# Can someone explain to me how to solve? I have a test tommorow and I am having problems with this question. Calculus and Vectors Unit/section - Vectors: intersection of two planes Determine the...

Can someone explain to me how to solve? I have a test tommorow and I am having problems with this question.

Calculus and Vectors

Unit/section - Vectors: intersection of two planes

Determine the Cartesian equation of the plane that is parallel to the line with equation x = -2y = 3z and that contains the line of intersection of the planes with equations x - y + z = 1 and 2y - z = 0.

Answer: 8x + 14y - 3z - 8 = 0

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### 2 Answers

plane x-y+z=1 has normal vector u = {1,-1,1} and

plane 3y-z=0 has normal vector v = {0,2,-1}.

Let vector w parallel to intersection of both planes,

w = u x v = {1,-1,1} x {0,2,-}

w = {-1,1,2}

Now find a vector (say ) m which perpendicular with respect to w and {1,-1/2,1/3}.

m = w x {1,-1/2,1/3}

m = {-1,1,2} x {1,-1/2,1/3}

={4/3,7/3,-1/2}

Two planes intersecting at point (1,0,0)

Put arbitrary coordinate (x,y,z) on plane where intersection line of planes {x-1 , y , z} which perpendicular to vector m.

{x-1 , y , z+2} • m = 0

{x-1,y,z}.{4/3,7/3,-1/2}=0

{x-1,y,z}.{8,14,-3}=0

8x-8+14y-3z=0

8x+14y-3z-8=0

It is required plane.

equation of the plane that is parallel to the line with equation x = -2y = 3z and that contains the line of intersection of the planes with equations x - y + z = 1 and 2y - z = 0.

Let equation of the plane containg ponit of intersection of plane

x-y+z-1=0 and 2y-z=0 be

`(x-y+z-1)+lambda(0x+2y-z)=0`

`x+(2lambda-1)y+(1-lambda)z-1=0` (i)

(i) is parallel to line `x/1=y/((-1/2))=z/(1/3)` ,therefore

`1.1+(2lambda-1).(-1/2)+(1-lambda).(1/3)=1`

`-lambda+1/2+1/3-lambda/3=0`

`-(4lambda)/3=-5/6`

`lambda=5/8`

Put `lambda=5/8` in (i) ,we get required equation.

8x+2y+3z-8=0