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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 ZX-plane.

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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
  • $\large\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
  • Equation of line passing through $ZX$ plane is $y=0$
Step 1:
We know that the equation of the line passing through $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is
$\large\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
The line passing through the points $(5,1,6),(3,4,1)$ is given by
Substituting for $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$
$\large\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}$
$\Rightarrow \large\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}$
Step 2:
Let this be equal to $k$
$\large\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=$$k$
Therefore $x=5-2k$,y=1+3k$,z=6-5k$
Let the coordinate of this point be
$(5-2k,1+3k,6-5k)$
Step 3:
Since the line passing through $ZX$ plane the equation is $y=0$
Therefore $1+3k=0$
$\Rightarrow k=-\large\frac{1}{3}$
Step 4:
Now substituting the value of $k$,we get the coordinates of the point is
$\big(5-2.(\large\frac{-1}{3})$$,1+3.(\large\frac{-1}{3}),$$6-5(\large\frac{-1}{3})\big)$
On simplifying we get,
$\Rightarrow \big(\large\frac{17}{3}$$,0,\large\frac{23}{3}\big)$
answered Jun 4, 2013 by sreemathi.v
 

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