Finding missing point of a vector when it is perpendicular to a line

[AJTheRed]

The question:
The line L₁ has equation \(\displaystyle r = \left( \begin{array}{ccc}4 \\-1 \\0\\\end{array} \right) + t\left( \begin{array}{ccc}1 \\1 \\-1\\\end{array} \right) \), and point A has coordinates (4, 8, -3).

Find the coordinates of point B on L₁, such that \(\displaystyle \overrightarrow{AB}\) is perpendicular to L₁.

My attempt to answer:
I know that \(\displaystyle \overrightarrow{AB} \cdot \left( \begin{array}{ccc}4 \\-1 \\0\\\end{array} \right) + t\left( \begin{array}{ccc}1 \\1 \\-1\\\end{array} \right) = 0\). I also know that this scalar product has to do with finding point B. I also know that \(\displaystyle \overrightarrow{AB}\) = \(\displaystyle B-A\) = \(\displaystyle B - \left( \begin{array}{ccc}4 \\8\\-3\\\end{array} \right)\) and I think that \(\displaystyle \left( \begin{array}{ccc}4 \\-1 \\0\\\end{array} \right) + t\left( \begin{array}{ccc}1 \\1 \\-1\\\end{array} \right) = \left( \begin{array}{ccc}4+t \\-1+t \\0-t\\\end{array} \right)\). If all of this is plugged into the scalar product, we get \(\displaystyle \left(B - \left( \begin{array}{ccc}4 \\8\\-3\\\end{array} \right)\right)\cdot \left( \begin{array}{ccc}4+t \\-1+t \\0-t\\\end{array} \right) = 0\), which can be rewritten as \(\displaystyle V_{1}W_{1} + V_{2}W_{2} + V_{3}W_{3} = 0\) or \(\displaystyle ((B_{1}-4)(4+t))+((B_{2}-8)(-1+t))+((B_{3}-(-3))(-t)) = 0\). Assuming that I've done everything above correctly, what I am struggling with is figuring out how to go about finding the B and t values so that the scalar product is equal to zero. Any hints as to a mistake or mistakes I may have made or what I should do next is very much appreciated and I thank you for taking the time to read through this whole question to lend some support.

 

[Prove It]

The question:
The line L₁ has equation \(\displaystyle r = \left( \begin{array}{ccc}4 \\-1 \\0\\\end{array} \right) + t\left( \begin{array}{ccc}1 \\1 \\-1\\\end{array} \right) \), and point A has coordinates (4, 8, -3).

Find the coordinates of point B on L₁, such that \(\displaystyle \overrightarrow{AB}\) is perpendicular to L₁.

My attempt to answer:
I know that \(\displaystyle \overrightarrow{AB} \cdot \left( \begin{array}{ccc}4 \\-1 \\0\\\end{array} \right) + t\left( \begin{array}{ccc}1 \\1 \\-1\\\end{array} \right) = 0\). I also know that this scalar product has to do with finding point B. I also know that \(\displaystyle \overrightarrow{AB}\) = \(\displaystyle B-A\) = \(\displaystyle B - \left( \begin{array}{ccc}4 \\8\\-3\\\end{array} \right)\) and I think that \(\displaystyle \left( \begin{array}{ccc}4 \\-1 \\0\\\end{array} \right) + t\left( \begin{array}{ccc}1 \\1 \\-1\\\end{array} \right) = \left( \begin{array}{ccc}4+t \\-1+t \\0-t\\\end{array} \right)\). If all of this is plugged into the scalar product, we get \(\displaystyle \left(B - \left( \begin{array}{ccc}4 \\8\\-3\\\end{array} \right)\right)\cdot \left( \begin{array}{ccc}4+t \\-1+t \\0-t\\\end{array} \right) = 0\), which can be rewritten as \(\displaystyle V_{1}W_{1} + V_{2}W_{2} + V_{3}W_{3} = 0\) or \(\displaystyle ((B_{1}-4)(4+t))+((B_{2}-8)(-1+t))+((B_{3}-(-3))(-t)) = 0\). Assuming that I've done everything above correctly, what I am struggling with is figuring out how to go about finding the B and t values so that the scalar product is equal to zero. Any hints as to a mistake or mistakes I may have made or what I should do next is very much appreciated and I thank you for taking the time to read through this whole question to lend some support.

$\displaystyle \begin{align*} B \left( b_1, b_2, b_3 \right) \end{align*}$ is a point on $\displaystyle \begin{align*} \mathbf{r} \left( t \right) \end{align*}$, so we can write $\displaystyle \begin{align*} b_1 = 4 + t , \, b_2 = -1 + t , \, b_3 = -t \end{align*}$. Thus the direction vector $\displaystyle \begin{align*} \vec{AB} = \left( 4 + t - 4 , -1 + t - 8 , -t - \left( - 3 \right) \right) = \left( t , -9 + t , 3 - t \right) \end{align*}$

Since we know $\displaystyle \begin{align*} \vec{AB} \end{align*}$ is perpendicular to $\displaystyle \begin{align*} \mathbf{r}\left( t \right) \end{align*}$, that means

$\displaystyle \begin{align*} \left( 4 + t, -1 + t , -t \right) \cdot \left( t , -9 + t , 3 - t \right) &= 0 \\ t\left( 4 + t \right) + \left( -9 + t \right) \left( -1 + t \right) + \left( 3 - t \right) \left( -t \right) &= 0 \\ 4\,t + t^2 + 9 - 9\,t - t + t^2 - 3 + t + 3\,t - t^2 &= 0 \\ t^2 - t + 6 &= 0 \\ \left( t - 3 \right) \left( t + 2 \right) &= 0 \\ t = 3 \textrm{ or } t &= 2 \end{align*}$

 

[I like Serena]

If $\vec{AB}$ is perpendicular to the line $L_1$, we should have that $\vec{AB}$ is perpendicular to the direction vector $\begin{pmatrix}1\\1\\-1\end{pmatrix}$, shouldn't we? (Wondering)
That means that we should have:
$$\vec{AB} \cdot \begin{pmatrix}1\\1\\-1\end{pmatrix} = 0$$