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anonymous
 5 years ago
A plane that contains (3,1,2) point and the line r(t)=<2,1,0>. What is the equation of that plane?
anonymous
 5 years ago
A plane that contains (3,1,2) point and the line r(t)=<2,1,0>. What is the equation of that plane?

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mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0Very useful notes: http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx (with great pics) Nice video also: http://patrickjmt.com/findingthescalarequationofaplane/ Hope this helps. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks! I'll check them out right now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops* the line r(t) should be = <2,1,0>+t<2,3,0>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I watched the video and went over the pauls notes section (both of which I've used and are great resources) but I am drawing a blank on the line part r(t).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I believe I need to do something to it to get a normal vector and then I cross that with the point given and then sub the values into the equation a(xx0)+b(yy0)+c(zz0)=0

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1You just need to take two vectors on the plane and cross product that to get the normal.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0<3,1,2> X <2,1,0> ?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Not quite, you can also think of in terms of points. Let A(3,1,2). Take another two points on the line (by plug in two values of t, up to you which t) to get B and C. Now you have Vector BA and BC and you can cross product that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this on the right track? i got <2+2t,1+3t,0> would this be correct to sub a number in for t?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1correct you substitute two values of t to get two points on the line. Then from the three points you have two vectors.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so I got these three points t(0) B= <2,1,0> , t(1) C=<4,2,0> now I take that <3,1,2> X <2,1,0> => Vect BA <2,1,0>X<4,2,0> => BC BA X BC => a point take that point and sub the x,y,z values into the equation a(xx0)+b(yy0)+c(zz0)=0

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Remember to get the vector say BC you do <42,2(1),00>=<2,3,0>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm, For BC i got 0,0,8 when I <2,1,0>X<4,2,0> I believe you used 1 for the sub while i used 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I went through and got <32,16,0> for the abc values and ended with: 32(x3)16(y+1)+0

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Hod on .... I used the points that you obtained earlier. Remember that if A(a,b,c) and B(d,e,f) Then the vector AB is given by <da,eb,fc>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhh, I was doing a+d. Let me retry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0<2t2,3t+1,2t2> does this look better?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Ok, lets use (a,b,c) for point and <a,b,c> for vector to avoid some confusion. Using the points A(3,1,2) and B(2,1,0) , C(4,2,0) that you obtained by plugin t=0,t=1 we have two vectors BA=<32,1(1),20>=<1,0,2> BC=<42,2(1),00>=<2,3,0> Then you cross product BA and BC to get the normal vector.
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