find shortest distance between the line :
(x - 8)/ 3 = (y + 19)/ -16 = (z - 10)/ 7
and
(x - 15)/ 3 = (y - 29) /8 = (z - 5) / -5
A(8,-19,10) is a point on the first line and
B(15,29,5) is a point on the second line
vector AB = (7,48,-5)
the second line has direction v = (3,8,-5)
so the projection of vectorAB on u = AB·v/|v|
= (21 + 384 + 25)/√(9 + 64 + 25) = 430/√98
= 430/(4√2)
|AB| = √(49 + 2304 + 25) = √2378
so we can use Pythagoras
let the distance between the lines be h
h^2 + (430/√98)^2 = (√2378)^2
h^2+ 184900/98 = 2378
h^2 = 48144/98 = 24072/49
h = √24072 /7 = 2√6018 /7 = appr 22.1645
better check my arithmetic on that one, easy to make typing errors.
To find the shortest distance between two lines, we can use vector equations for the lines and find the vector that is orthogonal (perpendicular) to both lines. Then, we can find the distance from a point on one line to the other line using this orthogonal vector.
Let's find the vector equations for the given lines:
Line 1:
(x - 8) / 3 = (y + 19) / -16 = (z - 10) / 7
Let's write it in the vector form by taking a parameter t:
r1(t) = (8, -19, 10) + t(3, -16, 7)
Line 2:
(x - 15) / 3 = (y - 29) / 8 = (z - 5) / -5
Let's write it in the vector form by taking a parameter s:
r2(s) = (15, 29, 5) + s(3, 8, -5)
Now, we need to find the vector that is orthogonal to both lines. We can take the cross product of the direction vectors of the lines to find this vector.
Direction vector of Line 1: d1 = (3, -16, 7)
Direction vector of Line 2: d2 = (3, 8, -5)
Now, let's take the cross product to find the orthogonal vector:
n = d1 x d2
n = (3, -16, 7) x (3, 8, -5)
Using the cross product formula:
n = ((-16)(-5) - (7)(8), (7)(3) - (3)(-5), (3)(8) - (-16)(3))
n = (56, 26, 72)
Now, we need to find the distance between the two lines using this orthogonal vector n. We can take any point on one line and find the perpendicular distance to the other line.
Let's take the point (8, -19, 10) on Line 1 and find the distance to Line 2.
Using the point-to-line distance formula:
Distance = |(x1 - x0, y1 - y0, z1 - z0) dot n| / |n|
Picking the variables:
(x1, y1, z1) = (8, -19, 10) (point on Line 1)
(x0, y0, z0) = (15, 29, 5) (point on Line 2)
n = (56, 26, 72)
Distance = |(8 - 15, -19 - 29, 10 - 5) dot (56, 26, 72)| / |(56, 26, 72)|
Calculating the dot product:
Distance = |(-7, -48, 5) dot (56, 26, 72)| / |(56, 26, 72)|
Calculating the magnitude and simplifying:
Distance = |(-7)(56) + (-48)(26) + (5)(72)| / sqrt((56)^2 + (26)^2 + (72)^2)
Simplifying the numerator:
Distance = |-392 - 1248 + 360| / sqrt(3136 + 676 + 5184)
Calculating the numerator:
Distance = |-1280| / sqrt(8996)
Simplifying the denominator:
Distance = 1280 / sqrt(8996)
Approximating the square root with a calculator:
Distance ≈ 1280 / 94.93
Final answer:
The shortest distance between the two given lines is approximately 13.49 units.