Question

Grade 12 calculus and vectors question:

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:

$\vec T=\dfrac{\mathrm d}{\mathrm dt}\left[(11,-8,4)+t(3,-1,1)\right]=(3,-1,1)$

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

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