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Given points of a triangle: $A(4,1,-2),B(2,0,0),C(-2,3,-5)$. Line $p$ contains point $B$, is orthogonal to $\overline{AC}$, and is coplanar with $ABC$. Intersection of $p$ and $\overline{AC}$ is the point $B_1$.

Find vector $\overrightarrow{B_1B}$.

EDIT: $$proj_{\overrightarrow{AC}}\overrightarrow{AB}=\overrightarrow{AB_1}=\frac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AC}|^2}\cdot \overrightarrow{AC}$$ $$\overrightarrow{AB}=[-2,-1,2],\overrightarrow{AC}=[-6,2,-3],|\overrightarrow{AC}|=7$$ $$\overrightarrow{AB}\cdot \overrightarrow{AC}=4$$ $$\Rightarrow proj_{\overrightarrow{AC}}\overrightarrow{AB}=\overrightarrow{AB_1}=\left[-\frac{24}{49},\frac{8}{49},-\frac{12}{49}\right]$$

From $\overrightarrow{AB_1}$ we can find the point $B_1\Rightarrow B_1=\left(\frac{172}{49},\frac{8}{49},-\frac{-110}{49}\right)$ $$\Rightarrow \overrightarrow{B_1B}=\left[-\frac{74}{49},-\frac{57}{49},\frac{110}{49}\right]$$

Is this correct?

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No, you are wrong. Replace the first equation with $proj_{\overrightarrow{AC}}\overrightarrow{AB}=\overrightarrow{AB_1}=\frac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AC}|^2}\cdot \overrightarrow{AC}$. Because the length of $\overrightarrow{AB_1}$ should be irrelevant to the length of $\overrightarrow{AC}$, there should have two $AC$s in the denominator.

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  • $\begingroup$ You are correct now. $\endgroup$ – lewton Mar 9 '16 at 12:41

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