $$(3\lambda -4,5\lambda -6,-2\lambda +1)$$. Since, this points lies on the plane

$3x-2y+z+5=0$

$3(3\lambda -4)-2(5\lambda -6)+(-2\lambda +1)+5=0$

$\Rightarrow 9\lambda -12-10\lambda +12-2\lambda +1+5=0$

$\Rightarrow -3\lambda +6=0$

$\Rightarrow \lambda =2$

$\text{So,the}\text{coordinates}\text{of}\text{the}\text{points}\text{are}$

$(3\lambda -4,5\lambda -6,-2\lambda +1)$

$=(3\left(2\right)-4,5\left(2\right)-6,-2\left(2\right)+1)$

$=(2,4,3)$

Substituting this point in another plane equation

$2x+3y+4z-4=0,$

We get,

$2\left(2\right)+3\left(4\right)+4(-3)-4=0$

$\Rightarrow 4+12-12-4=0$

$\Rightarrow 0=0$

So, the point $$(2,4,-3)$$ lies on another plane too.so, this is the intersection of both the lines.

Finding the equation of plane

Let the direction ratios be proportional to a,b,c.

Since the plane contains the line $$\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}$$.

It must passes through the point (-4,-6,1) and is parallel,

So, the equation of plane is

$a(x+4)+b(y+6)+c(z-1)=0.....\left(1\right)$

$\text{and}$

$3a+5b-2c=0......\left(2\right)$

$\text{Since}\text{the}\text{given}\text{contains}\text{planes}3x-2y+z+5=0=2x+3y+4z-4$

$3a-2b+c=0......\left(3\right)$

$2a+3b+4z=0......\left(4\right)$

$\text{Solving}\text{equation}\left(\text{3}\right)\text{and}\left(\text{4}\right)\text{to,and}\text{we}\text{get,}$

$\Rightarrow -45(x+4)+17(y+6)-25(z-1)=0$

$\Rightarrow 45(x+4)-17(y+6)+25(z-1)=0$

$\Rightarrow 45x-17y+25z+53=0$

Hence, this is the answer.