Question

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.

Answer 1

it sounds doable, but its hard to make out what its actually asking for

Answer 2

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.

Answer 3

@amistre64

Answer 4

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 ??

Answer 5

the line equation parts are

x = 4 + 3n
y = 1 + an
z = 3 +2n

Answer 6

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

Answer 7

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

Answer 8

\[(x,y,z)=(1+t-2s),2t+s, 1+3t+as)\]
\[x=1+t-2s\]
\[y=2t+s\]
\[z=1+3t+as\]

where \[t,s \in \]

Answer 9

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

Answer 10

(-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

Answer 11

I think I get the idea, will have to spend some time on i tough, thank you for your help! :)

Answer 12

good luck ;)