Find the projection of the line segment joining the points $(3,4,5)$ and $(4,6,3)$ on the straight line obtained by joining the points $(-1,2,4)$ and $(1,0,5)$
My try:
Direction ratios of the line joining the points $(3,4,5)$ and $(4,6,3)$ is $(1,2,-2)$ and direction ratios of the line joining the points $(-1,2,4)$ and $(1,0,5)$ is $(2,-2,1)$.
To find the projection I computed the angle between the the two lines which came out as $\cos \theta =\dfrac{4}{9}$ but the answer is not coming.
Where am I wrong?
The planes perpendicular to the line segment with endpoints $P=(-1,2,4)$ and $Q=(1,0,5)$ is of the form $2x-2y+z=c$ The plane containing $A$ is $\pi_1: 2x-2y+z=3$ and the plane containing $B$ is $\pi_2: 2x-2y+z = -1$
The line through the points $P=(-1,2,4)$ and $Q=(1,0,5)$ is
$\ell(t) = (2t-1,2-2t,4+t)$
We find $A' = \ell(t) \cap \pi_1$
\begin{align} (2t-1) + 2(2-2t) - 2(4+t) &= 3 \\ t &= -2 \\ A' &= \ell(-2) \\ A' &= (-5,6,2) \end{align}
We find $B' = \ell(t) \cap \pi_2$
\begin{align} (2t-1) + 2(2-2t) - 2(4+t) &= -1 \\ t &= -1 \\ B' &= \ell(-1) \\ B' &= (-3,4,3) \end{align}
The segment $\overline{A'B'}$ is the projection of the segment $\overline{AB}$ onto the line $\overleftrightarrow{PQ}$