Question

How do you find parametric equations for the line through the point (0,1,2) that is perpendicular to the line x =1 + t , y = 1 – t , z = 2t and intersects this line?

Answer 1

See below.

Explanation:

Given the line `L` and `p_1 = (0,1,2)` where

`L->p= p_0+t vec v`

where `p = (x,y,z)`, `p_0=(1,1,0)` and `vec v = (1,-1,2)`

The elements `p_1` and `L` define a plane `Pi` with normal vector `vec n` given by

`vec n = lambda_1 (p_0-p_1) xx vec v` where `lambda_1 in RR`

The sought line `L_1 in Pi` and is orthogonal to `L` so

`L_1->p = p_1+t_1 vec v_1` where `vec v_1 = lambda_2 vec v xx vec n` with `lambda_2 in RR`

Answer 2

Please see the helpful video and the explanation for my solution to the problem.

Explanation:

Here is a video that helped me to know how to do this problem. Helpful Video

Lets write the vector equation of the line:

`(x,y,z) = (1, 1, 0) + t(hati - hatj + 2hatk)`

A plane that is perpendicular to this line will have the general equation:

`x - y + 2z = c`

We make it contain the point by substituting in the point and solving for c:

`0 - 1 + 2(2) = c`

`c = 3`

The plane `x - y + 2z = 3` contains the point `(0,1,2)` and is perpendicular to the line.

To find the point where the line intersects the plane, substitute the parametric equations of the line into the equation of the plane:

`x - y + 2z = 3`

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

`1 + t - 1 + t + 4t = 3`

`6t = 3`

`t = 1/2`

`x = 1 + 1/2 = 3/2`

`y = 1 - 1/2 = 1/2`

`z = 2(1/2) = 1`

The line intersects the plane at the point (3/2, 1/2, 1)

check `x - y + 2z = 3`:

`3/2 - 1/2 + 2 = 3`

`3 = 3`

This checks.

The vector, `barv`, from the given point to the intersection point is:

`barv = (3/2 - 0)hati + (1/2 - 1)hatj + (1 - 2)hatk`

`barv = 3/2hati - 1/2hatj - hatk`

The vector equation of the line is:

`(x, y, z) = (0, 1, 2) + t(3/2hati - 1/2hatj - hatk)`

The parametric equations are:

`x = 3/2t`

`y = 1 - 1/2t`

`z = 2 - t`