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?

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?

Evelyn Agard · Accepted Answer

See below.

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#

Cameron Alexander · Answer

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

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#

Scarlett Allison · Answer

undefined To find parametric equations for the line passing through the point $ (0,1,2) $ that is perpendicular to the line given by $ x = 1 + t $, $ y = 1 - t $, $ z = 2t $, follow these steps:

1. **Direction Vector of the Given Line:**
The direction vector of the given line is given by the coefficients of $ t $ in the equations. 
$$ \mathbf{d} = \langle 1, -1, 2 \rangle $$

2. **Point-Direction Form:**
The equation of a line in point-direction form is:
$$ \mathbf{r}(t) = \mathbf{a} + t\mathbf{d} $$
where $ \mathbf{a} $ is a point on the line, and $ \mathbf{d} $ is the direction vector.

3. **Finding $ \mathbf{d'} $ - Direction Vector of the Perpendicular Line:**
For two lines to be perpendicular, their direction vectors must be orthogonal. 
To find a vector orthogonal to $ \mathbf{d} $, you can use the cross product with a vector like $ \langle 1, 1, 1 \rangle $ which is not parallel to $ \mathbf{d} $.
$$ \mathbf{d'} = \mathbf{d} \times \langle 1, 1, 1 \rangle $$

Calculating the cross product:
$$ \mathbf{d'} = \langle 1, -1, 2 \rangle \times \langle 1, 1, 1 \rangle $$
$$ \mathbf{d'} = \langle 1\cdot1 - (-1)\cdot1, (-1)\cdot1 - 2\cdot1, 2\cdot1 - 1\cdot1 \rangle $$
$$ \mathbf{d'} = \langle 2, -3, 1 \rangle $$

4. **Finding the Equation of the Perpendicular Line:**
Using the point-direction form:
$$ \mathbf{r}(t) = \langle 0, 1, 2 \rangle + t\langle 2, -3, 1 \rangle $$
$$ \mathbf{r}(t) = \langle 0 + 2t, 1 - 3t, 2 + t \rangle $$

So, the parametric equations for the line passing through the point $ (0,1,2) $ that is perpendicular to the given line are:
$$ x(t) = 2t $$
$$ y(t) = 1 - 3t $$
$$ z(t) = 2 + t $$