Question

Find the distance between the given skew lines.

x=1+7t, y=3+t, z=5−3t
x=4−t, y=6, z=7+2t

Answer 1

u=x1-x2
v=y2-y2
w=z2-z1

dist=sqrt(u^2+v^2+w^2)

Answer 2

Let the line one is \(L_1: \dfrac{x-1}{7}=y-3= \dfrac{z-5}{-3}\)

Then, the direction of \(L_1\) is \(\vec v_1=<7,1,-3>\) and it passes through P (1,3, 5) ok?

Answer 3

Do the same and give me \(\vec v_2\), please

Answer 4

To solve this problem, you need PQ and v1, v2
cross v1, v2 to get n
distance formula give you \(d =\dfrac{|PQ\bullet N|}{||N||}\)