Question

Obtenha, quando for o caso, a equação geral para o plano determinado pelas retas r e s.

a)- r: x=(1,-1,1)+λ(-2,1,-1)
s: x=(3,3,0)+λ(4,-2,2)

b)- r: x=(1,-1,0)+λ(1,0,1)
s: x=(1,-1,0)+λ(1,1,1)

c)- r: x=(-1,0,-1)+λ(2,3,2)
s: x=(0,0,0)+λ(1,2,0)

Answer 1

O vetor normal do plano será o produto vetorial dos vedores diretores das retas "s e r.

a)

dr = (-2, 1, -1)

ds = (4, -2, 2)


nπ = drXdx


$\\ drXds = \left[\begin{array}{ccc}i&j&k\\-2&1&-1\\4&-2&2\end{array}\right] \\ \\ drXds = i*1*2+j*-1*4+-2*-2*k-(4*1*k+-2*j*2+ \\ (-2)*-1*i) \\ \\ drXds = 2i-4j+4k-(4k-4j+4i) \\ \\ drXds = -2i+0j+0k$

Substituindo qualquer ponto da reta e igualando a zero

π = -2x + d      para P(1,-1,0)

0 = -2x +d

0 = -2*1+d

d = 2

∵

π = -2x+2
---------------------------------------------

B)


drXds = (1,0,1)X(1,1,1) = nπ



$\\ drXds = \left[\begin{array}{ccc}i&j&k\\1&0&1\\1&1&1\end{array}\right] \\ \\ drXds = 1*0*1+j*1*1+1*1*k-(1*0*k+1*1*i+1*j*1) \\ \\ drXds = 0i+j+k-(0k+i+j) \\ \\ drXds = -i+0j+k$

π = -x +z+d

Substituindo Ponto (1,01) e igualando a zero

-x+z+d=0

-1+1+d=0

d=0

∵

π = -x+z
--------------------------

C)

drXds = (2,3,2)X(1,2,0)



$\\ drXds = \left[\begin{array}{ccc}i&j&k\\2&3&2\\1&2&0\end{array}\right] \\ \\ drXds = i*3*0+j*2*1+2*2*k-(1*3*k+2*2*i+2*j*0) \\ \\ drXds = 0i+2j+4k-(3k+4i+0j) \\ \\ drXds = -4i+2j+k$

π = -4x+2y+z + d

Substituindo ponto (2,3,2) e igualando a zero

-4x+2y+z+d=0

-4*2+2*3+2+d=0

-8+6+2+d=0
d =0

π = -4x+2y+z