Question

Determine the equation of the line that is perpendicular to the lines r​(t) equalsleft angle6​t,1 plus3 ​t,4tright angle and R​(s)equalsleft angle 2 plus​s,negative 8 plus 3​s,negative 12 plus 4 sright angle and passes through the point of intersection of the lines r and​ R, where tau equals 0 corresponds to the point of intersection.

Answer 1

$\vec r(t)=\langle6t,1+3t,4t\rangle$

$\vec R(s)=\langle2+s,-8+3s,-12+4s\rangle$

Take the derivatives of each to get the tangent vectors:

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\langle6,3,4\rangle$

$\dfrac{\mathrm d\vec R(s)}{\mathrm ds}=\langle1,3,4\rangle$

Take the cross product of the tangent vectors to get a vector that is normal to both lines:

$\langle6,3,4\rangle\times\langle1,3,4\rangle=\langle0,-20,15\rangle$

The two given lines intersect when $\vec r(t)=\vec R(s)$:

$\langle6t,1+3t,4t\rangle=\langle2+s,-8+3s,-12+4s\rangle\implies t=1,s=4$

that is, at the point (6, 4, 4).

The line perpendicular to both of the given lines through the origin is obtained by scaling the normal vector found earlier by $\tau\in\Bbb R$; translate this line by adding the vector $\langle6,4,4\rangle$ to get the line we want,

$\vec\rho(\tau)=\langle6,4,4\rangle+\langle0,-20,15\rangle\tau$

$\boxed{\vec\rho(\tau)=\langle6,4-20\tau,4+15\tau\rangle}$