Finding equation of the line orthogonal to a plane & the intersection point?

[HardlyPuzzled]

I'm not sure I'm doing this correctly. If anyone could validate this it would be much appreciated!

1) Consider the line that is orthogonal to the plane 3x-2y+4z=6 and that passes through the point (-1,2,3).

a) Find the equation of this line.

<-1,2,3> + t<3,-2,4>
<-1+3t, 2-2t, 3+4t>

x= -1+3t
y= 2-2t
z= 3+4t

b) Determine where the line intersects the plane.

3(-1+3t) - 2(2-2t) + 4(3+4t) = 6
-3+9t-4+4t+12+16t = 6
29t = 1
t = 1/29

Plugging t into x,y, and z:

(-26/29, 56/29, 91/29)

 

[Dr.Peterson]

Looks good to me, and it's easy enough to check.

What part were you unsure of?

 

[HardlyPuzzled]

Looks good to me, and it's easy enough to check.

What part were you unsure of?

I guess I was just a little confused on the wording of the question. So basically when you're asked for the equation of a line in R3, it's the same as asking for the parametric equations?

 

[pka]

I guess I was just a little confused on the wording of the question. So basically when you're asked for the equation of a line in R3, it's the same as asking for the parametric equations?

No that is not necessarily true.
Now I do prefer the parametric form of the line. But the symmetric and vector forms are widely used.
Symmetric: \(\displaystyle \frac{x+1}{3}=\frac{y-2}{-2}=\frac{z-3}{4}\)
Vector: \(\displaystyle \ell: \left<-1+3t,~ 2-2t,~ 3+4t\right>\)
Are all useful and correct.