Answer to Question #211905 in Analytic Geometry for Faith

Question

1.find the vector form of the equation of the plane that passes through the point po=(1,-2,3) and has normal vector n=<3,1,-1>.

2.find an equation for the plane that contains the line x=-1+3t,y=5+3t,z=2+t and is parallel to the line of intersection of the planes x-2(y-1)+3z=-1 and y=-2x-1=0.

Answer 1

1. P(1, -2, 3)

normal vector n = 3i + j - k

Position vector of point P = i - 2j + 3k

The vector form of a plane that consists of a vector r₀ and a normal vector n is given by

n . ( r - r₀) = 0

So, the required equation of plane is

(3i + j - k ) . [r - ( i - 2j + 3k ) ] = 0

2.line x=-1+3t,y=5+3t,z=2+t

The direction ratios of the line are d₁= <3, 3, 1> and the point on the line is r₀ = <-1, 5, 2>

The point that lies on the line would also lie on the plane. So, the position vector of the point is

- i + 5j + 2k.

The plane is parallel to the line of intersection of the planes

 x - 2y + 3z = -3 and 2x + y = 1

To find the line of intersection of the given planes we let z = t.

So,

x - 2y = -3 -3t and 2x + y = 1

On solving the above equations we get the line as :

<x, y, z> = <$\dfrac{6}{5}$, $\dfrac{-7}{5}$, 0 > + t <$\dfrac{3}{5}, \dfrac{-6}{5}, 1$ >

The direction ratios of the line of intersection of the planes are d₂ = <$\dfrac{3}{5}, \dfrac{-6}{5}, 1$ >

Let the normal vector of the plane be n.

So,

n = d₁ X d₂

n = $\begin{vmatrix} i & j & k\\ 3 & 3 &1\\ 3/5&-6/5&1 \end{vmatrix}$ = $\dfrac{21}{5} i - \dfrac{12}{5}j - \dfrac{27}{5}k$

Hence the equation of the plane is

n.(r - r₀) = 0

( $\dfrac{21}{5} i - \dfrac{12}{5}j - \dfrac{27}{5}k$ ) . [ r - (-i + 5j + 2k)] = 0