vectors question: finding the equation of the plane and the distance to it from the origin..(adsbygoogle = window.adsbygoogle || []).push({});

The vertices of a quadrelateral OAB and C are (0,0,0) (1,4,1) (3,3,0) and (-1,1,4) respectively.

forces of magnitudes 3,2 and 4 newtons act along OA OB and OC respectively.

(a) Express each force as a vector and find the resultant force.

(b) find the equation of the plane containing A,B and C and show the distance from the origin to the plane is [itex]\frac{15}{\sqrt{29}}[/itex]

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My attempt:

(a)

find the unit vectors in the direction of each of the forces:

[itex]F_1 (OA) = (\frac{1}{\sqrt{18}})(1,4,1)[/itex]

[itex]F_2 (OB) = (\frac{1}{\sqrt{18}})(3,3,0)[/itex]

[itex]F_3 (OC) = (\frac{1}{\sqrt{18}})(-1,1,4)[/itex]

multiply the unit vectors by the magnitude of the respective force:

[itex]F_1 = (\frac{3}{\sqrt{18}}) (1,4,1)[/itex]

[itex]F_2 = (\frac{2}{\sqrt{18}}) (3,3,0)[/itex]

[itex]F_3 = (\frac{4}{\sqrt{18}})(-1,1,4)[/itex]

Add the forces together to find the resultant:

[itex]F_r = F_1 + F_2 + F_3[/itex]

[itex](\frac{1}{\sqrt{18}})(i(3+6-4)+j(12+6+4)+k(3+16))[/itex]

=[itex]\frac{1}{\sqrt{18}} (5i + 22j + 19k) N[/itex]

(b)

cross product 2 of the points to get the normal vector:

[itex](1,4,1)\times(3,3,0) = n[/itex]

=[itex]\left[\begin{array}{ccc}i&j&k\\1&4&1\\3&3&0\end{array} \right][/itex]

[itex]= i \left[\begin{array}{ccc}4&1\\3&0\end{array} \right] - j \left[\begin{array}{ccc}1&1\\3&0\end{array} \right] + k \left[\begin{array}{ccc}1&4\\3&3\end{array} \right][/itex]

[itex]n = -3i + 3j + 9k[/itex]

equation of a plane:

[itex](r-a).n[/itex]

[itex]-3x +3y +9z = (1\times -3) + (4\times 3) + (1\times 9) = 18[/itex]

i have gone wrong somewhere because this lot should be equal to 15 i think, anyone know where i screwed up?

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# Homework Help: Problem involving the equation of a plane

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