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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of a plane which bisects perpendicularly the line joining the points A(2,3,4) and B(4,5,8) at right angles.

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  • Vector equation of a plane is $[\overrightarrow r-\overrightarrow a].\overrightarrow n=0$ where $\overrightarrow n$ is normal and $\overrightarrow a$ is a point on the plane.
  • Mid point of AB where $A(x_1,y_1,z_1)\:and\:B(x_2,y_2,z_2)$ is given by $\bigg(\large\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\bigg)$
The given points are $A(2,3,4)\:\:and\:\:B(4,5,8)$
The line segement $AB$ is given by $(x_2-x_1),(y_2-y_1),(z_2-z_1)$
$(ie) (4-2),(5-3),(8-4)$
$=(2,2,4)=(1,1,2)$
Since the plane bisects $AB$ at rightangles, $\overrightarrow{AB}$ is the normal to the plane. which is $\overrightarrow {n}$
Therefore $ \overrightarrow n = \hat i+\hat j+2 \hat k$
Let C be the midpoint of AB.
$\bigg(\large\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\bigg)$
$=C\bigg(\large\frac{2+4}{2},\frac{3+5}{2},\frac{4+8}{2}\bigg)$
$=C\bigg(\large\frac{6}{2},\frac{8}{2},\frac{12}{2}\bigg)$=$(3,4,6)$
Let this be $\overrightarrow a=3 \hat i+4 \hat j+6 \hat k$
Hence the vector equation of the plane passing through C and $\perp$ AB is
$(\overrightarrow r-(3\hat i+4\hat j+6\hat k)).(\hat i+\hat j+2\hat k)=0$
We know $\overrightarrow r=x\hat i+ y\hat j+z \hat k$
=>$\bigg[(x\hat i+y\hat j+z\hat k).(3\hat i+4\hat j+6\hat k)\bigg].(\hat i+\hat j+2 \hat k)=0$
On simplifying we get,
$(x-3) \hat i+(y-4)\hat j+(z-6) \hat k).(\hat i+\hat j+ 2\hat k)=0$
Applying the product rule we get,
$(x-3) +(y-4)+2(z-6) =0$
On simplifying we get,
$x+y+2z=19$
This is the required equation of the plane
answered Jun 10, 2013 by meena.p
 

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