Find Vector \overrightarrow{B_1B} for Triangle ABC

[gruba]

Homework Statement

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}.

Homework Equations

-Vector projection
- Dot product
-Magnitude of a vector

The Attempt at a Solution

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?

 

[geoffrey159]

The reasoning is good, and the answer is correct if a scalar product evaluates to 0

 

[SammyS]

Homework Statement

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}.

Homework Equations

-Vector projection
- Dot product
-Magnitude of a vector

The Attempt at a Solution

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?

Notice that once you have, $\ \overrightarrow{AB_1}\$ and $\ \overrightarrow{AB}\$, you can get $\ \overrightarrow{B_1 B}\$ from $\ \overrightarrow{B_1 B}=\overrightarrow{B_1 A}+\overrightarrow{AB}\$