Question

Determine the slope of plane CDF. All points are referenced from the lowest point. The scale is 1:20. D is 700mm west, 600mm north and 450 mm above F. C is 1000mm west, 100 mm south, and 220mmabove F

Answer 1

Answer,

Well in 3D maths , straight lines in particular have a different concept for slope. We say they have direction ratios (3d slope) .This concept is similar to concept of vector.Direction cosines are an analogous extension of the usual notion of slope to higher dimensions.

So when we are talking about the slope of the plane we are talking about the Direction cosines of the Line Normal to the plane.

So, Let us consider the lowest point being point C which wil be the origin (0,0,0)

So Point D will be 700 mm west , 600 mm north , and 450 mm above i.e. D(,-700,600,450)

and point F will be 1000 mm west , 100 mm south , and 220 mm above i.e. F(-1000,-100,220)

Let consider 2 vectors P1 and P2,

P1=(-700-0)i+(600-0)j+(450-0)k =-700i+600j+450k

P2=(-1000-0)i+(-100-0)j+(220-0)k = -1000i-100j+200k

(These wil represent 2 line on the plane)

The Normal to these 2 lines could be found by cross croduct of theese 2 vectors

P3=P1xP2

P3=(-700i+600j+450k)X(-1000i-100j+200k)

P3=177000i-4654000j-53000k

Now we have to find Direction cosines of the equation

which will be

and for the we have to find the length of the line which can be found by

||P3|| =( $\sqrt{a^{2}+b^{2}+c^{2}}$ )

where a,b and c are the coeficcients of i,j and k respectively.

therefore, ||P3|| =$\left ( \sqrt{(177000 * 20)^{2} + (4654000 * 20)^{2} + (53000 * 20)^{2}} \right )$

Factor of 20 is multiplied as the scale is 1:20 (Given)

P3=93153323.08

Directions cosines will be given by:

Edll F=1

$l= \frac{177000 * 20}{\left \| P3 \right \|}$

I = -0.0381

$m= \pm \frac{b}{\left \| P3 \right \|}$

$m= \pm \frac{-4654000 * 20}{\left \| P3 \right \|}$

$m= \mp 0.99921$

$n= \pm \frac{c}{\left \| P3 \right \|}$

$n= \pm \frac{-530000 * 20}{\left \| P3 \right \|}$

$n= \mp 0.11379$

which are the required slopes.