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• Notes

Show that $P=(0,2,4), Q=(10,0,0), R=(5,1,2)$ are collinear

$\overrightarrow {P Q}=<10,-2,-4>$
$\overrightarrow {P R}=<5,-1,-2>$

$\overrightarrow{P Q}=2 \overrightarrow{P R}$

$P, Q, R$ are collinear

Given the vectors $u=(-2,1,-1)$ and $v=(0,1,1)$ find:
The dot product of $u-4 v$ and $2 v-u$

$(u-4 v) \cdot(2 v-u)=2 u \cdot v-u \cdot u-8 v \cdot v+4 v \cdot u$

$u \cdot v=v \cdot u$

$u \cdot u=\|u\|^{2}$

$u \cdot u=u_{1}^{2}+u_{2}^{2}+u_{3}^{2}$

$\|u\|^{2}=(\sqrt{u_{1}^{2}+u_{2}^{2}+u_{3}^{2}})^{2}$

$=u_{1}^{2}+u_{2}^{2}+u_{3}^{2}$

$=u \cdot u$

$u \cdot u=\|u\|^{2}$
$v \cdot v=\|v\|^{2}$

$=6 u \cdot v-\|u\|^{2}-8\|v\|^{2}$

Given the vectors $u=(-2,1,-1)$ and $v=(0,1,1)$ find:
The cross product of $u-4 v$ and $2 v-u$

$(u-4 v) \times(2 v-u)=2(u \times v)-(u \times u)-8(v \times v)+4(v \times u)$

$i \times i=0$
$j \times j=0$
$k \times k=0$
$u \times u=0$
$v \times v=0$

$|v \times v|=|v| |v| \sin \theta$

$\sin \theta=\sin 0=0$

$(u \times v)=-(v \times u)$

$=-2(v \times u)+4(v \times u)$
$=2(v \times u)=?$

Prove the Pythagoras' theorem: "For all vectors $u$ and $v$ in $R^{3}$ ,
$\|u+v\|^{2}=\|u\|^{2}+\|v\|^{2}$ if and only if $u$ and $v$ are orthogonal"

LHS

$(u+v) \cdot(u+v)=u \cdot u+u \cdot v+v \cdot u+v \cdot v$

$=\|u\|^{2}+2 u \cdot v+\|v\|^{2}$

$u \cdot v=0$

$=\|u\|^{2}+\|v\|^{2}=$RHS