Thread: vector help, straight lines

Relative to a fixed origin, the points A, B and C have position vectors (2i − j + 6k),
(5i − 4j) and (7i − 6j − 4k) respectively.
(a) Show that A, B and C all lie on a single straight line.

completely lost here, any help appreciated.

Hello, Tweety!

There are several ways to do this . . .

Relative to a fixed origin, the points $\displaystyle A, B, C$ have position vectors:
. . (2i − j + 6k), (5i − 4j), (7i − 6j − 4k) respectively.

(a) Show that $\displaystyle A, B, C$ all lie on a single straight line.

We have: .$\displaystyle \begin{Bmatrix}A\!: & (2,\text{-}1,6) \\ B\!: & (5,\text{-}4,0) \\ C\!: & (7.\text{-}6.\text{-}4) \end{Bmatrix}$

Then: . $\displaystyle \begin{array}{cccccc}\overrightarrow{AB} &=& \langle 3,\text{-}3,\text{-}6\rangle &=& 3\langle 1,\text{-}1,\text{-}2\rangle
\\ \overrightarrow{BC} &=& \langle 2,\text{-}2,\text{-}4 \rangle &=& 2\langle1,\text{-}1,\text{-}2\rangle \end{array}$


The vectors are parallel and contain point $\displaystyle B.$
Therefore, the points are collinear.




$\displaystyle \begin{array}{ccccccccc}
|AB| &=& \sqrt{9 + 9 + 36} &=& \sqrt{54} &=& 3\sqrt{6} \\
|BC| &=& \sqrt{4 + 4 + 16} &=& \sqrt{24} &=& 2\sqrt{6} \\
|AC| &=& \sqrt{25 + 25 + 100} &=& \sqrt{150} &=& 5\sqrt{6} \end{array}$


Since $\displaystyle |AB| + |BC| \:=\:|AC|$, points $\displaystyle A,B,C$ are collinear.




We have: .$\displaystyle \begin{array}{ccc}\overrightarrow{AB} &=& \langle 3,\text{-}3,\text{-}6\rangle \\ \overrightarrow{BC} &=& \langle 2,\text{-}2,\text{-}4\rangle \end{array}$

$\displaystyle \text{The angle }\theta \text{ between }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ is given by: }\; \cos\theta \;=\;\frac{\overrightarrow{AB}\cdot \overrightarrow{BC}}{|\overrightarrow{AB}|\,|\over rightarrow{BC}|} $

We have: .$\displaystyle \cos\theta \;=\;\frac{\langle 3,\text{-}3,\text{-}6\rangle\cdot\langle 2,\text{-}2,\text{-}4\rangle}{\sqrt{9+9+36}\,\sqrt{4+4+16}} \;=\;\frac{6+6+24}{\sqrt{54}\,\sqrt{24}} \;=\;\frac{36}{36}\;=\;1$


Since $\displaystyle \cos\theta \:=\:1$, then: .$\displaystyle \theta \:=\:0 \quad\Rightarrow\quad \overrightarrow{AB} \:\parallel\: \overrightarrow{BC}$

Therefore, points $\displaystyle A,B,C$ are collinear.