anonymous
  • anonymous
For which value on the real constant a were the line L intersects the point (4,1,3), with the direction vector (3,a,2), the plane \[(x,y,z)=(1+t-2s, 2t+s,1+3t+as), t,s \in \mathbb{R}\] in one point.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
it sounds doable, but its hard to make out what its actually asking for
anonymous
  • anonymous
New formulation: For which values of the real constant a does the line L ( L goes trough the point (4,1,3) and has the direction vector (3,a,2)) intersect the plane \[(x,y,z)=(1+t-2s, 2t+s,1+3t+as), t,s \in \mathbb{R}\] in one point.
anonymous
  • anonymous
@amistre64

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
we might wanna try to redefine the plane in a format thats easier to parse. (x,y,z)=(1+t−2s,2t+s,1+3t+as),t,s x = (1+t−2s,2t+s,1+3t+as) y = t z = s is there a way to construct that as: a(x-xo)+b(y-yo)+c(z-zo)=0 ??
amistre64
  • amistre64
the line equation parts are x = 4 + 3n y = 1 + an z = 3 +2n
anonymous
  • anonymous
I think it's x=1+t-2s, y= 2t+s, z=1+3t+as and the t,s is just to say that thet are Real
amistre64
  • amistre64
hmm, then we might be able to use x and y to solve for n, and use n to solve for a ... is my idea
anonymous
  • anonymous
\[(x,y,z)=(1+t-2s),2t+s, 1+3t+as)\] \[x=1+t-2s\] \[y=2t+s\] \[z=1+3t+as\] where \[t,s \in \]
amistre64
  • amistre64
x = 4 + 3n y = 1 + an z = 3 +2n and x = 1 + t - 2s y = 0 + 2t + s z = 1 + 3t + as (-3 + t - 2s)/3 = n (-1 + 2t + s)/a = n (-2 + 3t + as)/2 = n
amistre64
  • amistre64
(-3 + t - 2s)/3 = n (-1 + 2t + s)/a = (-3 + t - 2s)/3 (-2 + 3t + as)/2 = (-3 + t - 2s)/3 a = (-3 + 6t + 3s)/(-3 + t - 2s) -6 + 9t + 3s(-3 + 6t + 3s)/(-3 + t - 2s) = -6 + 2t - 4s 7t +4s + 3s(-3 + 6t + 3s)/(-3 + t - 2s) = 0 -21t + 7t^2 - 14ts -12s + 4st - 8s^2 + -9s +18st + 9s^2 = 0 -21t + 7t^2 -21s + s^2 +8st = 0 7t^2-21t + s^2 -21s+8st = 0 needs more shuffling, but i believe thats the right idea to take
anonymous
  • anonymous
I think I get the idea, will have to spend some time on i tough, thank you for your help! :)
amistre64
  • amistre64
good luck ;)

Looking for something else?

Not the answer you are looking for? Search for more explanations.