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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of the plane through the points (2,1,0),(3,-2,-2) and (3,1,7).

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Toolbox:
  • Normal vector $\perp$ to both $\overrightarrow a\:and\:\overrightarrow b$ is
  • $\overrightarrow a\times \overrightarrow b=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{array}\right|$
  • Vector equation of the plane passing through $\overrightarrow a$ and $\perp$ to $\overrightarrow n$ is $(\overrightarrow r-\overrightarrow a).\overrightarrow n=0$
Step 1:
Let the given points be
$A(2, 1, 0), B(3, -2, -2) C(3, 1, 7)$
The direction ratios of $ \overrightarrow{AB}$ is
$(x_2-x_1),(y_2-y_1),(z_2-z_1)$
$(ie) (3-2),(-2-1),(-2-0)$
$=(1, -3, -2)$
The direction ratios of $ \overrightarrow{BC}$ is
$(ie) (3-3),(1-(-2)),(7-(-2))$
$=(0, 3, 9)$
We know that normal vector $\overrightarrow n\:\:is\:\:\perp$ to $\overrightarrow {AB}\:\:and\:\:\overrightarrow {BC}$
Hence $ \overrightarrow n = \overrightarrow{AB}\times \overrightarrow{BC}$
$=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\1&-3&-2\\0&3&9\end{array}\right|$
$=(-27+6)\hat i-(9-0)\hat j+(3-0)\hat k$
$=-21\hat i - 9 \hat j+3 \hat k$
Step 2:
Vector equation of the plane is
$(\overrightarrow r-\overrightarrow a).\overrightarrow n=0$
We know $\overrightarrow r=x \hat i+y \hat j+z \hat k,\overrightarrow n=21 \hat i -9 \hat j+3 \hat k,$ and $\overrightarrow a=2 \hat i+ \hat j$
Substituting the respective values we get,
$\bigg[(x \hat i+y \hat j+z \hat k)-(2 \hat i+ \hat j)\bigg].(-21 \hat i -9 \hat j+3 \hat k)=0$
$\bigg[(x -2)\hat i+(y-1) \hat j+z \hat k\bigg].(-21 \hat i -9 \hat j+3 \hat k)=0$
On multiplying by using the dot product we get,
$(x-2)(-21)+(y-1)(-9)+3z=0$
$=>21 x+42-9y+9+3z=0$
On further simplifying we get,
$-21x-9y+3z=-51$
=>$21x+9y-3z=51$
This is the required equation of the plane
answered Jun 12, 2013 by meena.p
 

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