0
$\begingroup$

Question: In the pyramid $ABCDE$, the base is a square with a side of length $5$ and $\vec{AD} \perp \vec{DE}$ . The vector $\vec{AE}$ creates equal angles with the vectors $\vec{AD}$ and $\vec{AB}$. Let us denote $\vec{AB}=u$, $\vec{AD}=v$ and $\vec{AE}=w$ (See the figure): enter image description here

(a.) Showing all working, find the numerical value of $w⋅ v$ and the numerical value of $w⋅ u$

(b.) Show that $EDC$ is a right-angle triangle.

What I have done to start (a) we have that

$$\begin{cases} ||u||=5 \\ ||v||=5 \\ ||w||=? \\ \cos(\theta)=\frac{\vec{w}\cdot \vec{v}}{||w||||v||} \\ \cos(\theta)=\frac{\vec{w} \cdot \vec{u}}{||w||||u||} \\ \end{cases}$$

So since AE creates equal angles with AD and AB we can equate the angle of dot product:

$$ \cos(\theta) = \cos(\theta) \Longrightarrow \frac{\vec{w}\cdot \vec{v}}{||w||||v||}=\frac{\vec{w} \cdot \vec{u}}{||w||||u||} $$

Then i don't know how to continue :(

$\endgroup$
0
$\begingroup$

I dont know at the moment a proof for b but in a take origin at $A $ from triangle rule we have $AE=AD+DE $ we want $AD.AE=AD (AD+DE)=|AD|^2+0=25$ as $AD$ is perpendicular to $DE $

$\endgroup$
0
$\begingroup$

you know that $\underline v\cdot\vec{DE}=0$ so the way to find $\underline w$ goes like this: $$\underline w=\underline v+\vec{DE}\\\underline v\cdot\underline w=\underline v\cdot\left(\underline v+\vec{DE}\right)\\\underline v\cdot\underline w=\underline v\cdot\underline v+\underline v\cdot\vec{DE}\\\underline v\cdot\underline w=\underline v\cdot\underline v\\\underline v\cdot\underline w=\|\underline v\|\|\underline v\|\cos0\\\underline v\cdot\underline w=\|v\|^2\\\underline v\cdot\underline w=25$$ for $\underline u\cdot\underline w\,$ you are doing the same and get the same result.

for (b.) you have the following:$$\vec{EC}=-\underline w+\underline v+\underline u\\\implies\vec{EC}\cdot\underline u=\left(-\underline w+\underline v+\underline u\right)\cdot\underline u\\=-\underline w\cdot\underline u+\underline v\cdot\underline u+\underline u\cdot\underline u\\=-\|\underline u\|\|\underline w\|+0+\|u\|^2\\=-25+25\\=0$$ and you know that $\vec{EC}\cdot\underline u=\|\vec{EC}\|\|\underline u\|\cos\theta,\|\vec{EC}\|\ne0\ne\|\underline u\|\,$ therefore $\theta=90$

I see i came late but i'll answer nonetheless in case someone will see the question and will want to know the answer

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.