In a space with an orthonormal coordinate system consider the plane;
&: 4x-3y=12
(a)(i)Find the coordinates of the points of intersection of the plane & with the coordinates axes (=axes intersections???)
(ii)Find the parametric equations for the line of intersection of the plane & and the xy-plane.
Need to know the xy-plane, but how do you find that???
(b)(i)Find the coordinates of the point P, which is the reflection of the origin O in &.
(ii)Find cartesian equations of the planes which are parallel to & and at a distance of 4 from &.
I really don't know how to start any of this. Thank you for any help!
To solve this problem, let's break it down step by step.
(a)(i) Finding the coordinates of the points of intersection of the plane & with the coordinate axes:
To find the coordinates of the points of intersection with the coordinate axes, set the other two variables, y and z, equal to zero, one at a time. Then solve for x, y, and z.
First, set y and z equal to zero in the equation of the plane:
4x - 3(0) = 12
4x = 12
x = 3
So one point of intersection is (3, 0, 0).
Next, set x and z equal to zero in the equation of the plane:
4(0) - 3y = 12
-3y = 12
y = -4
So the other point of intersection is (0, -4, 0).
(a)(ii) Finding the parametric equations for the line of intersection of the plane & and the xy-plane:
To find the parametric equations for the line of intersection of the plane & and the xy-plane, we can set z = 0 in the equation of the plane.
4x - 3y = 12 (replace z with 0)
4x - 3y = 12
y = (4/3)x - 4
This equation represents a line in the xy-plane. We can write the parametric equations for this line as:
x = t
y = (4/3)t - 4
z = 0
The parameter t represents any real number, and by substituting different values of t, you can find different points on the line.
To address your question about the xy-plane, the xy-plane is the plane where the value of z is always zero. In other words, it's the plane formed by the coordinate axes x and y, with z value always equal to zero.
(b)(i) Finding the coordinates of the point P, which is the reflection of the origin O in &:
The reflection of a point in a plane is the point obtained by taking the original point and extending it by the same distance on the opposite side of the plane.
To find the reflection of the origin O in the plane &, we can start by finding the equation of the line through the origin perpendicular to the plane.
The equation of a line perpendicular to the plane & can be found by taking the coefficients of x, y, and z from the plane equation and treating them as the direction ratios of the line. So, the direction ratios of the line are (4, -3, 0).
The equation of the line perpendicular to the plane & passing through the origin (O) can be written as:
x = 0 + 4t
y = 0 - 3t
z = 0 + 0t
Now, we need to find the point where this line intersects the plane &. Substituting the parametric equations of the line into the plane equation, we have:
4(4t) - 3(-3t) = 12
16t + 9t = 12
25t = 12
t = 12/25
Substituting this value of t back into the parametric equations for the line, we can find the coordinates of the point P:
x = 4(12/25) = 48/25
y = -3(12/25) = -36/25
z = 0
So, the coordinates of the point P are (48/25, -36/25, 0).
(b)(ii) Finding the cartesian equations of the planes which are parallel to & and at a distance of 4 from &:
To find the cartesian equations of the planes parallel to & and at a distance of 4 from &, we can use the equation of the plane & and modify the constant term.
In the case of a plane parallel to &, the equation of the plane can be written as:
4x - 3y + dz = k
Here, d represents any non-zero real number, k is the constant term, and (x, y, z) are the coordinates of any point on the plane.
To find the value of k, we can use the fact that the distance between the planes is 4 units. The distance between two parallel planes can be computed by taking the absolute value of the difference between their constant terms.
In this case, the original plane has a constant term of 12. So, the new plane should have a constant term of either 12+4=16 or 12-4=8.
Therefore, the cartesian equations of the planes parallel to & and at a distance of 4 from & can be written as:
For k = 16: 4x - 3y + dz = 16
For k = 8: 4x - 3y + dz = 8
These equations represent planes that are parallel to & and are either 4 units above or 4 units below the original plane.