Question

Grade 12 calculus and vectors question:

“Determine the vector equation of a line through (4,5,5) that is perpendicular to the line (x,y,z) = (11,-8,4) + t(3,-1,1)”

Answer 1

The line

`(x,y,z)=(11,-8,4)+t(3,-1,1)`

runs in the direction of its tangent vector; we can get it by taking its derivative:

`\vec T=(\mathrm d)/(\mathrm dt)\left[(11,-8,4)+t(3,-1,1)\right]=(3,-1,1)`

Any line that runs perpendicular to this line will have a tangent vector that is orthogonal to
`\vec T` above. So construct some vector
`\vec v` that satisfies this.

`\vec v=(x,y,z)`

`\vec v\cdot\vec T=(x,y,z)\cdot(3,-1,1)=3x-y+z=0`

Suppose
`z=0`; then
`3x=y`, and we can pick any two values that satisfy this condition. For instance,

`\vec v=(1,3,0)`

And of course,

(1, 3, 0) • (3, -1, 1) = 3 - 3 + 0 = 0

so
`\vec v` and
`\vec T` are indeed orthogonal.

Now, the line running in the direction of
`\vec v` and passing through the origin can be obtained by scaling

`(x,y,z)=(4,5,5)+t(1,3,0)`

More generally, if you have a direction/tangent vector
`\vec v` and some point
`\vec p`, the line through
`\vec p` is given by

`(x,y,z)=\vec p+t\vec v`