Question

The distance of the point(3,0,5)from the line parallel to the vector (6i+j-2k) and passing through the point(8,3,1) is..?

Answer 1

We write the general distance between a point on the line and the given point, then complete the square to minimize it, giving a distance of `3 sqrt{2}`.

Explanation:

There are a few different ways to do this one. Let's write the squared distance from our point to a general point on the line, and find which of those points minimizes the distance.

Our line has direction vector `(6,1,-2)` and passes through `(8,3,1)` so has parametric equation

`(x,y,z) = t(6,1,-2) + (8,3,1)`

The distance `d` from `(x,y,z)` to `(3,0,5)` is given by the Pythagorean Theorem:

`d^2 = (x-3)^2 + (y-0)^2 + (z-5)^2`

`d^2 = (6t + 8 -3)^2 + (t+3)^2 + (-2 t + 1 - 5)^2`

`d^2 = 36 t^2 + 60 t + 25 + t^2 + 6t + 9 + 4t^2 + 16 t + 16`

`d^2 = 41t^2 + 82 t + 50`

We minimize `d^2` by completing the square,

`d^2 = 41(t^2 + 2t) + 50`

`d^2 = 41 (t^2 + 2t + 1) -41 + 50`

`d^2 = 41(t+1)^2 + 9`

Clearly this is minimized at t=-1 at which point

`d^2 = 9`

`d = 3`

Answer 2

`"Distance" = 3`

Explanation:

The scalar equation of the family of planes normal the vector `6hati+hatj-2hatk` is:

`6x+y-2z = C`

Find the value of C for the plane that contains the point `(3,0,5)`:

`6(3)+0-2(5) = C`

`C = 8`

The equation of the plane normal to the line that contains the point `(3,0,5)` is:

`6x+y-2z = 8`

The vector equation of the line is:

`(x,y,z) = (8,3,1)+t(6hati+hatj-2hatk)`

The parametric equations for the line are:

`x = 6t+8`
`y = t+3`
`z = 1-2t`

To find the value of t where the line intersects the plane substitute the parametric equations into the scalar equation for the plane:

`6(6t+8)+(t+3)-2(1-2t) = 8`

`t = -1`

Use the vector equation of the line to find the point on the plane at `t = -1`:

`(x,y,z) = (8,3,1)-6hati-hatj+2hatk`

`(x,y,z) = (2,2,3)`

Use the distance formula to find the distance from `(3,0,5)` to `(2,2,3)`:

`"Distance"= sqrt((2-3)^2+(2-0)^2+(3-5)^2)`

`"Distance" = 3`