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- anonymous

I have A(1,2,-1), B(1,-2,3), C(2,2,2)
pi: x-2*y+2*z = 0
d: (x-1)/2 = y/(-1) = (z-1)/(-2)
And i need to find:
- area of ABC
-the equation of the plane ABC
-distance from point A to plane pi
-distance from A to the line d
-equation of perpendicular line from A to the plane ABC

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- anonymous

- schrodinger

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- anonymous

calculate for vector AB = <0, -4,4>
calculate for vector BC = <1,4 -1>
calculate vector AB cross BC = <-12,4,4>
the area of ABC = 1/2 sqr ((-12)^2 +4^2+$^2) = 2 sqr(11)

- anonymous

the last sentence means areaABC = 1/2 the product of AB cross BC

- anonymous

to get the equation of plane ABC you need the normal vector n of the plane which is orthogonal with the plane. It is AB cross BC . As previous part we have, vector n = <-12,4,4> . so a = -12, b= 4, c = 4
then apply to formula for equation of the plane . I choose the point A(1,2,-1) to apply. that is a(x-1) + b (y-2) +c(z+1)=0 . At the end up, the equation is: _12(x-1) +4(y-2) +4 (z-1)=0 or you can expand it to get more simple form.

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- anonymous

i did not understood this part:
then apply to formula for equation of the plane . I choose the point A(1,2,-1) to apply. that is a(x-1) + b (y-2) +c(z+1)=0 . At the end up, the equation is: _12(x-1) +4(y-2) +4 (z-1)=0 or you can expand it to get more simple form.

- anonymous

another thing, by : "the last sentence means areaABC = 1/2 the product of AB cross BC"
you mean that the line l for example will have the coordinates (-6,2,2) ?

- anonymous

1st question: that means based on the point you choose to apply the formula, you have a distinguished equation for the plane. I choose the point A to apply the formula. you can choose B or C. we have so many equations for one plane. from my equation, I expand to -12x +12 +4y -8 +4z -4 = 0 . then simplify it to -12x + 4y + 4z = 0
2nd question: that the theorem from the lecture. review it from page 811 example 4 of the book calculus by James Stewart.
Hope this help

- anonymous

yup, thank you :D

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