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  • Find the coordinates of the point where the line through the points (3, – 4, – 5) and (2, – 3,1), crosses the plane determined by the points (1, 2, 3), (4, 2, – 3) and (0, 4, 3).  

Find the coordinates of the point where the line through the points (3, – 4, – 5) and
(2, – 3,1), crosses the plane determined by the points (1, 2, 3), (4, 2, – 3) and (0, 4, 3).

 

 

 

 
 
 
 
 

Answers (1)

Equation of the plane determined by the points (1,2,3) (4,2,-3) and (0,4,3) is
\begin{bmatrix} x-1 & y-2 &z-3 \\ 4-1& 2-2 &-3-3 \\ 0-1& 4-2 & 3-3 \end{bmatrix}= 0 \Rightarrow \begin{bmatrix} x-1 & y-2 &z-3 \\ 3& 0&-6 \\ -1& 2 & 0 \end{bmatrix}= 0
\left ( x-1 \right )\left ( 0+12 \right )-\left ( y-2 \right )\left ( 0-6 \right )+\left ( z-3 \right )\left ( 6-0 \right )= 0
\Rightarrow 12x-12+6y-12+6z-18= 0 \Rightarrow 12x+6y+6z-42= 0
\Rightarrow 2x+y+z-7= 0\; \; \Rightarrow 2x+y+z= 7 ---(i)
Now equation of line through (3,-4,-5) and (2,-3,1 )is
\frac{x-3}{2-3}= \frac{y+4}{-3+4}= \frac{z+5}{1+5}= \lambda
ie\; \frac{x-3}{-1}= \frac{y+4}{1}= \frac{z+5}{6}= \lambda
\therefore coordinates of any point on line is  : p\left ( -\lambda +3 ,\lambda -4,6\lambda -5\right )
Now the point p crosses the plane.
\therefore It satisfies the equation (1) of plane
2\left ( -\lambda +3 \right )+\left ( \lambda -4 \right )+\left ( 6\lambda -5 \right )= 7
\Rightarrow -2\lambda +6+\lambda -4+6\lambda -5= 7
\Rightarrow 5\lambda -3= 7\; \; \; \Rightarrow 5\lambda= 10
\therefore \lambda = 2 Hence the point of intersection is (1,-2,7)

Posted by

Ravindra Pindel

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