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  • Find the equation of the line passing through  and of the plane passing through <img a

Find the equation of the 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. 

 

 

 

 

 
 
 
 
 

Answers (1)

We know that equation of the line passing through points (x_1,y_1,z_1)\:and \: (x_2,y_2,z_2) is given by 

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

So the equation of line is given by 

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

Now equation of plane passing through points(2,0,3), ( 1,1,5)\:\:and\:\:(3,2,4) is given by 

\begin{vmatrix} x-x_1 &y-y_1 &z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 &z_3-z_1 \end{vmatrix}=0\\ \begin{vmatrix} x-2 & y-0 &z-3 \\ -1 &1 &2 \\ 1& 2 & 1 \end{vmatrix}=0\\ \Rightarrow (x-2)(-3)-y(-1-2)+(2-3)(-2-1)=0 \\ \Rightarrow -3x+6+3y-3z+9=0 \\ \Rightarrow -x+y-z+5=0 \:\:\: - (i)

This is the required equation of the plane 

Now, Let 

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

on putting these values in equation (i) we have, 

-(3k+2)4k-1-(2k+2)+5=0 \\ \Rightarrow -3k-2+4k-1-2k-2+5=0 \\ \Rightarrow -2k=0 \\ \Rightarrow k =0

Then intersection points are 

x=3k+2=2 \\ y= 4k-1 = -1 \\ z=2k+2=2 \\ \therefore points \: of \: intersection \: is \: (2,-1,2)

Posted by

Ravindra Pindel

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