Sketching three dimensional figures

Three dimensional shapes have many attributes such as faces, edges and vertices. The flat surfaces of the 3D shapes are called the faces. The line segment where two faces meet is called an edge. A vertex is a point where 3 edges meet.

In dimensional geometry, we use the right angled triangle. Joining the line to define the angle between the line and the plane that provides the least possible angle. Also, projection of one line to another on the plane is mostly used.

For the pyramid VABCD with VA=VB=VC=VD=5m, and ABCD a square of side 4cm; find the angle between VA and ABCD.

Calculated by dropping a perpendicular from V to ABCD. This meets ABCD at X, the centre of the square.

So, the projection of VA on plane ABCD is AX.

Given:

• AC is the diagonal of the square (or rectangle) ABCD.
• AC is calculated as:

$AC = \sqrt{AB^2 + BC^2}$

$= \sqrt{4^2 + 4^2}$

$= \sqrt{16 + 16}$

$= \sqrt{32} \approx 5.656$

Now:

• AX is half of AC:

$AX = \frac{1}{2} \times \sqrt{32}$

$AX = 2.828$

Next:

• Finding cos(VAX):

$\cos(\text{VAX}) = \frac{AX}{VA}$

$\cos(\text{VAX}) = \frac{2.828}{5}$

$\cos(\text{VAX}) = 0.5656$

Finally:

• Finding the angle VAX:

$\text{VAX} = \cos^{-1}(0.5656) \approx 55.6^\circ$

$AC = \sqrt{4^2 + 4^2} = \sqrt{32} \approx 5.656$

$AX = \frac{1}{2} \times \sqrt{32} = 2.828$

$\cos(\text{VAX}) = \frac{2.828}{5} = 0.5656$

$\text{VAX} \approx 55.6^\circ$

There are infinite possible lines that could be drawn on planes, making different angles with each other. The angle between planes is the angle between lines within those planes, Must be the lines which are at the middle of the plane for non rectangular planes and any other lines for rectangular planes. Then Right angled triangles are used to find the angles between those planes.

Determine the angle between the following planes:

$\pi_1 = 2x - y + z - 1 = 0 \quad \pi_2 = x + z + 3 = 0$

$\vec{r}_1 = (2, -1, 1) \quad \vec{r}_2 = (1, 0, 1)$

$\alpha(\pi_1, \pi_2) = \alpha(\vec{r}_1, \vec{r}_2) = \arccos \frac{\left|2 \cdot 1 + (-1) \cdot 0 + 1 \cdot 1\right|}{\sqrt{2^2 + (-1)^2 + 1^2} \cdot \sqrt{1^2 + 0^2 + 1^2}} = \frac{3}{\sqrt{6} \cdot \sqrt{2}}$

$(\pi_1, \pi_2) = \arccos \frac{\sqrt{3}}{2} = 30^\circ$