logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
0 votes

Find the coordinates of the point where the line through $(3, – 4, – 5)$ and $(2, – 3, 1)$ crosses the plane $2x + y + z = 7.$

$\begin{array}{1 1}(1,-2,7) \\ (5,-6,-17) \\ (0,-1,13) \\(6,-7,-23) \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • The equation of line passing through the line passing through $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is
  • $\large\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
Step 1:
We know that the equation of the line passing through the point $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is
$\large\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
Now the given line passes through the points $(3,-4,-5)$ and $(2,-3,1)$
Hence the equation of the line is
$\large\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}$
(i.e) $\large\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}$
Step 2:
Let the above equation be equal to $k$
Therefore $\large\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}$$=k$
On simplifying we get
$x=3-k$
$y=-4+k$
$z=-5+6k$
Therefore the coordinates of the points are $(3-k,-4+k,-5+6k)$
Step 3:
The point lies on the line $2x+y+z=7$
Now substituting for $x$,$y$,$z$ we get,
$2(3-k)+(-4+k)+(-5+6k)=7$
On simplifying we get,
$6-2k-4+k-5+6k=7$
$5k-3=7$
$\Rightarrow 5k=10$
$k=2$
Step 4:
Now substituting the value of $k$,we get the coordinates of the point is
$(3-2),(-4+2),(-5+6\times 2)$
On simplifying we get
$\Rightarrow (1,-2,7)$
Hence the required point is $(1,-2,7)$
answered Jun 4, 2013 by sreemathi.v
edited Jun 4, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...