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The following problem appears at MIT OCW Course 18.02 multivariable calculus.

The top extremity of a ladder of length $L$ rests against a vertical wall, while the bottom is being pulled away.

Find parametric equations for the midpoint $P$ of the ladder, using as a parameter the angle $\theta$ between the ladder and the ground (i.e., the $x$-axis).

And here is a sketch of the diagram for the problem.

enter image description here

We can find the parametric equations for the midpoint $P$ by finding the vector $OP.$ I write $OP$ as a sum of two vectors $$OP = OB + BP.$$

We know how to calculate the two vectors $OB$ and $BP.$ Using the Pythagorean Theorem, we can find the first component of $OB$ and the second one is obviously zero, hence $$OB = \langle L \cos \theta, 0 \rangle.$$

And by assuming that $BP$ is the radius of a circle with center at $B,$ we can find that $$BP = \left \langle \frac{L}{2} \cos \theta, \frac{L}{2} \sin \theta \right \rangle.$$

So, we find that $$OP = \left \langle \frac{3L}{2} \cos \theta, \frac{L}{2} \sin \theta \right \rangle.$$

But the professor's solution is $$OP = \biggl \langle -\frac{L}{2} \cos \theta,\frac{L}{2} \sin \theta \biggr \rangle.$$

What's wrong with my solution?

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When you write a vector $\vec v = (v_1,v_2)$ you're implicitly writing $$ \vec v = v_1 \vec i + v_2 \vec j$$ for some linearly independent vectors $\vec i ,\vec j.$ You wrote

$$\vec {OB} = (L \cos \theta,0)$$

Depending on which vectors $i,j$ you are choosing this can be either true or false.

It seems to me quite natural to consider the line representing the wall as the $y$ axis and the $x$ axis as the line drawn by the floor along with the usual coordinate system.

You've correctly identified the distance between $O$ and $B$ as $L \cos \theta$. In the usual coordinates we then have $$ \vec {OB} = (-L \cos \theta,0) \quad \text { and not } \quad (L \cos \theta ,0)$$ This gives you the desired result of $\vec {OB} + \vec {BP} = (-L/2 \cos \theta , L/2 \sin \theta).$

Regardless of all this, it seems to me that it is much easier to find the coordinates of $B$ and $Q$ (where the ladder touches the wall) and to take the average component wise than to determine $\vec{BP}.$

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  • $\begingroup$ Thank you, that's really a silly mistake, I know the easier solution but I wanted to solve it using this way but I made this mistake. $\endgroup$ – Kais Hasan Jun 3 at 17:56
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By the Pythagorean Theorem, the length of the base of the triangle in the diagram is $\langle L \cos \theta, 0 \rangle;$ however, we must take into account that the vector $OB$ points in the negative $x$-direction, hence we have that $OB = \langle -L \cos \theta, 0 \rangle.$ Everything else you have written is correct, so we have that $$OP = OB + BP = \langle -L \cos \theta, 0 \rangle + \biggl \langle \frac L 2 \cos \theta, \frac L 2 \sin \theta \biggr \rangle = \biggl \langle -\frac L 2 \cos \theta, \frac L 2 \sin \theta \biggr \rangle.$$

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