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Find the equation of the plane through the point whose $p.v. $ is $\overrightarrow{2i}-\overrightarrow{j}+\overrightarrow{k}$ and perpendicular to the vector $\overrightarrow{4i}+\overrightarrow{2j}-\overrightarrow{3k}$

1 Answer

  • Equation of the plane through a point $A(x_1, y_1, z_1) [pv=\overrightarrow a]$ and perpendicular to vector $ \overrightarrow n = a \overrightarrow i+b \overrightarrow j+c \overrightarrow k$
  • $ \overrightarrow r. \overrightarrow n = \overrightarrow a. \overrightarrow n $ (vector equation )
  • $ a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ (cartesian equation)
Step 1
The point $A$ with $PV$ $ \overrightarrow a=2\overrightarrow i-\overrightarrow j+\overrightarrow k$ lies in the plane and $\overrightarrow n=4\overrightarrow i+2\overrightarrow j-3\overrightarrow k$ is normal the plane.
Step 2
The vector equation of the plane is
$ \overrightarrow r.\overrightarrow n=\overrightarrow a.\overrightarrow n$
$ \overrightarrow r.(4\overrightarrow i+2\overrightarrow j-3\overrightarrow k)=(2\overrightarrow i-\overrightarrow j+\overrightarrow k).(4\overrightarrow i+2\overrightarrow j-3\overrightarrow k)$
$\overrightarrow r.(4\overrightarrow i+2\overrightarrow j-3\overrightarrow k)=8-2-3$
$ \overrightarrow r.(4\overrightarrow i+2\overrightarrow j-3\overrightarrow k)=3$
$ \overrightarrow r=x\overrightarrow i+y\overrightarrow j+z\overrightarrow k$
Step 3
The cartesian equation is
$ 4x+2y-3y=3$
answered Jun 14, 2013 by thanvigandhi_1

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