The plane $Π_1$ has equation
$$r=i+2j+k+ \theta (2j-k) + \phi (3i+2j-2k) $$
Which is $$2x+3y+6z=14$$
The line $l$ has equation
$$r = 3i + 8j + 2k + t(4i + 6j + 5k)$$
The point on $l$ where $t = λ$ is denoted by P. Find the set of values of λ for which the perpendicular distance of P from $Π_1$ is not greater than 4.
This is a diagram I drew to picture this :
where $n=2i+3j+6k$
What I tried:
I thought of finding the critical value at which distance is 4
$$3i+8j+2k + \lambda (4i+6j+5k) + \frac {4}{u} (2i+3j+6k)=(i+2j+k)+ \theta (2j-k) + \phi (3i+2j-2k)$$
Where $u = \sqrt{(2^{2}+3^{2}+6^{2})}=7$
Which is very long method, and still didn't get right.
My book did a simple way, which I don't understand, Please help me understand how the book did it:
Book Method
Perpendicular distance, p, of P from l in terms of 1 parameter
$$p =\frac{1}{7}2(3+4λ)+3(8+6λ)+6(2+5λ)−14$$
$$= |4+8λ|$$
$$p\leq 4 \implies –1 \leq λ \leq 0$$
It seems to me that they have taken the plane equation and replaced 0 with p in:
$$2x+3y+6z-14=0$$
To:
$$2x+3y+6z-14=p$$
And then substituted the line onto this, but how can they do that. The line does not meet/on the plane to substitute into this formula. Please help. I don't understand.
If we let $P=(x_1,y_1,z_1)$, the distance from P to the plane is given by
$d=\left|\text{comp}_{\vec{n}}\vec{v}\right|$ where $\vec{v}$ is a vector from P to any point Q in the plane and $\vec{n}$ is a normal vector for the plane.
This gives $d=\displaystyle\frac{\left|2x_1+3y_1+6z_1-14\right|}{\sqrt{4+9+36}}=\frac{\left|2(3+4\lambda)+3(8+6\lambda)+6(2+5\lambda)-14\right|}{7}$
$\hspace{.78 in}\displaystyle=\frac{\left|28+56\lambda\right|}{7}=\left|4+8\lambda\right|=4\left|1+2\lambda\right|$.