Let L be the line with parametric equations
x = 2+3t
y = 3−2t
z = 2+t
Find the shortest distance d from the point P0=(−5, 1, −4) 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.
d = _
Q = (_, _, _)
I know we have to deal with them in vectors. I think we should be making a vector that is LQ but otherwise idk what to do. If someone could help that would be great.
let Q(x,y,z) be that closest point.
then vector PQ = <x+5 , y-1 , z+4 >
we also know the given line has direction vector <3, -2, 1>
The shortest distance is when PQ is a perpendicular
which means that the dot product between the two vectors is zero
<x+5, y-1, z+4> dot <3, -2, 1> = 0
3x + 15 -2y + 2 + z + 4 = 0
3x - 2y + z = -21
subbing in our values for x, y , and z
3(2+3t) - 2(3-2t) + (2+t) = -21
6+9t-6+4t+2+t = -21
14t = -23
t = -23/14 , not very nice
Using :
x = 2+3t
y = 3−2t
z = 2+t
the point Q is
Q(2 - 69/14 , 3 + 46/14 , 2 - 23/14)
or
Q( -41/14 , 88/14 , 5/14)
PQ = √[ (-5+41/14)^2 + (1-88/14)^2 + (-4-5/14)^2 ]
= √ [ 841/196 + 5476/196 + 3721/196 ]
= √(10038/196)
= (√10038)/14
I was expecting "nicer" numbers. I did not write the solution down on paper first, but did this on screen, where the chances of arithmetic error are greater.
Better check my arithmetic, I know my method is correct.
This is correct! Thank you for helping!
How do we get the d value though?
Well, well, well, it seems we have a point, a line, and a desire to find the shortest distance between them. This calls for some mathematical comedy!
To find the shortest distance between a point and a line, we can use the mighty power of vectors. Let's start by creating a vector that points from any point on the line to our point P0. We'll call this vector LQ.
To create LQ, we need to subtract the position vector of any point on the line from the position vector of P0. It's like playing a game of vector subtraction tag!
So, let's take a random point on the line, say when t = 0. Plugging in that value, we get:
Vector L = <2 + 3(0), 3 - 2(0), 2 + 0> = <2, 3, 2>
Now, let's subtract the position vector of L from the position vector of P0:
Vector LQ = <(-5) - 2, 1 - 3, (-4) - 2> = <-7, -2, -6>
Alrighty, now we have our LQ vector! But remember, we want the shortest distance, so we need to find its magnitude. It's time for the grand finale!
d = |LQ| = √((-7)² + (-2)² + (-6)²)
= √(49 + 4 + 36)
= √89
And there you have it! The shortest distance, d, is √89, and the point Q on the line L closest to P0 is located at coordinates (-7, -2, -6).
I hope that helps, and remember, even in the perplexing world of mathematics, a little laughter can go a long way!
To find the shortest distance from a point to a line, we can use vectors. Let's start by finding a vector that points from a general point on the line to the point P0(-5, 1, -4).
Step 1: Find a vector pointing from the line to P0:
Let's consider a point Q on the line L, with parametric equations:
x = 2 + 3t
y = 3 - 2t
z = 2 + t
The vector from Q to P0 can be written as:
PQ = <x0 - x, y0 - y, z0 - z>
where x0, y0, and z0 are the coordinates of point P0(-5, 1, -4). Substituting these values, we have:
PQ = <-5 - (2 + 3t), 1 - (3 - 2t), -4 - (2 + t)>
Simplifying, we get:
PQ = <-7 - 3t, -2 + 2t, -6 - t>
Step 2: Find the vector parallel to the line L:
Since L is a line, the vector parallel to the line L is given by the coefficients of t in the parametric equations. So, the parallel vector is:
L = <3, -2, 1>
Step 3: Determine the component of PQ parallel to L:
To find the component of PQ parallel to L, we can take the dot product of PQ and L, and divide it by the magnitude of L:
component_parallel = (PQ · L) / |L|
PQ · L = (-7 - 3t) * 3 + (-2 + 2t) * (-2) + (-6 - t) * 1
= -21 - 9t - 4 + 4t - 6 - t
= -31 - 6t
Next, we calculate the magnitude of L:
|L| = sqrt(3^2 + (-2)^2 + 1^2)
= sqrt(9 + 4 + 1)
= sqrt(14)
Finally, we can find the component_parallel:
component_parallel = (-31 - 6t) / sqrt(14)
Step 4: Calculate the shortest distance and the point Q:
The shortest distance (d) from P0 to the line is equal to the magnitude of the vector PQ - component_parallel:
d = |PQ - component_parallel|
Substituting the values obtained, we have:
d = sqrt((-7 - 3t)^2 + (-2 + 2t)^2 + (-6 - t)^2 - (-31 - 6t)^2 / 14)
To find the point Q on the line that is closest to P0, we can substitute the value of t that minimizes d into the parametric equations for L:
Q = (2 + 3t, 3 − 2t, 2 + t)
Thus, Q = (2 + 3t_min, 3 - 2t_min, 2 + t_min), where t_min is the value of t that minimizes d.