How do I find a line $ r ( x, y, z)$ if I only have one point $P(1, 5, -18) $?

Question

How do I find a line given only one point? I do have the information that the given line $R$ is perpendicular to a plane with equation $x+y+z=6$.

Answer 1

If the equation of a plane takes the form $$Ax + By + Cz = D$$ Then the vector $(A,B,C)$ is perpendicular to the plane, and in fact any vector perpendicular to this plane is a scalar multiple of $(A,B,C)$. In this case, we have the plane $$x+y+z=6$$ which has normal vector $(1,1,1)$.

We generally write lines in parametric form as $$(x(t),y(t),z(t)) = (B_x, B_y, B_z) + t \cdot (D_x, D_y, D_z)$$ where $(B_x, B_y, B_z)$ is some constant base point and $(D_x, D_y, D_z)$ is the direction vector of the line. To say that a line is perpendicular to a plane is to say that the direction vector of the line is normal to the plane. Our desired line takes the form $$(B_x, B_y, B_z) + t \cdot (1,1,1)$$ We can choose the base point $(B_x, B_y, B_z)$ to be any point on our line (this is equivalent to shifting $t$ by a constant). In our case, we choose $(B_x, B_y, B_z) = (1,5,-18)$, which gives us the line $$(1,5,-18) + t \cdot (1,1,1)$$

Answer 2

Since the line $R$ must be perpendicular to the plane defined by the equation $x+y+z=6$, a parametric equation of the line $\mathbf{r}(t)$ must point in the direction $\mathbf{v} = \langle 1,1,1\rangle$. Note that the vector $\langle 1,1,1\rangle$ is perpendicular to any vector parallel to the plane.

Also, the line $R$ must pass through the point $P(1,5,-18)$. So its position vector is $\mathbf{p} = \vec{OP} = \langle 1,5,-18\rangle$.

Putting these two together, a parametric equation of the line is given by $$ \boxed{\mathbf{r}(t) = \langle 1,5,-18\rangle + t \langle 1,1,1 \rangle, \mbox{ where } -\infty < t< \infty}. $$