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 $\sqrt{a^2+ c^2}$ so the angle between plane 0ABC and plane ODEF (and so both all three $\alpha$, $\beta$, and $\gamma$) are $arctan\left(\frac{l}{\sqrt{a^2+ b^2}}\right)$.

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

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

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

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

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

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