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How could I find the exact coordinates of the point N for example which is the point of intersection of the line $L=\begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix} +t\begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}$ and a perpendicular line that passes through a point C(1,2,3) not on the line.

What I have done so far is find the value of point N in terms of $t$ to find the vector $\overrightarrow{CN}$: $$\longrightarrow N\begin{pmatrix} -3t \\ -2t-1 \\ -3t+2 \end{pmatrix}$$ $$\longrightarrow \overrightarrow{CN}=\begin{pmatrix} -3t \\ -2t-1 \\ -3t+2 \end{pmatrix}-\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$ $$=\begin{pmatrix} -3t-1 \\ -2t-3 \\ -3t-1 \end{pmatrix}$$

After this I made the scalar product eual zero: $$\overrightarrow{CN} \cdot \begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}=0$$

With this I find that, $$t=\frac{-6}{11}$$ But my textbook says that $$t=\frac{5}{11}$$

Can anyone confirm which answer for $t$ is right and if it is $t=\frac{5}{11}$ then what did I do wrong?

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    $\begingroup$ For me, your answer it correct. $\endgroup$
    – Bernard
    Jul 30 '20 at 12:32
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    $\begingroup$ WA agrees. Maybe there's the line $L$ points swapped in the solution? $\endgroup$ Jul 30 '20 at 12:42
  • $\begingroup$ Thank you for the confirmation $\endgroup$
    – mikejacob
    Jul 30 '20 at 12:42
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Your derivation seems correct indeed, by a direct check, for $t=-\frac 6{11}$ we obtain

$$N=\begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix} -\frac 6{11}\begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}=\begin{pmatrix} \frac {18}{11} \\ \frac {1}{11} \\ \frac {40}{11} \end{pmatrix}$$

then

$$\overrightarrow{CN}=\begin{pmatrix} \frac {7}{11} \\ -\frac {21}{11} \\ \frac {7}{11} \end{pmatrix}$$

and

$$\begin{pmatrix} \frac {7}{11} \\ -\frac {21}{11} \\ \frac {7}{11} \end{pmatrix} \cdot \begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}= -\frac {21}{11}+\frac {42}{11}-\frac {21}{11}=0$$

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