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Chapter 11: Q. 60 (page 862)

If 伪, 尾, and 纬are nonzero constants, the graph of a vector function of the formrole="math" localid="1649577570077" rt=伪t,尾t2,纬t3 is called a twisted cubic. Prove that a twisted cubic intersects any plane in at most three points.

Short Answer

Expert verified

A twisted cubic intersects any plane in at most three points.

Step by step solution

01

Step 1. Given information

We have been given definition of twisted cubic.

We have to prove that a twisted cubic intersects any plane in at most three points.

02

Step 2. Proof of the given question.

The twisted cubic in the form of vector valued function is rt=伪t,尾t2,纬t3.

Assume equation of a general plane is ax+by+cz=1

When a vector valued function intersects a plane it will satisfy the equation of a plane. That is

a伪t+b尾t2+c纬t3=1c纬t3+b尾t2+a伪t-1=0

On solving equation three values of tis obtained.

Thus, a twisted cubic intersects any plane in at most three points.

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