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?

2 Answers
Dec 11, 2016

See below.

Explanation:

Given the line LL and p_1 = (0,1,2)p1=(0,1,2) where

L->p= p_0+t vec vLp=p0+tv

where p = (x,y,z)p=(x,y,z), p_0=(1,1,0)p0=(1,1,0) and vec v = (1,-1,2)v=(1,1,2)

The elements p_1p1 and LL 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

Dec 11, 2016

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