Vectors and the Geometry of Space

Question

Find the distance of the point (4,6,−4) from the line r(t)=⟨−1+3t,1+3t,5−3t⟩.

Answer:

Answer 1

Find the distance of the point (4,6,−4) from the line r(t)=⟨−1+3t,1+3t,5−3t⟩.

SOLUTION

Lets call P the given point and let us get any point Q on the line .

If we let t = 1 then x = 2, y = 4 and z = 2 That is Q ( 2, 4 , 2 )

Then the vector PQ = < -2, -2, 6 >

Then the length of PQ is || PQ|| = √(4 +4+36 = 2√11

Now let us find the projection of vector PQ on the directing vector of the given line being u=< 3, 3 -3 >

Next we need to calculate the projection of PQ on the vector u.

|| proj_u PQ || = | PQ • u | / || u || = 30 /(3√3)

Then using the Pythagorean Theorem ( || PQ|| ) ² = ( || proj_u PQ || )² + d ² where d the distance in question

( || PQ|| ) ² = ( || proj_u PQ || )² + d ²

( 2√11 )² = ( 30 /(3√3) )² + d²

44 = 100 /3 + d²

d= √(32/3)

Answer 2

If you are looking for the minimum distance between the point and the line, then you can extend the previous answer slightly. We can build a distance function between the given point and points along the line specified by the parameter t as using the usual distance function only in three dimensions as follows:

d(t) = √[(4 - (-1+3t))²+(6 - (1+3t))²+(-4 - (5-3t))²]

= √[(5-3t)²+(5-3t)²+(-9+3t)²]

If we want to minimize this distance we can take the derivative with respect to our parameter t and set equal to zero, then solve for t. Using the chain rule and the power rule, we can take the derivative of the distance function. After multiplying out unnecessary factors we find:

0 = 19 - 9t,

which leads directly to the parameter value

t = 19/9

Plugging this into the vector, we find the point along the line closest to the given point is

<16,22,-4> / 3

The distance between these two points is found to be

d(19/9) = 4√(6) / 3 ≈ 3.27.

I hope this helps!

Answer 3

First, we take the parametric equation to x, y, and z

x=-1+3t

y=1+3t

z=5-3t

then

since x is coefficient for i, y is for j, and z is k

we reorganize the components with i, j, and k, which would look like

(-1+3t)i+(1+3t)j+(5-3t)k

then expand the equation, to separate the coefficient that has t.

= -i+3ti+j+3tj+5k-3tk

= -i+j+5k+(3i+3j-3k)t

Coefficients with t represent the point which the line passes and coefficients without t give direction numbers. We will call direction number as vector a=<3,3,-3>

So we know that (based on the coefficient without t), the line passes through (-1, 1, 5).

Next, we will find Vector B, which gives the vector between the given point (4,6,-4) and (-1, 1, 5).

b=<4-(-1), 6-1, -4-5>

=<5, 5, -9>

We find the distance by using d

d=|a x b|/|a|

|a x b| can be determined using the cross product

The determinant is -12i+12j, which translates to <-12,12,0>

|a| is sqrt(3²+3²+(-3)²)=sqrt(27)

so the distance is

d= <-12,12,0>/sqrt(27) or d=<-12,12,0> /(3*sqrt(3))