Consider the plane P with equation −2x−2y−z = −11 and line L with equations
x = −3−4t
y = −2−3t
z = −6−4t
Find two distinct points Q1 and Q2 on the line L at distance 3 to the plane P.
Q1 = (0, 0, 0)
Q2 = (0, 0, 0)
Let Q be a general point on the line
Q is (-3-4t, -2-3t, -6-4t)
Plane : -2x -2y -z + 11 = 0
Using the distance from a point to a plane formula
| -2(-3-4t) - 2(-2-3t) - (-6-4t) + 11|/ √(4+4+1) = 3
± (6 + 8t + 4 + 6t + 6 + 4t + 11)/3 = 3
using the positive
18t + 27 = 9
18t = -18
t = -1
then Q is (-3+4, -2+3, -6+4) ,
Q1 is (1, 1, -2)
using the negative result from above:
-(6 + 8t + 4 + 6t + 6 + 4t + 11)/3 = 3
-18t - 27 = 9
-18t = 36
t = -2
Q2 is (-3+8, -2+6, -6+8)
= Q2(5, 4, 2)
check:
midpoint of Q1Q2 = (3, 5/2, 0) , this point should lie on our plane
LS = -2(3) -2(5/2) - (0)
= -6 -5 - 0 = -11
RS = -11
looks good!
It seems like I made a mistake, and Q1 and Q2 are actually the same point, the origin (0, 0, 0). I apologize for the confusion. Let me try again.
To find two distinct points on the line L that are at a distance of 3 from the plane P, we need to consider the normal vector of the plane.
The normal vector of the plane P is (-2, -2, -1).
Let's calculate the distance between a point (x, y, z) on the line L and the plane P using the distance formula:
distance = |(-2x - 2y - z + 11) / sqrt((-2)^2 + (-2)^2 + (-1)^2)|
Substituting the equations of the line L into the distance formula, we get:
distance = |(-2(-3-4t) - 2(-2-3t) - (-6-4t) + 11) / sqrt(9 + 9 + 1)|
Simplifying this expression, we have:
distance = |(6 + 8t + 4 + 6t + 6 + 4t + 11)| / sqrt(19)
Now we need to solve the equation distance = 3. However, for this specific plane and line, it turns out that there are no distinct points Q1 and Q2 on the line L at a distance of 3 from the plane P.
Looks like it's a "no-Q1-and-Q2" situation. Keep in mind that this is a mathematical result and not a joke.
To find two distinct points Q1 and Q2 on the line L that are at a distance of 3 units from the plane P, we can set up the equation of the line L and use the distance formula between a point and a plane.
Given: Plane P with equation −2x − 2y − z = −11 and line L with equations:
x = −3 − 4t
y = −2 − 3t
z = −6 − 4t
1. Substitute the equations of the line L into the equation of the plane P to find the value of "t" at the intersection point(s) between the line and the plane:
-2(−3 − 4t) − 2(−2 − 3t) − (−6 − 4t) = −11
Simplify the equation:
6 + 8t + 4 − 6t + 6 + 4t + 6 + 4t = -11
20t + 22 = -11
Solve for "t":
20t = -33
t = -33/20
2. Substitute the value of "t" back into the equations of the line to find the coordinates of the intersection point(s) Q1 and Q2:
Q1:
x = −3 − 4(-33/20) = -3 + (132/20) = -3 + 33/5 = -12/5
y = −2 − 3(-33/20) = -2 + (99/20) = -2 + 99/20 = -1/20
z = −6 − 4(-33/20) = -6 + (132/20) = -6 + 33/5 = -12/5
Therefore, Q1: (-12/5, -1/20, -12/5)
Q2 can be found by substituting t = -33/20 into the equations of the line:
Q2:
x = −3 − 4(-33/20) = -3 + (132/20) = -3 + 33/5 = -12/5
y = −2 − 3(-33/20) = -2 + (99/20) = -2 + 99/20 = -1/20
z = −6 − 4(-33/20) = -6 + (132/20) = -6 + 33/5 = -12/5
Therefore, Q2: (-12/5, -1/20, -12/5)
So, the two distinct points on the line L at a distance of 3 units from the plane P are:
Q1 = (-12/5, -1/20, -12/5)
Q2 = (-12/5, -1/20, -12/5)
To find two distinct points Q1 and Q2 on the line L at a distance of 3 from the plane P, we need to find two values of t for which the point (x, y, z) on the line L is 3 units away from the plane P.
First, let's find the equation of the plane P in the form ax + by + cz + d = 0. Given that the equation of the plane is −2x − 2y − z = −11, we can rewrite it as −2x − 2y − z + 11 = 0.
The distance between a point (x, y, z) and a plane ax + by + cz + d = 0 is given by the formula:
distance = |ax + by + cz + d| / sqrt(a^2 + b^2 + c^2)
Substituting the values from the equation of plane P, we have:
distance = |-2x - 2y - z + 11| / sqrt((-2)^2 + (-2)^2 + (-1)^2)
= |-2x - 2y - z + 11| / sqrt(4 + 4 + 1)
= |-2x - 2y - z + 11| / sqrt(9)
= |-2x - 2y - z + 11| / 3
Now let's substitute the parametric equations of line L into the formula for distance to the plane P:
|-2(-3 - 4t) - 2(-2 - 3t) - (-6 - 4t) + 11| / 3 = 3
Simplifying the expression:
|6 + 8t + 4 + 6t + 6 + 4t + 11| / 3 = 3
|18t + 27| / 3 = 3
|6t + 9| = 3
Now we can solve this equation to find the values of t that satisfy it:
6t + 9 = 3 or 6t + 9 = -3
Solving the first equation:
6t = -6
t = -1
Solving the second equation:
6t = -12
t = -2
So, we have two distinct points on line L at a distance of 3 from plane P when t = -1 and t = -2.
Using the parametric equations of line L, we can find the corresponding points as follows:
When t = -1:
x = -3 - 4(-1) = -3 + 4 = 1
y = -2 - 3(-1) = -2 + 3 = 1
z = -6 - 4(-1) = -6 + 4 = -2
Therefore, Q1 = (1, 1, -2).
When t = -2:
x = -3 - 4(-2) = -3 + 8 = 5
y = -2 - 3(-2) = -2 + 6 = 4
z = -6 - 4(-2) = -6 + 8 = 2
Therefore, Q2 = (5, 4, 2).
So, the two distinct points Q1 and Q2 on line L at a distance of 3 from plane P are Q1 = (1, 1, -2) and Q2 = (5, 4, 2).