1. ## Question regarding Vectors

Hi All!

I have this nagging problem that I've been at it for a long time. I have attached a scanned sheet with the question. I really hope someone can help me out or atleast push me in the right direction.

Akshay

2. One difficulty is that it is only the point E that is height "l" above B. D, for example, cannot be height "l" above A without "warping" the plane. But we can calculate the angle. By the Pythagorean theorem, the distance from O to B is $\displaystyle \sqrt{a^2+ c^2}$ so the angle between plane 0ABC and plane ODEF (and so both all three $\displaystyle \alpha$, $\displaystyle \beta$, and $\displaystyle \gamma$) are $\displaystyle arctan\left(\frac{l}{\sqrt{a^2+ b^2}}\right)$.

3. ## 3D Trigonometry

Hello bollsakkie

Welcome to Math Help Forum!
Originally Posted by bollsakkie
Hi All!

I have this nagging problem that I've been at it for a long time. I have attached a scanned sheet with the question. I really hope someone can help me out or atleast push me in the right direction.

Akshay
I'm still thinking about the rest of the question, but here's a start.

The attached diagram shows the vertical plane containing the line $\displaystyle OB$. It is in this plane that the $\displaystyle z$-axis - and therefore also the line $\displaystyle OB$ - is rotated through an angle $\displaystyle \gamma$ about $\displaystyle O$. Note that $\displaystyle E$, the new position of the point $\displaystyle B$, is not vertically above $\displaystyle B$, but above a point $\displaystyle D$ lying on $\displaystyle OB$. $\displaystyle ED = l$.

Now $\displaystyle OB = OE = \sqrt{a^2+c^2}$

So $\displaystyle \gamma = \arcsin\left(\frac{l}{\sqrt{a^2+c^2}}\right)$

As far as the unit vector along $\displaystyle Ow$ is concerned, its $\displaystyle z-$component is clearly $\displaystyle \cos\gamma = \sqrt{\frac{a^2 +c^2 - l^2}{a^2+c^2}}$.

For the $\displaystyle x-$ and $\displaystyle y-$components, consider the unit vector along $\displaystyle OB$: it is $\displaystyle \frac{a}{\sqrt{a^2+c^2}}\vec{i} + \frac{c}{\sqrt{a^2+c^2}}\vec{j}$. Then multiply this by $\displaystyle -\sin\gamma\, (= - \frac{l}{\sqrt{a^2+c^2}})$ for the component of the unit vector of $\displaystyle Ow$ along the line $\displaystyle OB$.

So I reckon the unit vector along $\displaystyle Ow$ is $\displaystyle -\frac{la}{a^2+c^2}\vec{i} - \frac{lc}{a^2+c^2}\vec{j}+ \sqrt{\frac{a^2 +c^2 - l^2}{a^2+c^2}}\vec{k}$