3d plane equation!?

Question

An implicit equation for the plane passing through the point (-4,2,-3) that is perpendicular to the line L(t)=<1-2t,-3-2t,4+2t> is ___________

i'd appreciate it if you guys would answer and also explain the process because i really want to get this concept. thank you.

Answer 1

1 step. From equation of the line you can see that it is parallel to vector (-2, -2, 2) which is taken from coeffitients in front of t.

Note: If you need to understand why it is so, take 2 points A(t) and B(t+1) on this line. A has coordinates 1-2t, -3-2t, 4+2t, while B has coordinates 1-2(t+1), -3-2(t+1), 4+2(t+1). Now to find coordinates of vector from A to B just subtract coordinates (X from X, Y from Y and Z from Z) and you get (-2, -2, 2).

2. Equation of plane perpendicular to vector (-2, -2, 2) is -2x - 2y + 2z + C = 0.

Note: To understand why it is so take to points U=(x1, y1, z1) and V=(x2, y2, z2) on this plane, i.e. -2x1 - 2y1 + 2z1 + C = 0 and -2x2 - 2y2 + 2z2 + C = 0. Vector UV is difference of these points, i.e. (x2-x1, y2-y1, z2-z1). If you subtract 2 equations written before you get -2(x2-x1) - 2(y2-y1) + 2(z2-z1) = 0. But expression in left part is expression for dot multiplication of 2 vectors with coordinates (-2, -2, 2) and (x2-x1, y2-y1, z2-z1). The fact that the product is 0 means that vectors are perpendicular. But UV is arbitrary vector in our plane, so (-2, -2, 2) perpendicular to ANY vector in our plane, so it is perpendicular to plane.

3. Substitute your point (-4, 2, -3) into plane equation and find C: -2(-4) - 2(2) + 2(-3) + C = 0.
C = 2
Plane equation is -2x -2y - 2z + 2 = 0. Now you can do nearly everything you want with it:) I prefer to divide it by -2 and get x + y - z - 1 = 0, or x + y - z = 1

Check my calculations.

Answer 2

Find the equation of the plane passing thru the
point P(-4, 2, -3) that is perpendicular to the line
L(t) = <1 - 2t, -3 - 2t, 4 + 2t>

The directional vector v, of the line L is:
v = <-2, -2, 2>

This is also the normal vector of the perpendicular plane thru point P(-4, 2, -3). With a point in the desired plane and the normal vector we can write the equation of the desired plane.

-2(x + 4) - 2(y - 2) + 2(z + 3) = 0

Divide thru by -2 to simplify the coefficients.

(x + 4) + (y - 2) - (z + 3) = 0
x + 4 + y - 2 - z - 3 = 0
x + y - z - 1 = 0

Answer 3

given x = 2y = 3z, parametric equation of the line is x = t y = t/2 z = t/3 and direction vector of the line is (a million, a million/2, a million/3) or equivalently, (6, 3, 2) any airplane that consists of the line could have a classic vector perpendicular to the line, which comprise (2,-2,-3), so 2x - 2y - 3z = d, and to get d, plug interior the element -- 2(a million) - 2(-a million) - 3(a million) = d d = 2 + 2 - 3 = a million so 2x - 2y - 3z = a million considering the fact that what we've is a line and a element no longer on the line, there ought to be in effortless terms a million airplane that consists of the two.