# Equation of plane

## Homework Statement

Find the equation of the plane which passes through M(2,3,1) and the line

$$\left\{\begin{matrix} 2x-y+3z-1=0 & \\ 5x+4y-z-7=0 & \end{matrix}\right.$$

## Homework Equations

$$\frac{x-x_1}{\begin{vmatrix} B_1 & C_1\\ C_1 & C_2 \end{vmatrix}}=\frac{y-y_1}{\begin{vmatrix} C_1 & A_1\\ C_2 & A_2 \end{vmatrix}}=\frac{z-z_1}{\begin{vmatrix} A_1 & B_1\\ A_2 & B_2 \end{vmatrix}}$$

## The Attempt at a Solution

first I find the equation of the line:

$$\frac{x}{-1}=\frac{y+7}{-5}=\frac{z+3}{-3}$$

What should I do next?

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Are you familiar with cross products? (You must be, to some extent, as you have posted determinants. A solution of this form I have never seen before.)
If yes, then $$N1 \\ X \\ N2$$ will be the normal to your plane, where N1 and N2 are the normals to the lines given.

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$n_1 x n_2$ will give me vector perpendicular to this ones, but how it will help me?

from that, you will have a, b, and c of the following equation for a plane.

$$a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=d$$

In other words, you will have the normal to the plane you're trying to solve for. Substitution of the point they give you will solve it.

$$n_1 x n_2=(-1,-5,-3)[/itex] A(x+1)+B(y+5)+c(z+3)=D Hmm... And what about x,y,z, should I substitute for M? HallsofIvy Science Advisor Homework Helper Are you familiar with cross products? (You must be, to some extent, as you have posted determinants. A solution of this form I have never seen before.) If yes, then [tex] N1 \\ X \\ N2$$ will be the normal to your plane, where N1 and N2 are the normals to the lines given.
Unfortunately, any line, R3 has an normals in an infinite number of directions- it doesn't make sense to talk about "the" normal to a line.

HallsofIvy
Homework Helper

## Homework Statement

Find the equation of the plane which passes through M(2,3,1) and the line

$$\left\{\begin{matrix} 2x-y+3z-1=0 & \\ 5x+4y-z-7=0 & \end{matrix}\right.$$

## Homework Equations

$$\frac{x-x_1}{\begin{vmatrix} B_1 & C_1\\ C_1 & C_2 \end{vmatrix}}=\frac{y-y_1}{\begin{vmatrix} C_1 & A_1\\ C_2 & A_2 \end{vmatrix}}=\frac{z-z_1}{\begin{vmatrix} A_1 & B_1\\ A_2 & B_2 \end{vmatrix}}$$

## The Attempt at a Solution

first I find the equation of the line:

$$\frac{x}{-1}=\frac{y+7}{-5}=\frac{z+3}{-3}$$

What should I do next?
Choose any two points on the line. For example, if you choose x= 0, you can solve the two equations 2(0)-y+3z-1=0 and 5(0)+4y-z-7=0 for y= 2 and z= 1: (0, 2, 1) is on the line. If you take x= 1, you can solve 2(1)- y+ 3z- 1= 0 and 5(1)+ 4y- z- 7= 0 for y= -23/11 and z= -2/11. The cross product of the vectors from (2, 3, 1) to those points will give you the normal to the plane.

I retract having said 'normals' to the line. I got 'normal-happy' , what with the terminology for planes.

But the plane don't necessarily need to intersect the line in 2 points. It can intersect it in one point...

dx
Homework Helper
Gold Member
The plane intersects the line at an infinite number of points, i.e. the line is contained in the plane.

dx
Homework Helper
Gold Member
Otherwise, you would need more information than you are given to determine the plane uniquely.

HallsofIvy
Homework Helper
But the plane don't necessarily need to intersect the line in 2 points. It can intersect it in one point...
What plane are you talking about? You were told to find the plane that contains that line. Every point on the line is in that plane.

Yes, sorry. My fault.

Will it be correct if I show it like this:

$$a \circ n=0$$

$$(a_1,a_2,a_3) \circ (A,B,C)=0$$

So one of the conditions is:

-A-5B-3C=0

the second condition is:

2A+3B+C+D=0

and the third condition:

0*A-7B-3C+D=0

$$\left\{\begin{matrix} -A-5B-3C=0 & \\ 2A+3B+C+D=0 & \\ -7B-3C+D=0 & \end{matrix}\right.$$

The result is:

$$-5x-y-7=0$$

right?

Just to mention that in the first post the line is:

$$\left\{\begin{matrix} x+y-2z+1=0 & \\ 2x-y+z-4=0 & \end{matrix}\right.$$

Sorry, for mistake, I type the wrong line...

And can I ask you something?

Two planes to be parallel, must this condition be satisfied?

$$A_1x+B_1y+C_1z+D=0$$

$$A_2x+B_2y+C_2z+D=0$$

$$\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}$$, since in my book doesn't look like that...