Let L be the line with parametric equations
x = 1−2t
y = 3+3t
z = −1−3t
Find the shortest distance d from the point P0=(1, −4, −2) to L, and the point Q on L that is closest to P0. Use the square root symbol '√' where needed to give an exact
value for your answer.
Find
d=?
Q=(_,_,_)
This is easiest with vectors, as discussed here:
http://math.stackexchange.com/questions/371649/distance-of-a-3d-point-from-the-parametric-form-of-a-line
Using the distance formula and calculus is a bit more involved.
or
http://geomalgorithms.com/a02-_lines.html
To find the shortest distance between a point and a line in three-dimensional space, we can use vector projection.
Step 1: Find a vector parallel to the line L.
The direction vector of the line L can be obtained by taking the coefficients of t in the parametric equations. In this case, the direction vector is given by:
d = <coefficient of t in x, coefficient of t in y, coefficient of t in z>
= <-2, 3, -3>
Step 2: Find a vector between any point on the line L and the given point P0.
We can pick any point on the line L to find a vector between that point and P0. Let's use the point on L when t=0:
P = <1-2(0), 3+3(0), -1-3(0)>
= <1, 3, -1>
Now, we can find the vector between P0 and P:
v = P0 - P
= <1, -4, -2> - <1, 3, -1>
= <0, -7, -1>
Step 3: Project v onto the direction vector d.
The projection of v onto d can be calculated using the dot product of the two vectors:
proj_d(v) = (v · d) / ||d||^2 * d
Where ||d|| represents the magnitude of the vector d.
Let's compute the projection:
||(d)|| = √((-2)^2 + 3^2 + (-3)^2)
= √(4 + 9 + 9)
= √(22)
(v · d) = <0, -7, -1> · <-2, 3, -3>
= (0 * (-2)) + (-7 * 3) + (-1 * (-3))
= -6 - 21 + 3
= -24
proj_d(v) = (-24) / (22) * <-2, 3, -3>
= <-48/22, 72/22, -72/22>
= <-24/11, 36/11, -36/11>
Step 4: Find the point Q on the line that is closest to P0.
To find Q, we take any point on the line and add the projection vector to it. Let's use the point on L when t=0 as before:
Q = P + proj_d(v)
= <1, 3, -1> + <-24/11, 36/11, -36/11>
= <11/11 - 24/11, 33/11 + 36/11, -11/11 - 36/11>
= (-13/11, 69/11, -47/11)
Step 5: Calculate the distance d between P0 and Q.
The distance can be found using the Euclidean distance formula:
d = √[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]
= √[(-13/11 - 1)^2 + (69/11 - (-4))^2 + (-47/11 - (-2))^2]
= √[(-24/11)^2 + (73/11)^2 + (-35/11)^2]
= √[(576/121) + (5329/121) + (1225/121)]
= √[(576 + 5329 + 1225)/121]
= √[7130/121]
=√[58.851]
Therefore,
d = √[58.851]
Q = (-13/11, 69/11, -47/11)
To find the shortest distance d from the point P0 to the line L, we can use the concept of vector projection. The point Q on the line L, which is closest to P0, can be obtained by finding the point on the line that is perpendicular to the vector formed between P0 and any point on the line L.
1. Calculate the direction vector of the line L:
The direction vector of the line L is obtained by taking the coefficients of t in the parametric equations. In this case, the direction vector of L is represented by vector v = <−2, 3, −3>.
2. Find a point on the line L:
We can select any t value (in this case, let's say t = 0) and substitute it into the parametric equations to find a point on L. Let's choose t = 0, so we have:
x = 1 - 2(0) = 1
y = 3 + 3(0) = 3
z = -1 - 3(0) = -1
So, the point P on L is P(1, 3, -1).
3. Calculate the vector between P0 and P:
The vector between P0 and P is given by vector u = <x2 - x1, y2 - y1, z2 - z1> where (x1, y1, z1) = P0 and (x2, y2, z2) = P.
Substituting the values, we have:
u = <1 - 1, -4 - 3, -2 - (-1)> = <0, -7, -1>
4. Project vector u onto vector v:
The projection of vector u onto vector v is given by the dot product of u and the unit vector of v, multiplied by vector v. The unit vector of v is obtained by dividing v by its magnitude.
First, let's calculate the magnitude of v:
|v| = √((-2)^2 + 3^2 + (-3)^2) = √(4 + 9 + 9) = √22
The unit vector of v is then: u_v = v / |v| = <-2/√22, 3/√22, -3/√22>
Calculating the projection of u onto v, we have: proj_v(u) = (u ∙ u_v) * v
proj_v(u) = ((0)(-2/√22) + (-7)(3/√22) + (-1)(-3/√22)) * <-2, 3, -3>
proj_v(u) = (-21/√22) * <-2, 3, -3> = <-42/√22, 63/√22, -63/√22>
5. Find the point Q on the line L closest to P0:
Add the components of the point P and the projected vector proj_v(u) to get the coordinates of Q:
Q = (1 - 42/√22, 3 + 63/√22, -1 - 63/√22)
Finally, to find the shortest distance between P0 and the line L, we can calculate the length of the vector between P0 and Q:
d = √((1 - 1 + 42/√22)^2 + (-4 - 3 - 63/√22)^2 + (-2 + 1 + 63/√22)^2)
Simplifying the expression should give you the exact value for d.
Therefore,
d = √((42/√22)^2 + (-7 - 63/√22)^2 + (1 + 63/√22)^2)
Q = (1 - 42/√22, 3 + 63/√22, -1 - 63/√22)