Koi toh batado
This is the locus of the point P.
Now, for parabola, the plane must be either X = cont, or Y = const type of graphs (basically, i meant vertical planes)
For circle, the plane must be of Z = const. type (meaning horizontal planes)
For ellipse, the plane must be similar to the circle, but a bit tilted
And for hyperbola, the plane must intersect both the cones, and should not be planes for the parabola
with all that, i think we can solve the problem
(please do check and correct, might have said something wrong)
(sorry, im lazy, dont want to try with options… but if someone wants me to, i will try)
How exactly did you plot this graph since we need to find graph from eqn of a specific plane and here its not necessary that P and Q have same z coordinate either
They have given the coordinates of Q, and now try to find a point which makes a fixed angle with the Z-axis. It will be of this type itself, and extends both up and down…
its a really cool qn… i remember in school, before they actually start conic sections, in NCERT, they describe how to form the conics, which is with the help of a cone
For 50, is the ans BC?
A is wrong (just try subbing a = i or try to visualise the taking of the cross product, and you will find that v1, v2, v3, v4 etc are just like a windmill, and, v1 = v5, v2 = v6, and so on…
and also v1 = -v3, v2 = -v4 and so on
Now, with the sigmas… you will find that after every three iterations, a vector of form 2i + j is formed… and in the next three iterations, the vector rotates 90degrees… so after every 12 iterations, the summation become 0…
So, 2016 is a multiple of 12…
at 2017, just sub it, and you will get -v1
And at 2021, you also get -3v1
Could you pls send a written solution @Curious_Monkey when you have time ?