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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the coordinates of the point where the line through $(5, 1, 6)$ and $(3, 4, 1)$ crosses the YZ-plane.

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the XY-plane.

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Toolbox:
  • Equation of a line passing through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is
  • $\large\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
  • Equation of $YZ$ plane is $x=0$
Step 1:
We know that the equation of line passing through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is
$\large\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
The line passing through the points $(5,1,6)$ and $(3,4,1)$ is given by,
Substituting for $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$
$\large\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}$
(i.e) $\large\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}$
Step 2:
Let this be equal to $k$
$\large\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=$$k$
Therefore $x=5-2k$
$\qquad\quad\;\; y=3k+1$
$\qquad\quad\;\;z=6-5k$
Let the coordinates of this point be $(5-2k,3k+1,6-5k)$
Step 3:
The equation of $YZ$ plane,$x=0$
Since the line passes through the $YZ$ plane.
$5-2k=0$
$\Rightarrow k=\large\frac{5}{2}$
Step 4:
Now substituting for $k$ we get the coordinates as
$(5-2.\large\frac{5}{2},$$3.\large\frac{5}{2}$$+1,6-5.\large\frac{5}{2})$
On simplifying we get
$(0,\large\frac{17}{2}$$,\large\frac{-13}{2})$
answered Jun 4, 2013 by sreemathi.v
 

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