Question

https://gyazo.com/c529640d0bde8c60e6098716aa124e19

a)

(x,y,z) = (-3/2,0,0) + s(3/2,-1,0)+t(0,1,3) = Parametic Equation 

Vector Equation: 

x = -3/2+3/2s 

y = -s+t 

z = 3t 

b) 

1 = -3/2 + 3/2s 

5 = -s+t 

6 = 3t 

But I don't know how to do it from here. Can someone help? 

Answer 1

@Angle

Answer 2

honestly, I haven't properly learned this topic before

but I think that the equation of a plane is of the form: ax + by + cz = d
and the parametric form is: x = n + mt, y = # + #t, z = # + #t etc

Answer 3

I know my equations are right, but I don't know how to determine if that point is on the plane.

Answer 4

ok

in that case,
1 = -3/2 + 3/2s [<-- you can solve for s]
6 = 3t [<-- you can solve for t]
5 = -s+t [<-- you can plug in s and t]

Answer 5

Oh right.

t = 2

5=-s+2
s = -3

So now what?

Answer 6

you can plug in s too?

Answer 7

if 5 = 5 then it's true
if 5 =/= # then it's not true

Answer 8

5 = -(-3)+2
5 = 3+2
5=5
So the point is on the plane.

Tysm!! You're so helpful