Find the distance between the skew lines P(t)=(2,0,2)+t⟨−2,−4,5⟩ and Q(t)=(4,5,−4)+t⟨3,5,4⟩. Hint: Take the cross product of the slope vectors of P and Q to find a vector normal to both of these lines.

Get the slope/tangent vectors for each line:

dP/dt = ⟨-2, -4, 5⟩

dQ/dt = ⟨3, 5, 4⟩

Take their cross product - I'll call it v :

v = ⟨-2, -4, 5⟩×⟨3, 5, 4⟩ = ⟨-41, 23, 2⟩

Normalize this vector by multiplying it by 1/||v|| :

n = v/||v|| = ⟨-41, 23, 2⟩/(3√246)

Then the distance d between P and Q is

d = |n • (⟨-2, -4, 5⟩ - ⟨4, 5, -4⟩)|

d = 1/(3√246) |⟨-41, 23, 2⟩ • ⟨-6, -9, 9⟩|

d = 1/(3√246) |246 - 207 + 18|

d = 57/(3√246)

d = 19/√246

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

< 3

step-by-step explanation:

shell's the art director at her local community center. she paid $10 and for each child that attends she earns $4.

for each child she ears $4 that is the slope m = 4

she paid $10, that is the initial amount paid. so 10 is the y intercept b

m =4   and b = 10

y intercept is 10 so the point is (0,10)

we start graph from y intercept (0,10)

slope = 4 . we know slope = rise / run

slope = 4   or 4/1   are same

rise = 4   and run = 1

for rise we move 4 units up and for run we move 1 unit to the right

the graph is attached below

x axis is the number of child she attends

y axis is the amount   she earns