cant find the orthogonal proyection of the line on a plane. plane: 10x-6y-12z=7, line: (8-15t,9t,5+18t).

Question

i have the following: when replacing x,y,z values of the line on the plane equation: 10(8-15t)-6(9t)-12(5+18t)=7, t=13/420.Then if we replace "t" in the equation of the line, we obtain the intersection point (211/28,39/140,389/70). now how do i find the orthogonal proyection of the line on the plane?

Answer 1

The normal vector to the plane is $n = [10 , -6, -12]$ and the direction vector of the line is $d = [-15,9, 18]$. The direction vector of the projected line (let's call it $v$ ) satisfies

$ v \cdot n = 0 \tag{1}$

$ ( v \times d ) \cdot n = 0 \tag{2} $

From vector algebra, we know that $ (a \times b) \cdot c = c \cdot (a \times b) = b \cdot (c \times a) = a \cdot (b \times c) $

Thus $ (v \times d ) \cdot n = 0 $ implies that

$ v \cdot (d \times n) = 0 \tag{3}$

From $(1)$ and $(3)$ it follows that we can take

$ v = n \times (d \times n) $

using the triple vector product identity, this becomes

$ v = d (n \cdot n) - n (n \cdot d) $

And this evaluates to

$ v = [-15, 9, 18] (100 + 36 + 144) - [10, -6, -12] (-150 - 54 - 216) $

which simplifes to

$ v = [-15, 9, 18] (280) + [10, -6, -12] (420) $

Dividing this vector quantity by $(140)$ does not change its direction, so we can take

$ v = [-15, 9, 18] (2) + [10 , -6, -12 ] (3) = [0,0,0] $

Since we get a direction vector of zero, this means the projection of the given line is a single point, which is the intersection point of the line with the plane which you found.