Vector geometry question?

Question

Find the distance from the point P(-5,2,-11) to the line which passes through the points Q(3,-8,-7) and R(-7,12,3).

I cant figure out this question can anyone help me out

Answer 1

You have all the three points P Q and R
get the area of triangle PRQ

get the distance PQ

now use, area = 1/2 * PQ * distance of P from PQ

to calculate the desired distance....

Answer 2

Find the distance from the point P(-5,2,-11) to the line which passes through the points Q(3,-8,-7) and R(-7,12,3).

Define the directional vector v of the line L.

v = QR = = <-7-3, 12-(-8), 3-(-7)> = <-10, 20, 10>
Since v is just the directional vector of the line any non-zero multiple of v will do as well. Divide by 10.
v = <-1, 2, 1>

The equation of the line L containing Q and R is:

L(t) = Q + tv = <3, -8, -7> + t<-1, 2, 1>
where t is a scalar ranging over the real numbers

Now define the vector u from the point P off the line to an arbitrary point on the line. Let's choose point Q.

u = PQ = = <3-(-5), -8-2, -7-(-11)> = <8, -10, 4>

The magnitude of vector v is:

|| v || = √[(-1)² + 2² + 1²] = √(1 + 4 + 1) = √6

The distance d from P to the line L is:

d = || u X v || / || v || = || u X v || / √6 = || -18i - 12j + 6k || / √6

d = √[(-18)² + (-12)² + 6²] / √6 = √(324 + 144 + 36) / √6

d = √(504 / 6) = √84 = 2√21 ≈ 9.1651514
___________________

As a check, the nearest point on the line is X(-1, 0, -3).

L(t) = <3, -8, -7> + t<-1, 2, 1>
Plug in t = 4

L(4) = <3, -8, -7> + 4<-1, 2, 1> = X(-1, 0, -3)

Let w = PX = = <1-(-5), 0-2, -3-(-11)> = <4, -2, 8>

If X is the closest point on the line to P then w = PX should be perpendicular to v, the directional vector of the line. And if they are perpendicular the dot product is zero.

v • w = <-1, 2, 1> • <4, -2, 8> = - 4 - 4 + 8 = 0

So the vectors are normal and X is the closest point on the line to P.

|| PX || = || w || = √(4² + (-2)² + 8²) = √(16 + 4 + 64)

|| w || = √84 = 2√21.

This is the same distance we derived above so it checks.

Answer 3

The vector equation of the line through the points Q and R is
r(t) = Q + t(R-Q)
So in this case we get
r(t) = 3i - 8j - 7k + t(-10i + 20j + 10k)

To make the vector from P to r(t) perpendicular to R-Q, we need the dot product to be zero:
(r(t) - P)*(R-Q) = 0
(3i - 8j - 7k + t(-10i + 20j + 10k) - (-5i + 2j - 11k)) * (-10i + 20j + 10k) = 0
((8i - 10j +4k) + t(-10i + 20j + 10k)) * (-10i + 20j + 10k) = 0
-80 - 200 + 40 + t(100 + 400 + 100) = 0
600t = 240
t = 240/600 = 2/5

Substituting this value of t into the function r(t) we get
r(2/5) = (-5i + 2j - 11k) + (2/5)(-10i + 20j + 10k)
= -5i + 2j - 11k - 4i + 8j + 4k
= -9i -2j - 7k

The magnitude of (r(2/5) - P) is the distance from P to the line.
distance = | (-9i -2j - 7k) - (-5i + 2j - 11k) |
= | -4i - 4j + 4k |
= sqrt (16 + 16 + 16)
= 4 sqrt(3)

So the answer is
4 sqrt(3)

Answer 4

who cares, noo body likes me.