Question

find the equation of line passing through (2,-1,2)and (5,3,4)and of the plane passing through (2,0,3),(1,1,5)and (3,2,4) also find their point of intersection​

Answer 1

Answer:

Therefore the equation of required line is

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$

The equation of plane is

x-y+z-5=0

Therefore the intersection of point is (8,7,6)

Step-by-step explanation:

If a straight line passes through the points (x₁,y₁,z₁) and (x₂,y₂,z₂) then the equation of the straight line is

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Here x₁=2,y₁=-1,z₁=2,x₂=5,y₂=3,z₂=4

Therefore the equation of line which passes through the points (2,-1,2) and (5,3,4) is

$\frac{x-2}{5-2}=\frac{y-(-1)}{3-(-1)}=\frac{z-2}{4-2}$

$\Rightarrow \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$

If a plane passes through the points  (x₁,y₁,z₁) , (x₂,y₂,z₂) and (x₃,y₃,z₃).

Then the equation of plane

$\left|\begin{array}{ccc}x-x_1&y-y_1&z-z_1\\(x_1-x_2)&(y_1-y_2)&(z_1-z_2)\\(x_1-x_3)&(y_1-y_3)&(z_1-z_3)\end{array}\right|=0$

Here x₁=2,y₁=0,z₁=3,x₂=1,y₂=1,z₂=5,x₃=3,y₃2,z₃=4

The equation of plane is

$\left|\begin{array}{ccc}x-2&y-0&z-3\\(2-1)&(0-1)&(3-5)\\((2-3)&(0-2)&(3-4)\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{ccc}x-2&y&z-3\\(1)&(-1)&(-2)\\(-1)&(-2)&(-1)\end{array}\right|=0$

⇒(x-2)(1-4)+y(2+1)+(z-3)(-2-1)=0

⇒-3(x-2)+3y-3(z-3)=0

⇒-3(x-2-y+z-3)=0

⇒x-y+z-5=0

The equation of the straight line is

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2} =r$

Any point of the straight line be (3r+2,4r-1,2r+2)

[$\frac{x-2}{3} =r \rightarrow x-2=3r \rightarrow x=3r+2$,  similarly for  y and z]

Let P(3r+2,4r-1,2r+2) be the intersection point of the line and the plane.

Then the point will also lie on the plane. Then the point will be satisfy the equation of the plane.

Putting x=3r+2,y=4r-1,z=2r+2 in the equation of plane

3r+2-4r+1+2r+2-5=0

⇒r-2=0

⇒r=2

Therefore the intersection of point is (3.2+2,4.2-1,2.2+2)=(8,7,6).