# Doubt from geometry

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51

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For 51

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 optionsâ€¦ 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

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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 downâ€¦

its a really cool qnâ€¦ 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

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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 onâ€¦
and also v1 = -v3, v2 = -v4 and so on

Now, with the sigmasâ€¦ you will find that after every three iterations, a vector of form 2i + j is formedâ€¦ and in the next three iterations, the vector rotates 90degreesâ€¦ so after every 12 iterations, the summation become 0â€¦

So, 2016 is a multiple of 12â€¦
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 ?

yeeyee

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