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Question Number 114443 by ARVIND990 last updated on 19/Sep/20

Commented by ARVIND990 last updated on 19/Sep/20

$$\mathrm{fine}\mathrm{the}\mathrm{projection}\mathrm{of}\mathrm{point}\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{p}}\mathrm{in}\mathrm{the}$$$$\mathrm{line}\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}$$

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

$$\overrightarrow{AP}=\mathit{p}-\mathit{a}$$$$say\overrightarrow{AN}=\beta \mathit{b}$$$$\beta \mid \mathit{b}\mid =\overrightarrow{AP}\cdot \mathit{b}$$$$\Rightarrow \beta =\frac{(\mathit{p}-\mathit{a})\cdot \mathit{b}}{\mid \mathit{b}\mid }$$$$\overrightarrow{ON}=\overrightarrow{OA}+\overrightarrow{AN}=\mathit{a}+\frac{(\mathit{p}-\mathit{a})\cdot \mathit{b}}{\mid \mathit{b}\mid }\mathit{b}$$$$\overrightarrow{PN}=\mathit{a}+\frac{(\mathit{p}-\mathit{a})\cdot \mathit{b}}{\mid \mathit{b}\mid }\mathit{b}-\mathit{p}$$$$\overrightarrow{OQ}=\overrightarrow{OP}+2\overrightarrow{PN}$$$$\Rightarrow \mathit{q}=2\mathit{a}+\frac{2(\mathit{p}-\mathit{a})\cdot \mathit{b}}{\mid \mathit{b}\mid }\mathit{b}-\mathit{p}$$

Commented by ARVIND990 last updated on 19/Sep/20

$$\mathrm{i}\mathrm{dont}\mathrm{understand}\mathrm{pls}\mathrm{make}\mathrm{a}\mathrm{yours}$$$$\mathrm{digram}\mathrm{and}\mathrm{deep}\mathrm{solution}\mathrm{like}\mathrm{how}$$$$\mathrm{AP}=\mathrm{p}-\mathrm{a}\mathrm{pls}$$

Commented by ARVIND990 last updated on 19/Sep/20

$$\mathrm{and}\mathrm{my}\mathrm{question}\mathrm{is}\mathrm{projection}\mathrm{of}\overrightarrow{\mathrm{OP}}$$$$\mathrm{and}\mathrm{this}\mathrm{is}\mid \overrightarrow{\mathrm{OP}}\mid \cos \theta \mathrm{maybe}$$

Commented by ARVIND990 last updated on 19/Sep/20

$$\mid \overrightarrow{\mathrm{OP}}\mid \cos \theta \mathrm{is}\mathrm{equal}\mathrm{to}\frac{\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{p}}}{\mid \overrightarrow{\mathrm{r}}\mid }=\overrightarrow{\mathrm{p}}.\hat{\mathrm{r}}$$

Commented by ARVIND990 last updated on 19/Sep/20

$$\mathrm{and}\mathrm{pls}\mathrm{show}\overrightarrow{\mathrm{r}}\mathrm{in}\mathrm{daigram}$$

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

$$\overrightarrow{OA}+\overrightarrow{AP}=\overrightarrow{OP}$$$$\mathit{a}+\overrightarrow{AP}=\mathit{p}$$$$\Rightarrow \overrightarrow{AP}=\mathit{p}-\mathit{a}$$

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

$$yourquestionisabouttheprojection$$$$ofpointPinline\mathit{a}+\lambda \mathit{b},thatisthe$$$$pointN,whichisgivenby$$$$\overrightarrow{ON}=\mathit{a}+\frac{(\mathit{p}-\mathit{a})\cdot \mathit{b}}{\mid \mathit{b}\mid }\mathit{b}$$$$QistheimageofpointPintheline$$$$\mathit{a}+\lambda \mathit{b}.$$

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

$$\mathit{r}=\mathit{a}+\lambda \mathit{b}$$$$(-\mathrm{\infty }<\lambda <+\mathrm{\infty })$$$$whichisthelineinyourdiagram$$$$wherethevector\mathit{b}isshown.$$

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

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