line of intersection of 2 planes.
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HallsofIvy
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\(\displaystyle \vec{r}\), I take it, is the vector \(\displaystyle x\vec{i}+ y\vec{j}+ z\vec{k}\) so that
\(\displaystyle \vec{r}\cdot (2\vec{i}- 3\vec{j}+ 4\vec{k}))= 2x- 3y+ 4z= -8\). That is the form I am more familar with. Of course the normal vector to that plane is \(\displaystyle 2\vec{i}- 3\vec{j}+ 4\vec{k}\) so that the line through (8, 2, 10) perpendicular to the plane is given by x= 2t+ 8, y= -3t+ 2, z= 4t+ 10. That line intersects the plane where 2(2t+ 8)- 3(-3t+ 2)+ 4(4t+ 10)= 4t+ 16+ 9t- 6+ 16t+ 40= 29t+ 50= -8. 29t= -58, t= 2. The perpendicular intersects the plane at (2(2)+ 8, -3(2)+ 2, 4(2)+ 10)= (12, -4, 18). The distance between the plane and the point is the distance between (12, -4, 10) and (8, 2, 10), \(\displaystyle \sqrt{16+ 36}= \sqrt{52}= 2\sqrt{13}\). I presume that is what you got,
The second plane is given by \(\displaystyle \vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}= \lambda(\vec{i}+ 3\vec{j})+ \vec{k})+ \mu(2\vec{i}- \vec{j}+\vec{k})\) so that \(\displaystyle x= \lambda+ 2\mu\), \(\displaystyle y= 3\lambda- mu\), and \(\displaystyle z= \lambda+ \mu\). To get the form I am more accustomed to, note that \(\displaystyle Ax+ By+ Cz= A\lambda+ 2A\mu+ 3B\lambda- B\mu+ C\lambda - C\mu= (A+ 3B+ C)\lambda+ (2A- B- C)\mu\). Since that must be a number, not a function of parameters, we must have A+ 3B+ C= 0 and 2A- B- C= 0. That is only two equations in three unknowns, as it should be, since, in fact, any multiple of (A, B, C) would give the same equations. Adding the two equations, C cancels and we have 3A+ 2B= 0. Taking A= 2 we have 6+ 2B= 0 so B= -3. The 2A- B- C= 4+ 3- C= 0 so C= 7. This second plane is 2x- 3y+ 7z= 0.
The problem now is to find the line of intersection of the planes 2x- 3y+ 4z= -8 and 2x- 3y+ 7z= 0. The coefficients of x and y are the same so an obvious thing to do is to subtract the first equation from the second: 3z= -8 so z= -8/3. Then 2x- 3y+ 4z= 2x- 3y- 32/3= -8. 2x- 3y= -8+ 32/3= -24/3+ 32/3= 8/3. Multiplying by 3, the line of itersection is given by 6x- 9y= 8, z= -8/3.
If you prefer parametric equations, x= (3/2)y+ 8/3 so if we take y= 2t, x= 3t+8/3, Parametric equations for the line of intersection of the two planes are
x= 3t+ 8/3
y= 2t
z= -8/3.
Yes, the normals to the two planes are given by \(\displaystyle N_2= 2\vec{i}- 3\vec{j}+ 4\vec{k}\) and \(\displaystyle N_2= 2\vec{i}- 3\vec{j}+ 7\vec{k}\). The dot product can be written \(\displaystyle N_1\cdot N_2= |N_1||N_2| cos(\theta)\), where \(\displaystyle \theta\) is the angle between the vectors, so we have \(\displaystyle 4+ 9+ 28= 41= (\sqrt{29})(\sqrt{62})cos(\theta)\). \(\displaystyle cos(\theta)= \frac{41}{(\sqrt{29})(\sqrt{62})}\).
\(\displaystyle \vec{r}\cdot (2\vec{i}- 3\vec{j}+ 4\vec{k}))= 2x- 3y+ 4z= -8\). That is the form I am more familar with. Of course the normal vector to that plane is \(\displaystyle 2\vec{i}- 3\vec{j}+ 4\vec{k}\) so that the line through (8, 2, 10) perpendicular to the plane is given by x= 2t+ 8, y= -3t+ 2, z= 4t+ 10. That line intersects the plane where 2(2t+ 8)- 3(-3t+ 2)+ 4(4t+ 10)= 4t+ 16+ 9t- 6+ 16t+ 40= 29t+ 50= -8. 29t= -58, t= 2. The perpendicular intersects the plane at (2(2)+ 8, -3(2)+ 2, 4(2)+ 10)= (12, -4, 18). The distance between the plane and the point is the distance between (12, -4, 10) and (8, 2, 10), \(\displaystyle \sqrt{16+ 36}= \sqrt{52}= 2\sqrt{13}\). I presume that is what you got,
The second plane is given by \(\displaystyle \vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}= \lambda(\vec{i}+ 3\vec{j})+ \vec{k})+ \mu(2\vec{i}- \vec{j}+\vec{k})\) so that \(\displaystyle x= \lambda+ 2\mu\), \(\displaystyle y= 3\lambda- mu\), and \(\displaystyle z= \lambda+ \mu\). To get the form I am more accustomed to, note that \(\displaystyle Ax+ By+ Cz= A\lambda+ 2A\mu+ 3B\lambda- B\mu+ C\lambda - C\mu= (A+ 3B+ C)\lambda+ (2A- B- C)\mu\). Since that must be a number, not a function of parameters, we must have A+ 3B+ C= 0 and 2A- B- C= 0. That is only two equations in three unknowns, as it should be, since, in fact, any multiple of (A, B, C) would give the same equations. Adding the two equations, C cancels and we have 3A+ 2B= 0. Taking A= 2 we have 6+ 2B= 0 so B= -3. The 2A- B- C= 4+ 3- C= 0 so C= 7. This second plane is 2x- 3y+ 7z= 0.
The problem now is to find the line of intersection of the planes 2x- 3y+ 4z= -8 and 2x- 3y+ 7z= 0. The coefficients of x and y are the same so an obvious thing to do is to subtract the first equation from the second: 3z= -8 so z= -8/3. Then 2x- 3y+ 4z= 2x- 3y- 32/3= -8. 2x- 3y= -8+ 32/3= -24/3+ 32/3= 8/3. Multiplying by 3, the line of itersection is given by 6x- 9y= 8, z= -8/3.
If you prefer parametric equations, x= (3/2)y+ 8/3 so if we take y= 2t, x= 3t+8/3, Parametric equations for the line of intersection of the two planes are
x= 3t+ 8/3
y= 2t
z= -8/3.
Yes, the normals to the two planes are given by \(\displaystyle N_2= 2\vec{i}- 3\vec{j}+ 4\vec{k}\) and \(\displaystyle N_2= 2\vec{i}- 3\vec{j}+ 7\vec{k}\). The dot product can be written \(\displaystyle N_1\cdot N_2= |N_1||N_2| cos(\theta)\), where \(\displaystyle \theta\) is the angle between the vectors, so we have \(\displaystyle 4+ 9+ 28= 41= (\sqrt{29})(\sqrt{62})cos(\theta)\). \(\displaystyle cos(\theta)= \frac{41}{(\sqrt{29})(\sqrt{62})}\).
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I found a simpler method on youtube and it works. You find the cross product of the normals, it gives you the direction vector of the line and then you find a point of intersection. You can set x to 0, the that gives you 2 simul equations with 2 variables, so you find the x and y.
so you have a point and a direction vector and thats your ans. i will post the workings later but its easy.
so you have a point and a direction vector and thats your ans. i will post the workings later but its easy.
Cubist
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Note:-
[math] \left(\begin{array}{c} 2\\ -3\\ 4 \end{array}\right)\times\left(\begin{array}{c} 4\\ 1\\ -7 \end{array}\right)\neq\left(\begin{array}{c} -21\\ 7\\ 16 \end{array}\right) [/math]
HallsofIvy
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The cross product of \(\displaystyle \begin{pmatrix}2 \\ -3 \\ 4 \end{pmatrix}\) and \(\displaystyle \begin{pmatrix} 4 \\ 1 \\ -7 \end{pmatrix}\) is
\(\displaystyle \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 2 & -3 & 4 \\ 4 & 1 & -7 \end{array}\right|=\)\(\displaystyle \vec{i}\left|\begin{array}{cc}-3 & 4 \\ 1 & -7 \end{array}\right|-\)\(\displaystyle \vec{j}\left|\begin{array}{cc}2 & 4 \\ 4 & -7 \end{array}\right|+\vec{k}\left|\begin{array}{cc} 2 & -3 \\ 4 & 1 \end{array}\right|\)\(\displaystyle = (21- 4)\vec{i}- (-14- 16)\vec{j}+ (2+ 12)\vec{k}= 17\vec{i}+ 30\vec{j}+ 14\vec{j}\).
(Check my arithmetic! I never did get the hang of arithmetic!)
\(\displaystyle \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 2 & -3 & 4 \\ 4 & 1 & -7 \end{array}\right|=\)\(\displaystyle \vec{i}\left|\begin{array}{cc}-3 & 4 \\ 1 & -7 \end{array}\right|-\)\(\displaystyle \vec{j}\left|\begin{array}{cc}2 & 4 \\ 4 & -7 \end{array}\right|+\vec{k}\left|\begin{array}{cc} 2 & -3 \\ 4 & 1 \end{array}\right|\)\(\displaystyle = (21- 4)\vec{i}- (-14- 16)\vec{j}+ (2+ 12)\vec{k}= 17\vec{i}+ 30\vec{j}+ 14\vec{j}\).
(Check my arithmetic! I never did get the hang of arithmetic!)
HallsofIvy
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Calculators are the work of the devil!
pka
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