How do you find the (shortest) distance from the point P(1, 1, 5) to the line whose parametric equations are x = 1 + t, y = 3 - t, and z = 2t?

Question

Ezekiel Alexander · Accepted Answer

#= sqrt(5) #

The line can be re-written as follows

#vec l = ((1+t),(3-t),(2t)) = ((1),(3),(0)) + t ((1),(-1),(2))#
#= ((1),(3),(0)) + t' 1/sqrt(6) ((1),(-1),(2))#
#=vec (OQ) + t' hat u#

where

#Q = ((1),(3),(0))# is a point on the line

and

#hat u = 1/sqrt(6) ((1),(-1),(2)) # is the unit vector along its direction.

From the sketch, the shortest distance, #d#, between point P and line, is a straight line that runs orthogonal to the line. Distance d is given by

#d = abs(vec(QP)) sin theta = abs (vec(QP) times hat u)#

(NB # abs (vec(QP) times hat u )=abs( abs(vec(QP)) abs ( hat u) sin theta )# but #abs( hat u) = 1#)

#vec(QP) = vec(OP) - vec (OQ)#

#= ((1),(1),(5)) - ((1),(3),(0)) = ((0),(-2),(5)) #

# vec(QP) times hat u = 1/sqrt(6) det ((hat i, hat j, hat k),(0,-2,5),(1,-1,2))#

#=1/sqrt(6)( hat i + 5 hat j + 2 hat k) = vec n#

The vector cross product is also vector, #vec n#, which is perpendicular to both #vec (QP)# and #hat u#. We can check this first by evaluating the scalar dot products to zero:

#hat u * vec n =1/sqrt(6) (1,-1,2)*(1,5,2) = 0#!!

#vec(QP) * vec n = (0,-2,5)*(1,5,2) = 0#!!

So finally,

#abs (vec(QP) times hat u) = abs(vec n)#

#= abs (1/sqrt(6)( hat i + 5 hat j + 2 hat k)) #

#= 1/sqrt(6) * sqrt( 1^2 + 5^2 + 2^2)#

#= sqrt(5) #

William Abdalla · Answer

undefined To find the shortest distance from point $ P(1, 1, 5) $ to the line with parametric equations $ x = 1 + t $, $ y = 3 - t $, and $ z = 2t $, you can use the formula for the distance between a point and a line in three dimensions:

$$ \text{Distance} = \frac{\lvert \vec{AP} \times \vec{v} \rvert}{\lvert \vec{v} \rvert} $$

Where:
- $ \vec{AP} $ is the vector from any point on the line to point $ P $,
- $ \vec{v} $ is the direction vector of the line.

First, find $ \vec{AP} $ by subtracting the coordinates of a point on the line from the coordinates of point $ P $:

$$ \vec{AP} = \langle 1 - (1 + t), 1 - (3 - t), 5 - 2t \rangle $$

Simplify to get:

$$ \vec{AP} = \langle -t, 2 + t, 5 - 2t \rangle $$

The direction vector of the line is $ \vec{v} = \langle 1, -1, 2 \rangle $.

Now, compute the cross product $ \vec{AP} \times \vec{v} $:

$$ \vec{AP} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -t & 2 + t & 5 - 2t \ 1 & -1 & 2 \end{vmatrix} $$

$$ = \mathbf{i}(2 + t)(2) - \mathbf{j}(-t)(2 - 5 + 2t) + \mathbf{k}(-t)(-1 - 2 - t) $$

$$ = \langle 4 + 2t + 2t, 2t - 5 + 2t, t^2 + 3t \rangle $$

$$ = \langle 4 + 4t, 4t - 5, t^2 + 3t \rangle $$

Now, find the magnitude of this vector:

$$ \lvert \vec{AP} \times \vec{v} \rvert = \sqrt{(4 + 4t)^2 + (4t - 5)^2 + (t^2 + 3t)^2} $$

Finally, divide the magnitude of the cross product by the magnitude of the direction vector $ \vec{v} $:

$$ \text{Distance} = \frac{\lvert \vec{AP} \times \vec{v} \rvert}{\lvert \vec{v} \rvert} $$

$$ = \frac{\sqrt{(4 + 4t)^2 + (4t - 5)^2 + (t^2 + 3t)^2}}{\sqrt{1^2 + (-1)^2 + 2^2}} $$