Consider the planes 2x + 1y + 5z = 1 and 2x + 5z = 0. (A) Find the unique point P on the y-axis which is on both planes. ( ? , ? , ? ) (B) Find a unit vector u with positive first coordinate that is parallel to both planes. ? I + ? J + ? K (C) Use the vectors found in parts (A) and (B) to find a vector equation for the line of intersection of the two planes,r(t) = ? I + ? J + ? K
ok (A) is a linear equations system with 2x + 1y + 5z = 1 2x + 5z = 0 distance to y axis = 0 could translate in sqrt of x²+z²= 0, there might be another way solve it to find the point
B is easy, given 2x + 1y + 5z = 1 and 2x + 5z = 0, normal vectors of those planes are (2,1,5) and (2,0,5), cross product will give the parallel vector, divide by its length so it gives the unit vector
!!!
you can multiply by -1 in order to get the first coordinate positive
u said devide it by its length ! what length ?
cause the product will be (5,0,2)
compute the lenght of that one sqrt(25+0+4)
so the product devided by sqrt29 ?!
yes , btw, forget what I said about A, I misuderstood the problem
ok ! then what about A ?!
ok, 2x + 1y + 5z = 1 2x + 5z = 0 being in tthe y axis means that x is 0 and z is 0
since 2x + 1y + 5z = 1 is the only one with a y , we get 2*0+1y+5*0=1, the point is (0,1,0)
of course note that the point also works for the other equation 2*0+0*1+5*0=0
C is simply the parametric equation of the line that pases the point we just found, also \[\left(\begin{matrix}0 \\ 1 \\0\end{matrix}\right) + t \times \left(\begin{matrix}5 \\0 \\2\end{matrix}\right)\]
can u simplify that please !?
which part?
C ! cause the other ones i understood very well how u explained
ok, consider the point we got from A, and the vector we got from B, the point is in the line of intersection, since it is in both planes, and since the vector is parallel to the planes, it is also parallel to the intersection line, I'll draw It, it wil seem obvious
ok
but how do i get the numerical value for it !?
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