# Vector question?

May 8, 2018

See process below

#### Explanation:

i) ABC are in straight line, then are linearly dependents $\vec{A B} = k \vec{B C}$

Calculate $\vec{A B} = \left(2 - 0 , 5 - 2 , - 2 + 3\right) = \left(2 , 3 , 1\right)$ and
$\vec{B C} = \left(1 , p - 5 , q + 2\right)$. Then

$\left(2 , 3 , 1\right) = k \left(1 , p - 5 , q + 2\right) = \left(k , k p - 5 k , k q + 2 k\right)$ from here

$2 = k$
$3 = k p - 5 k = 2 p - 10 =$ then $p = \frac{13}{2}$
$1 = q + 2$ then $q = - 1$

ii) If BAC is 90º, then vec(AB)·vec(AC)=0

$\vec{A B} = \left(2 , 3 , 1\right)$
$\vec{A C} = \left(3 , p - 2 , q + 3\right)$

vec(AB)·vec(AC)=6+3p-6+q+3=3p+q+3=0 then

$q = - 3 - 3 p$

iii)If $p = 3$ the mod$\vec{A B} = \sqrt{4 + 9 + 1} = \sqrt{14}$

$\vec{A C} = \left(3 , 1 , q + 3\right)$ then mod $\vec{A C} = \sqrt{9 + 1 + {\left(q + 3\right)}^{2}}$

Then squaring both sides $14 = 10 + {\left(q + 3\right)}^{2} = 10 + {q}^{2} + 6 q + 4$ or

${q}^{2} + 6 q = q \left(q + 6\right) = 0$ which give $q = 0$ and $q = - 6$

May 8, 2018

#### Explanation:

$Q u e s t i o n \left(i\right)$

If $A B C$ is a straight line, this means that $A , B \text{ and } C$ are colinear then

$\vec{A B} = k \vec{B C}$ where $k \in \mathbb{R}$

$\left(\begin{matrix}2 - 0 \\ 5 - 2 \\ - 2 - \left(- 3\right)\end{matrix}\right) = k \cdot \left(\begin{matrix}3 - 2 \\ p - 5 \\ q + 2\end{matrix}\right)$

$\left(\begin{matrix}2 \\ 3 \\ 1\end{matrix}\right) = k \cdot \left(\begin{matrix}1 \\ p - 5 \\ q + 2\end{matrix}\right)$

$\left\{\begin{matrix}k = 2 \\ 3 = 2 p - 10 \\ 1 = 2 q + 4\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}k = 2 \\ p = \frac{13}{2} \\ q = - \frac{3}{2}\end{matrix}\right.$

$Q u e s t i o n \left(i i\right)$

If the angle $\hat{B A C} = {90}^{\circ}$, then

The dot product is

$\vec{A B} . \vec{A C} = 0$

$\left(\begin{matrix}2 - 0 \\ 5 - 2 \\ - 2 - \left(- 3\right)\end{matrix}\right) . \left(\begin{matrix}3 - 0 \\ p - 2 \\ q - \left(- 3\right)\end{matrix}\right) = 0$

$\left(\begin{matrix}2 \\ 3 \\ 1\end{matrix}\right) . \left(\begin{matrix}3 \\ p - 2 \\ q + 3\end{matrix}\right) = 0$

$2 \cdot 3 + 3 \left(p - 2\right) + \left(q + 3\right) = 0$

$6 + 3 p - 6 + q + 3 = 0$

$3 p + q = - 3$

*$Q u e s t i o n \left(i i i\right)$*

$\vec{A B} = \left(\begin{matrix}2 \\ 3 \\ 1\end{matrix}\right)$

$\vec{A C} = \left(\begin{matrix}3 \\ p - 2 \\ q + 3\end{matrix}\right)$

If $A B = A C$, then

$\sqrt{{2}^{2} + {3}^{2} + {1}^{2}} = \sqrt{{3}^{2} + {\left(p - 2\right)}^{2} + {\left(q + 3\right)}^{2}}$

$14 = 9 + {\left(p - 2\right)}^{2} + {\left(q + 3\right)}^{2}$

${\left(p - 2\right)}^{2} + {\left(q + 3\right)}^{2} = 5$

If $p = 3$, then

$1 + {\left(q + 3\right)}^{2} = 5$

${q}^{2} + 6 q + 10 - 5 = 0$

${q}^{2} + 6 q + 5 = 0$

Solving this quadratic equatio in $q$

$q = \frac{- 6 \pm \sqrt{36 - 20}}{2} = \frac{- 6 \pm \sqrt{16}}{2}$

$= - 3 \pm 2$

The possible values of $q$ are

$S = \left\{- 1 , - 5\right\}$