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A straight line satisfies the following condition.

  • $p>0$ is the length of the perpendicular on the line from the origin
  • $\alpha$ is the angle made by this perpendicular with the positive $x$-axis.

Vectorically derive that the equation of this straight line is $$x \cos\alpha + y \sin\alpha =p$$

enter image description here

With some ideas I have made a figure. But not sure that the figure is correct. Please help me with this and the further derivation.

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Let $P$ denote the point on the line that is closest to the origin.

Then $\vec{OP}=\begin{pmatrix} p\cos \alpha \\ p \sin \alpha\end{pmatrix}$

Let $R$ be an arbitrary point on the line, let $\vec{OR}=\begin{pmatrix} x \\ y\end{pmatrix}$

$\vec{PR}$ and $\vec{OP}$ are perpendicular, hence their inner product (or dot product) should give us $0$.

$$\vec{PR}.\vec{OP}=0$$ $$(\vec{OR}-\vec{OP}).\vec{OP}=0$$ $$\vec{OR}.\vec{OP}=\vec{OP}.\vec{OP}$$

$$xp\cos (\alpha)+yp\sin(\alpha)=p^2$$ $$x\cos(\alpha)+y\sin(\alpha)=p$$

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  • $\begingroup$ where is 'R' in the figure? Could you show me please $\endgroup$ – pi-π Dec 25 '16 at 3:46
  • $\begingroup$ $R$ is an arbitrary point on the line $AB$. $\endgroup$ – Siong Thye Goh Dec 25 '16 at 4:11
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Firstly, you haven't marked the angle $\alpha$ in your diagram though it is implicit from your question.

Once you mark $\alpha$ express $\cos(\alpha)$ and $\sin(\alpha)$ in terms of the distances in the figure.

Hint: $$\cos(\alpha) = \frac{PO}{OA}$$

You should get your proof.

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  • $\begingroup$ How has the vector method used here? $\endgroup$ – pi-π Dec 4 '16 at 15:31
  • $\begingroup$ What do you mean by vector method? Using $\hat{i}$ etc? @user354073 $\endgroup$ – Rama Dec 4 '16 at 15:48
  • $\begingroup$ Yes. Using the algebraic vector forms like $\vec {i}$ and $\vec {j}$. $\endgroup$ – pi-π Dec 4 '16 at 16:33

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