Answer to Question #104986 in Analytic Geometry for Jacky

Question

a) Find a non-zero unit vector u with terminal point Q(-7,1,4) such that u has the opposite direction as v=(3,2,1).

b) Find a non-zero unit vector u with terminal point Q(-7,1,4) such that u has the same direction as v=(1,2,3).

Answer 1

a) $\overrightarrow{PQ}=|\vec{u}|=1$

$Q(-7,1,4); P(x,y,z)$

$\vec{v}=(3,2,1)$

$\vec{u}\uparrow\downarrow\vec{v} \rArr \vec{u}=\lambda\vec{v}, \lambda<0$

$\begin{cases} |\vec{u}|=\sqrt{(-7-x)^2+(1-y)^2+(4-z)^2}=1 \\ (-7-x,\ 1-y, \ 4-z)=\lambda(3,\ 2,\ 1) \end{cases} \hArr$

$\begin{cases} (-7-x)^2+(1-y)^2+(4-z)^2=1 \\ -7-x=3\lambda\\ 1-y=2\lambda\\ 4-z=\lambda \end{cases} \hArr$

$\begin{cases} (-7-x)^2+(1-y)^2+(4-z)^2=1 \\ (-7-x)^2=9\lambda^2\\ (1-y)^2=4\lambda^2\\ (4-z)^2=\lambda^2 \end{cases} \hArr$

$\begin{cases} 9\lambda^2+4\lambda^2+\lambda^2=1 \\ (-7-x)^2=9\lambda^2\\ (1-y)^2=4\lambda^2\\ (4-z)^2=\lambda^2 \end{cases} \hArr \begin{cases} \lambda^2=\frac{1}{14} \\ (-7-x)^2=\frac{9}{14}\\ (1-y)^2=\frac{4}{14}\\ (4-z)^2=\frac{1}{14} \end{cases} \hArr$

$\lambda<0 \rArr\lambda=-\frac{1}{\sqrt{14}}$

$\begin{cases} \lambda=-\frac{1}{\sqrt{14}} \\ -7-x=-\frac{3}{\sqrt{14}}\\ 1-y=-\frac{2}{\sqrt{14}}\\ 4-z=-\frac{1}{\sqrt{14}} \end{cases} \hArr \begin{cases} \lambda=-\frac{1}{\sqrt{14}} \\ x=\frac{3}{\sqrt{14}}-7\\ y=\frac{2}{\sqrt{14}}+1\\ z=\frac{1}{\sqrt{14}}+4 \end{cases} \hArr \vec{u}=(-\frac{3}{\sqrt{14}},-\frac{2}{\sqrt{14}},-\frac{1}{\sqrt{14}})$

b) $\overrightarrow{PQ}=|\vec{u}|=1$

$Q(-7,1,4); P(x,y,z)$

$\vec{v}=(1,2,3)$

$\vec{u}\upuparrows\vec{v} \rArr \vec{u}=\lambda\vec{v}, \lambda>0$

$\begin{cases} |\vec{u}|=\sqrt{(-7-x)^2+(1-y)^2+(4-z)^2}=1 \\ (-7-x,\ 1-y, \ 4-z)=\lambda(1,\ 2,\ 3) \end{cases} \hArr$

$\begin{cases} (-7-x)^2+(1-y)^2+(4-z)^2=1 \\ -7-x=\lambda\\ 1-y=2\lambda\\ 4-z=3\lambda \end{cases} \hArr$

$\begin{cases} (-7-x)^2+(1-y)^2+(4-z)^2=1 \\ (-7-x)^2=\lambda^2\\ (1-y)^2=4\lambda^2\\ (4-z)^2=9\lambda^2 \end{cases} \hArr$

$\begin{cases} \lambda^2+4\lambda^2+9\lambda^2=1 \\ (-7-x)^2=\lambda^2\\ (1-y)^2=4\lambda^2\\ (4-z)^2=9\lambda^2 \end{cases} \hArr \begin{cases} \lambda^2=\frac{1}{14} \\ (-7-x)^2=\frac{1}{14}\\ (1-y)^2=\frac{4}{14}\\ (4-z)^2=\frac{9}{14} \end{cases} \hArr$

$\lambda>0 \rArr\lambda=\frac{1}{\sqrt{14}}$

$\begin{cases} \lambda=\frac{1}{\sqrt{14}} \\ -7-x=\frac{1}{\sqrt{14}}\\ 1-y=\frac{2}{\sqrt{14}}\\ 4-z=\frac{3}{\sqrt{14}} \end{cases} \hArr \begin{cases} \lambda=\frac{1}{\sqrt{14}} \\ x=-\frac{1}{\sqrt{14}}-7\\ y=-\frac{2}{\sqrt{14}}+1\\ z=-\frac{3}{\sqrt{14}}+4 \end{cases} \hArr \vec{u}=(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}})$