find the vector and the cartesian equation of the line passing through (1,2,3) and parallel to the planes x-y+2z=5 and 3x+y+z=6
that "parallel to the planes" part is a little confusing. The planes themselves have different normals so they are not even parallel to each other. And also, there are alot of lines that pass thru a point that are all parallel to a given plane. For example, take a plane that contains the given point; how many lines can you form in that plane that go thru the point?
First find the equation of the plane parallel to x-y+2z=5 and that passes through (1,2,3) Then also find equation of the plane that is parallel to 3x+y+z=6 and that passes through (1,2,3) Then the intersection of those two plane is the required line
ahhh, thats what they meant; the line of intersection of the 2 planes .... good eye on that one :)
how..
@sauravshakya
Actually there are infinite lines that is parallel to plane and that passes thorough a point. So, first find equation of the plane that is parallel to given plane and that passes throughh a point
Do similar for another given plane
Then intersection of those two plane will be our required line tha will be parallel to both the given planes
would u plse solve one
@sauravshakya plse help me
@sauravshakya plse help me
find the equation of the plane parallel to x-y+2z=5 and that passes through (1,2,3)
Can u find it?
i can't find it..
srry
Let the equation of the plane parallel to x-y+2z=5 be x-y+2z=d
Now, using the give point find d
d = 5
if you plug in the point values into both plane equations you will discover that the point is common to both planes; which can only happen if that point is on the line of the intersection of both planes. if we cross the normals of the planes, we create a vector that is parallel to the line of intersection. Then just apply that vector to the given point to define the line itself
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