Find an equation of the plane that passes through the line of intersection of the planes x-z=1 and y+2z=3 and?

Question

Find an equation of the plane that passes through the line of intersection of the planes x-z=1 and y+2z=3 and is perpendicular to the plane x+ y-2z=1...

Answer 1

First find the line L, of intersection of the two planes.

x - z - 1 = 0
y + 2z - 3 = 0

The directional vector v, of the line will be normal to the normal vectors of the two planes. Take the cross product.

n1 = <1, 0, -1>
n2 = <0, 1, 2>

v = n1 X n2 = <1, -2, 1>

Find a point P on the line L.

x - z - 1 = 0
y + 2z - 3 = 0

Let z = 0.

From the first equation.
x = z + 1 = 0 + 1 = 1

From the second equation.
y = -2z + 3 = 0 + 3 = 3

A point on the line is therefore P(1, 3, 0).

The equation of the line L is:

L = P + tv = <1, 3, 0> + t<1, -2, 1>
where t is a constant ranging over the real numbers
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The plane containing L is perpendicular to the plane

x + y - 2z - 1 = 0

The normal vector n3, of this plane is:

n3 = <1, 1, -2>

Therefore the normal vector n3, of the given plane is a directional vector of the desired plane. Take the cross product of the directional vectors v and n3 to get the normal vector n4, of the desired plane.

n4 = v X n3 = <1, -2, 1> X <1, 1, -2> = <3, 3, 3>

Any non-zero multiple of n4 is also a normal vector of the desired plane. Divide by 3.

n4 = <1, 1, 1>

With the normal vector n4 and the point P(1, 3, 0) in the plane we can write the equation of the desired plane.

1(x - 1) + 1(y - 3) + 1(z - 0) = 0
x - 1 + y - 3 + z = 0
x + y + z - 4 = 0
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As a test the dot product of the normal vector of the given plane n3, and the desired plane n4 should be zero if they are orthogonal.

n3 • n4 = <1, 1, -2> • <1, 1, 1> = 1 + 1 - 2 = 0

So the planes are orthogonal.

Answer 2

First discover the directional vector v, of the line of intersection of the 1st 2 given planes. The directional vector will lie in the two planes and so be orthogonal to the traditional vectors n1 and n2 of the given planes. Take the circulate product. v = n1 X n2 = X <0, a million, 2> = Now we would desire to discover a factor interior the wished airplane. to try this we are able to come across a factor on the line of intersection. this would be a factor in the two between the 1st 2 given planes. enable z = 0 and remedy for x and y. x = a million y = 3 the factor interior the wished airplane is P(a million, 3, 0). by way of fact the wished airplane is perpendicular to the 0.33 given airplane, the traditional vector n3, of the 0.33 given airplane is a directional vector of the wished airplane. to discover the traditional vector n, of the wished airplane, take the circulate manufactured from its 2 directional vectors v and n3. n = v X n3 = X = <3, 3, 3> Any non-0 diverse of n is likewise a common vector to the wished airplane. Divide with the aid of 3. n = With a factor P(a million, 3, 0) interior the airplane and the traditional vector to the airplane we are able to write the equation of the wished airplane. keep in mind, the traditional vector to the airplane is orthogonal to each vector that lies interior the airplane. And the dot manufactured from orthogonal vectors is 0. To create a vector that lies interior the airplane enable R(x, y, z) be an arbitrary factor interior the airplane. n • PR = 0 n • = 0 • = 0 a million(x - a million) + a million(y - 3) + a million(z - 0) = 0 x - a million + y - 3 + z = 0 x + y + z - 4 = 0