Find an equation for the plane that (a) is perpendicular to f{v} = (1, 1, 1f{v}=(1,1,1) and passes through (1,0,0). (b) is perpendicular to f{v} = (1, 2, 3f{v}=(1,2,3) and passes through (1,1,1). (c) is perpendicular to the line f{l}(t) = (5, 0, 2)t + (3, −1, 1f{l}(t)=(5,0,2)t+(3,−1,1) and passes through (5, −1, 0)(5,−1,0). (d) is perpendicular to the line f{l}(t) = (−1, −2, 3)t + (0, 7, 1f{l}(t)=(−1,−2,3)t+(0,7,1) and passes through (2, 4, −1)(2,4,−1).

Question

Find an equation for the plane that  (a) is perpendicular to f{v} = (1, 1, 1f{v}=(1,1,1) and passes through (1,0,0).  (b) is perpendicular to f{v} = (1, 2, 3f{v}=(1,2,3) and passes through (1,1,1).  (c) is perpendicular to the line f{l}(t) = (5, 0, 2)t + (3, −1, 1f{l}(t)=(5,0,2)t+(3,−1,1) and passes through (5, −1, 0)(5,−1,0).  (d) is perpendicular to the line f{l}(t) = (−1, −2, 3)t + (0, 7, 1f{l}(t)=(−1,−2,3)t+(0,7,1) and passes through (2, 4, −1)(2,4,−1).

sonorous9n · Accepted Answer

a) n→=(1,1,1) where n→ is the normal to the plane that passes through (1,0,0)  Using the equation of the plane we get:  1(x−1)+1(y−0)+1(z−0)=0  x+y+z−1=0  b) n→=(1,2,3) where n→ is the normal to the plane that passes through (1,1,1)  Using the equation of the plane we get  1(x−1)+2(y−1)+3(z−1)=0  x+2y+3z−6=0

Raymond Foley · Accepted Answer

c) n→= direction of line =(5,0,2) where n→ is the normal to the plane that passes through (5,−1,0)  Using the equation of the plane we get:  5(x−5)+0(y+1)+2(z−0)=0  5x+2z−25=0  d) n→= direction of line =(−1,−2,3) where n→ is the normal to the plane that passes through (2,3,-1)  Using the equation of the plane we get:  −1(x−2)−2(y−4)+3(z+1)=0  −x−2y+3z+13=0

Find an equation for the plane that (a) is perpendicular

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