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}\$