Respuesta :
Answer:
The equation is:
(1/a)x + (1/b)y + (1/c)z = 1
Step-by-step explanation:
The direction vector between the points (a, 0, 0) and (0, b, 0) is given as:
<0 - a, b - 0, 0 - 0>
<-a, b, 0> .....................(1)
The direction vector between (0, 0, c) and (0, b, 0) is given as:
<0 - 0, b - 0, 0 - c>
= <0, b, -c> .....................(2)
To obtain the direction vector that is normal to the surface of the plane, we take the cross product of (1) and (2).
Doing this, we have:
<-a, b, 0> × <0, b, -c> = <-bc, -ac, -ab>
To find the scalar equation of the plane we can use any of the points that we know. Using (0, b, 0), we have:
(-bc)x + (-ac)y + (-ab)z = (-bc)0 + (-ac)b + (-ab)0
(bc)x + (ac)y + (ab)z = (ac)b
Dividing both sides by abc, we have:
(1/a)x + (1/b)y + (1/c)z = 1
We will call π, the plane with points P, Q, R ( the intercepts )
The solution is:
π : -bc×x + ac×y -ab×z - bac = 0
c) d = | bac|/√ ( -bc)² + (-ac)² + ( -ab)²
The intercepts P Q R are three points of the plane, according to that
P ( 0, b, 0 ) intercept with y-axis
Q ( a, 0, 0 ) intercept with x-axis
R ( 0, 0, c ) intercept with z-axis
We will find the vectors PQ and PR both are on the plane
PQ = [ a, 0, 0 ] - [ 0, b, 0 ] ⇒ PQ = ( a , -b, 0 )
PR = [ 0, 0, c ] - [ 0, b, 0 ] ⇒ PR = ( 0, -b c )
The vectorial product of these two vector will give us, one normal vector to the plane.
i j k
PQ * PR = a -b 0 = i (-bc) - j ( ac - 0 ) + k ( -ab - 0)
0 -b c
PQ * PR = [ -bc, -ac, -ab ]
we call this vector n = [ -bc, -ac, -ab ]
Let´s now find a vector between P, and a general point T ( x , y , z ) on the plane.
PT = [ x, y, z ] - [ 0, b, 0 ] ⇒ PT = [ x-0 , y - b , z - 0 ]
PT = [ x, y-b, z ]
Vector PT is perpendicular to vector n, then their dot product must be 0
Then:
PT × n = 0
[ x, y-b, z ] × [ -bc, -ac, -ab ] = 0
x × (-bc) + ( y - b )× (ac) + z × (- ab) = 0
-bc×x + ac×y -bac - ab×z = 0
-bc×x + ac×y -ab×z = bac (1)
Finally, we got the implicit equation for plane π as:
π : -bc×x + ac×y -ab×z - bac
c) The distance (d) between the plane and the origin is:
d = | D | / | n |
| n | = √ ( -bc)² + (-ac)² + ( -ab)²
|D| = | bac| the right side of the equation (1)
Then
d = | bac|/√ ( -bc)² + (-ac)² + ( -ab)²
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