How Do Parallel and Orthogonal Orientations Affect Line and Plane Equations?

[Offlinedoctor]

I having trouble understanding the difference between parallel and orthogonal in relation to finding the relevant line or plane equations.

Example;

Determine the vector and Cartesian line if:
a) passes through (2,1,-3) and is parallel to v=(1,2,2)
What would happen if it was perpendicular though?

And the other one pertains to planes;
Plane passes through point (1,4,5) and perpendicular to (7,1,4).

I'm having confusion on what would happen if the opposites happened for both cases.

 

[HallsofIvy]

I having trouble understanding the difference between parallel and orthogonal in relation to finding the relevant line or plane equations.

Example;

Determine the vector and Cartesian line if:
a) passes through (2,1,-3) and is parallel to v=(1,2,2)
What would happen if it was perpendicular though?

Then it would not be "well defined". In two dimensions, there would be exactly one line thorugh a given point perpendicular to a given line but in three dimensions, there are an infinite number of lines perpendicular to the given line- all lying in the plane perpendicular to the given line containing the given point.

And the other one pertains to planes;
Plane passes through point (1,4,5) and perpendicular to (7,1,4).

I assume you mean the vector (7, 1, 4). That's why I prefer to write vectors as <x, y, z> rather than (x, y, z)- to avoid confusing them with points. If (x, y, z) is a point in that plane then <x- 1, y- 4, z- 5> is a vector lying in the plane. It must be perpedicular to the vector <7, 1, 4> so the dot product <7, 1, 4>.<x-1, y- 4, z- 5>= 7(x- 1)+ 1(y- 4)+ 4(z- 5)= 0.

I'm having confusion on what would happen if the opposites happened for both cases.

What "opposite" do you mean for the second case?
A line is one dimensional. In two- dimensions, there is only "other" coordinate so we can write y= ax+ b. In three dimensions there are two "other" coordinates so, to describe a line, we must use two equations, y= ax+ b, z= cx+ d, or use three "parametric equations".

 

[Offlinedoctor]

Is it right to assume that,
To find vector equations for lines in r2 and r3, we'd generally use the equation,
R=Ro+tV, where Ro represents a point the line pass and V presents the vector parallel to the line, what if that vector was perpendicular instead? That's what confuses me the most.

And for plane equations I'm having trouble understanding the relationship of the normal vector to finding a planes equation, despite seeing proofs that relate them.

 

[haruspex]

Is it right to assume that,
To find vector equations for lines in r2 and r3, we'd generally use the equation,
R=Ro+tV, where Ro represents a point the line pass and V presents the vector parallel to the line,

Yes.

what if that vector was perpendicular instead?

First, find a vector orthogonal to V. In 2D, swap the x and y values in V and negate one of them. In 3D, as Halls says, there are multiple solutions. there is an entire plane orthogonal to V and any vector in it would do.

And for plane equations I'm having trouble understanding the relationship of the normal vector to finding a planes equation, despite seeing proofs that relate them.

In 3D, there is a unique line through a point orthogonal to a given plane, and a unique plane through a point orthogonal to a given line. How would you represent a plane?

 

[Offlinedoctor]

I've just ot one more thing that's concerning me, for a question such as, find Cartesian and vector equations of a plane perpendicular to (1,0-2) and containing the point (1,-1,-3), my professor seemed to just use formula A(X-Xo)+B(Y-Yo)+C(Z-Zo)=o, and then just assigned parameters to determine the vector equation.

Yet the textbook seems to use the idea that, X=Xo+sV+tU, where V and U are parallel to the plane and not each other...