Question

Find the equation of the plane that is parallel to a line
and that passes through the intersection line of two planes

Answer 1

Answer:

suppose.

X = -2 and Z = 4

1, X - y + 2 = 0

-2 + 2 = y

0= y

2, y + z -4 = 0

y +4 -4 = 0

y = 0

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Answer 2

Our plane is parallel to the line,

$\longrightarrow\dfrac{x-2}{-1}=\dfrac{y+7}{3}=\dfrac{z}{-2}$

whose direction ratio is $\left<1,\ -3,\ 2\right>.$ So this vector is parallel to the plane.

Our plane contains the intersection line of the two planes,

$\longrightarrow x-y+2=0$

$\longrightarrow x+2=y\quad\quad\dots(1)$

and,

$\longrightarrow y+z-4=0$

$\longrightarrow y=4-z\quad\quad\dots(2)$

Combining (1) and (2),

$\longrightarrow x+2=y=4-z$

or,

$\longrightarrow\dfrac{x+2}{1}=\dfrac{y}{1}=\dfrac{z-4}{-1}$

This is the equation of the intersection line. Its vector form is,

$\longrightarrow\vec{r}=\left<-2.\ 0,\ 4\right>+\lambda\left<1,\ 1,\ -1\right>$

As this line is contained in our plane, the vector $\left<1,\ 1,\ -1\right>$ is parallel to the plane and the point $(-2,\ 0,\ 4)$ lies on the plane.

Let $(x,\ y,\ z)$ be a point on the plane so that the vector $\left<x+2,\ y,\ z-4\right>$ lies along the plane.

Now we have three vectors parallel to our required plane.

• $\vec{b_1}=\left<1,\ -3,\ 2\right>$

• $\vec{b_2}=\left<1,\ 1,\ -1\right>$

• $\vec{b_3}=\left<x+2,\ y,\ z-4\right>$

Taking cross product of $\vec{b_1}$ and $\vec{b_2},$

$\longrightarrow\vec{b_1}\times\vec{b_2}=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\1&-3&2\\1&1&-1\end{array}\right|$

$\longrightarrow\vec{b_1}\times\vec{b_2}=\left<1,\ 3,\ 4\right>$

This vector is perpendicular to our plane, or to $\vec{b_3}.$

$\Longrightarrow\vec{b_3}\cdot\left(\vec{b_1}\times\vec{b_2}\right)=0$

$\longrightarrow\left<x+2,\ y,\ z-4\right>\cdot\left<1,\ 3,\ 4\right>=0$

$\longrightarrow x+2+3y+4(z-4)=0$

$\longrightarrow\underline{\underline{x+3y+4z-14=0}}$

This is the equation of our plane.