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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

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