Verifying Point (3,1,6) Lies on Both Planes ∏1 and ∏2

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Homework Statement



The planes ∏1 and ∏2 have equations 3x - y - z = 2 and x + 5y + z = 14 respectively. Show that the point (3,1,6) lies on both planes.

The Question:
By finding the coordinates of another point lying in both planes, or otherwise, show that the line of intersection of ∏1 and ∏2 has equation r = 3i + j + 6k + t ( i -j + 4k ).


Homework Equations



Show point (3,1,6) lies in both planes:
Substitute point (3,1,6) into the r of both plane equations.


The Attempt at a Solution



The first part:

Let point (3,1,6) be a

Substitute a into r of both plane equations.

The dot product of a and r = ''d" of the plane equations.
LHS=RHS [Shown]

How do I attempt the second part of the question? I do not quite understand the second part of the question.
Please help me. :rolleyes:
Thanks!
 
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rainez said:

Homework Statement



The planes ∏1 and ∏2 have equations 3x - y - z = 2 and x + 5y + z = 14 respectively. Show that the point (3,1,6) lies on both planes.

The Question:
By finding the coordinates of another point lying in both planes, or otherwise, show that the line of intersection of ∏1 and ∏2 has equation r = 3i + j + 6k + t ( i -j + 4k ).


Homework Equations



Show point (3,1,6) lies in both planes:
Substitute point (3,1,6) into the r of both plane equations.


The Attempt at a Solution



The first part:

Let point (3,1,6) be a

Substitute a into r of both plane equations.

The dot product of a and r = ''d" of the plane equations.
LHS=RHS [Shown]

How do I attempt the second part of the question? I do not quite understand the second part of the question.
Please help me. :rolleyes:
Thanks!

You just need to find x,y, and z that work in both equations. You have 2 equations in 3 unknowns, so it should be easy with an extra unknown. Try something like letting y = 0 and see if you can find an x and z that work.
 


rainez said:

Homework Statement



The planes ∏1 and ∏2 have equations 3x - y - z = 2 and x + 5y + z = 14 respectively. Show that the point (3,1,6) lies on both planes.

The Question:
By finding the coordinates of another point lying in both planes, or otherwise, show that the line of intersection of ∏1 and ∏2 has equation r = 3i + j + 6k + t ( i -j + 4k ).

Homework Equations



Show point (3,1,6) lies in both planes:
Substitute point (3,1,6) into the r of both plane equations.

The Attempt at a Solution



The first part:

Let point (3,1,6) be a

Substitute a into r of both plane equations.

The dot product of a and r = ''d" of the plane equations.
LHS=RHS [Shown]

How do I attempt the second part of the question? I do not quite understand the second part of the question.
Please help me. :rolleyes:
Thanks!
You are given planes 3x - y - z = 2 and x + 5y + z = 14 and want to show that their line of intersection is r = 3i + j + 6k + t ( i -j + 4k ).
Since you are given a line you don't have to solve for it, just check as you did for the point.

You have already shown that (3, 1, 6) is in both planes so all you now need to do is show that the vector i -j + 4k lies in both planes and so in the line of intersection. And you can do that by showing it is perpendicular to the normal vectors to both planes: 3i- j- k and i+ 5j+ k.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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