The line of intersection of the planes 2x+x-3z=3 and x-2y+z=-1 is L.

Determine the parametric equations for L.

The line of intersection of the planes 2x+x-3z=3 and x-2y+z=-1 is L.

Determine the parametric equations for L.

What I did is I took the second equation and multiplied it by 2 to get 2x-4y+2z=-2.

Then I used the process of elimination from the first equation and the adapted second equation to get -3y-z=5. I then set **y=t.

-3y-z=5

**z=-3t-5

x-2t-3t-5=3

x=3+5+2t+3t

**x=8+5t

Did I do this correctly? I'm aware that depending which value you use to define as a parameter (I used y=t), the answer can vary, but I want to know if using y=t, my answer is correct. Please try not to give me another way how to do it, this is the way I was taught and another way will just confuse me :( Thanks!

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You should find two points that lie on the line, hence you need to set x = 0 and z = 0 such that:

`x = 0 =gt y - 3z = 3 and -2y + z = -1`

You need to solve for y and z the simultaneous equations such that:

`y = 3z + 3`

You need to substitute 3z + 3 for y in equation `-2y + z = -1 ` such that:

`-2(3z+3) + z = -1`

`-6z - 6 + z = -1 =gt -5z = 6-1 =gt -5z = 5 =gt z = -1`

`y = -3 + 3 =gt y =0`

Hence, the first point on the line is (0,0,-1).

You need to determine the second point, hence you should set z = 0 such that:

`2x + y = 3 `

`x - 2y = -1 =gt x = 2y - 1`

`2(2y - 1) + y = 3 =gt 4y - 2 + y = 3 =gt 5y = 5 =gt y = 1 `

`x = 2 - 1 =gt x = 1`

Hence, the second point on the line is (1,1,0).

You may write the parametric equations of the line of intersection of the given planes such that:

`x = x_1 + (x_2 - x_1)t `

`x =0 + (1 - 0)t `

`x = t`

`y = y_1 + (y_2 - y_1)t `

`y =0 + (1 - 0) t `

`y = t`

`z= z_1 + (z_2 - z_1)t `

`z = -1 + (0 + 1)t `

`z = -1 + t`

**Hence, evaluating the parametric equations of the line of intersection of the given planes yields x = t ; y = t ; z = -1 + t.**

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