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Question Number 104035 by ajfour last updated on 19/Jul/20

$$Findtheimageofpoint\overrightarrow{OP}=\overline{p}in$$$$theline\overline{r}=\overline{a}+\lambda \overline{b}.$$

Answered by mr W last updated on 19/Jul/20

Commented by mr W last updated on 19/Jul/20

$$AN=\lambda b$$$$\frac{(p-a)\cdot b}{\mid b\mid }=\lambda \mid b\mid$$$$\lambda =\frac{(p-a)\cdot b}{\mid b{\mid }^2}$$$$PN=a+\lambda b-p$$$$OQ=q=p+2PN=p+2(a+\lambda b-p)$$$$=2(a+\lambda b)-p$$$$\Rightarrow \mathit{q}=2(\mathit{a}+\frac{(\mathit{p}-\mathit{a})\cdot \mathit{b}}{\mid \mathit{b}{\mid }^2}\mathit{b})-\mathit{p}$$

Commented by ajfour last updated on 19/Jul/20

$$Ibelieve{this}^{\prime }llbetruefor3Deven$$$$Sir?$$

Commented by mr W last updated on 19/Jul/20

$$basicallyyes.$$

Commented by ajfour last updated on 19/Jul/20

$$\frac{p+q}{2}=a+\lambda b\Rightarrow q=2(a+\lambda b)-p$$$$Now(q-p).b=0$$$$[(a+\lambda b)-p].b=0$$$$\lambda =\frac{(a-p).b}{\mid b{\mid }^2}$$$$\overline{q}=(2\overline{a}-\overline{p})+\left(\frac{2(\overline{a}-\overline{p}).\overline{b}}{\mid \overline{b}{\mid }^2}\right)\overline{b}$$

Commented by mr W last updated on 19/Jul/20

$$yes,thanks!$$

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