Let plane have eqn x - 2y + z = 5: choose point in plane and

[koalamath]

Problem:

Let the plane ? have the equation: x - 2y + z = 5
a) Choose a point in the plane ? and write its Cartesian coordinates.
b) Find the equation of a plane passing through the point (1,1,1) that is parallel to this plane.
c) Give an example of a plane that is perpendicular on this plane (there are an infinite number, just find one).
d) Write the symmetric equations of the line that is perpendicular on this plane and passes through the origin

 

[Deleted member 4993]

Problem:

Let the plane ? have the equation: x - 2y + z = 5
a) Choose a point in the plane ? and write its Cartesian coordinates.
b) Find the equation of a plane passing through the point (1,1,1) that is parallel to this plane.
c) Give an example of a plane that is perpendicular on this plane (there are an infinite number, just find one).
d) Write the symmetric equations of the line that is perpendicular on this plane and passes through the origin

Please show us your work - indicating exactly where you are stuck - so that we know where to begin to help you.

 

[koalamath]

so for a I plugged in 1 and 0 for y and z and found coordinates (7,1,0)

b) I took cross product and found 2x-4y+2z

c) 3x-4y+5z because the dot product equals 0

d is where im stuck I know that the equation for symmetric equations but I don't understand what the question is asking. How do I find the line that is perpendicular and passes through the origin do I just use my answer from c and then find the direction variable from that and the original equation?

Thank you

 

[Deleted member 4993]

so for a I plugged in 1 and 0 for y and z and found coordinates (7,1,0)<--------Correct

b) I took cross product and found 2x-4y+2z <----- This is not an equation

c) 3x-4y+5z <----- This is not an equationbecause the dot product equals 0

d is where im stuck I know that the equation for symmetric equations but I don't understand what the question is asking. How do I find the line that is perpendicular and passes through the origin do I just use my answer from c and then find the direction variable from that and the original equation?

Thank you

I'll show you another problem:

What is the symmetric equation of the line through point (4,-5,20) & perpendicular to the plane x+3y-6z-8=0.?

The normal vector n, to the given plane

x + 3y - 6z - 8 = 0
is
n = <1, 3, -6>

The vector n is also the directional vector of the line thru point P(4, -5, 20) and perpendicular to the given plane.

The equation of the line L is:

L = P + tn = <4, -5, 20> + t<1, 3, -6>
L = <4 + t, -5 + 3t, 20 - 6t>

Now convert the equation of the line to symmetric form by solving for t.

L:
x = 4 + t
y = -5 + 3t
z = 20 - 6t

Symmetric form of equation.

(x - 4)/1 = (y + 5)/3 = (z - 20)/(-6)

 

[koalamath]

how do you know The vector n is also the directional vector of the line thru point P(4, -5, 20) and perpendicular to the given plane
would that mean that in my equation the vector (1,-2,1) is the directional vector and perpendicular?

 

[Deleted member 4993]

how do you know The vector n is also the directional vector of the line thru point P(4, -5, 20) and perpendicular to the given plane
would that mean that in my equation the vector (1,-2,1) is the directional vector and perpendicular?

Parallel planes have normals with same direction cosines (or unit vectors).