What is the relationship of the given plane 2x-10y-2z+2=0 to each of the planes below? Is it coincident, parallel or intersecting? 1. 7x-35y-7z = 112

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lemjay's profile pic

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

The equation of plane 1 is:

`2x -10y -2z + 2 =0`

`2x-10y-2z=-2`    (EQ1)

And the equation of plane 2 is:

`7x-35y-7z=112`   (EQ2)

To determine if the two planes intersect, parallel or coincident, let's use elimination method of system of equations.

Let's try to eliminate x. To do so, multiply plane1 by 7 and plane2 by -2. 

`7*EQ1:`

 `7(2x-10y-2z)=-2*7`

`14x - 70y - 14z=-14`

`-2*EQ2:`

`-2(7x-35y-7z)=112*(-2)`

`-14x+70y+14z=-224`

And add 7EQ1 and -2EQ2 .

         `14x - 70y - 14z=-14`

`+`   `-14x+70y+14z=-224`

`---------------------`

        `0x + 0y + 0z =-238`

                          `0 = -238`    (False)

Eliminating one of the variables result to 0=-238, which is a False condition. This means that the two equations has no solution.

Hence, the two planes are parallel.                                             

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to find the normal vectors to the given planes, such that:

`bar (n_1) = <2,-10,-2>`

`bar (n_2) = <7,-35,-7>`

You need to evaluate the cross product of normal vectors, such that:

`bar (n_1) X bar (n_2) = [(bar i, bar j, bar k),(2,-10,-2),(7,-35,-7)]`

`bar (n_1) X bar (n_2) = 70bar i - 70 bar k - 14 bar j + 70 bar k - 70 bar i + 14 bar j`

`bar(n_1) X bar(n_2) = 0`

Since the cross product of normal vectors to the given planes yields 0, then the plane `2x-10y-2z+2=0` is parallel to `7x-35y-7z = 112` .

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