Complete the parametric equation for the line where the planes cross

[MP97]

Homework Statement

Complete the parametric equation for the line where the planes cross.

Relevant Equations

Plane 1: x-12z=-36
Plane 2: -11x16y-14z=-30

Note: the first parametric equation is given as x(t)=-72t

First, I use the unit vector of each plane, and I compute their cross-product to obtain a vector parallel to the line of interception.

Then, I algebraically use x=0 to obtain the coordinates of the point in the line of interception. However, not having a y coordinate in plane one is confusing me.

 

[WWGD]

It seems you're missing a sign in Plane 2. Also, please use Latex for your Math content.

 

[Mark44]

Homework Statement: Complete the parametric equation for the line where the planes cross.
Relevant Equations: Plane 1: x-12z=-36
Plane 2: -11x16y-14z=-30

Note: the first parametric equation is given as x(t)=-72t

You haven't mentioned how this equation will come into play.

First, I use the unit vector of each plane, and I compute their cross-product to obtain a vector parallel to the line of interception.

You don't need a unit vector for each plane, but you do need a normal vector for each one.

Then, I algebraically use x=0 to obtain the coordinates of the point in the line of interception. However, not having a y coordinate in plane one is confusing me.

You can use the given equation of x as a function of t to solve for z in the first plane, and then use your equations of x and z as functions of t to solve for y in the second plane.

In any case, please show us what you have done.

 

[Bosko]

Complete the parametric equation for the line where the planes cross.
Plane 1: x-12z=-36

Ok the first one is a plane ...

$x-12z=-36$
$x+36=12z$
$\frac{x}{12}+\frac{36}{12}=z$
$z=\frac{x}{12}+3$

... parallel to the y axis

Plane 2: -11x16y-14z=-30

An second "plane" is ...

$-11x16y-14z=-30$
$-176xy+30=14z$
$z= \frac{-176xy}{14}+\frac{30}{14}$

... is not a plane at all, but a saddle ( hyperbolic paraboloid )....

Wolfram Alpha -> https://www.wolframalpha.com/input?i=z=+\frac{-176xy}{14}+\frac{30}{14}

Note: the first parametric equation is given as x(t)=-72t

First, I use the unit vector of each plane, and I compute their cross-product to obtain a vector parallel to the line of interception.

Then, I algebraically use x=0 to obtain the coordinates of the point in the line of interception. However, not having a y coordinate in plane one is confusing me.

... I don't understand the goal?

 

[SammyS]

Ok the first one is a plane ...

$x-12z=-36$

... parallel to the y axis

An second "plane" is ...

$-11x16y-14z=-30$
$-176xy+30=14z$
$z= \frac{-176xy}{14}+\frac{30}{14}$

... is not a plane at all, but a saddle ( hyperbolic paraboloid )....

Very likely, there is a typographical error in the equation for the second plane. Probably it should be
$\displaystyle -11x+16y-14z=-30 \ \$ or $\displaystyle \ \ -11x-16y-14z=-30$.

If OP had shown us any significant amount of his work on this problem, the correct equation would have been evident.

 

[WWGD]

Very likely, there is a typographical error in the equation for the second plane. Probably it should be
$\displaystyle -11x+16y-14z=-30 \ \$ or $\displaystyle \ \ -11x-16y-14z=-30$.

If OP had shown us any significant amount of his work on this problem, the correct equation would have been evident.

I had asked for clarification in that respect too.

 

[Mark44]

$-11x16y-14z=-30$

I'm 99.44% certain that the equation above is missing a '+' sign between -11x and 16y.

 

[Bosko]

I'm 99.44% certain that the equation above is missing a '+' sign between -11x and 16y.

Me to, either + or - sign.