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

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all