Question

Q: x = y = z −1 does not intersect the cone x^2+y^2+z^2+2(xy+yz+zx)=0
Is the statement true? Give reason for your answer, either with a short proof or a counterexample.

Answer 1

Answer on Question #78516 – Math – Analytic Geometry

$$\left\{ \begin{array}{l} x = y = z - 1 \\ x ^ {\wedge} 2 + y ^ {\wedge} 2 + z ^ {\wedge} 2 + 2 (x y + y z + z x) = 0 \end{array} \right.$$ $$\left\{ \begin{array}{l} x = y = z - 1 \\ x ^ {\wedge} 2 + x ^ {\wedge} 2 + (x + 1) ^ {\wedge} 2 + 2 (x ^ {*} x + x ^ {*} (x + 1) + x ^ {*} (x + 1)) = 0 \end{array} \right.$$ $$\left\{ \begin{array}{l} x = y = z - 1 \\ 9 ^ {*} x ^ {\wedge} 2 + 6 ^ {*} x + 1 = 0 \end{array} \right.$$ $$(3 x + 1) ^ {\wedge} 2 = 0.$$

The system has one solution $A(-1/3; -1/3; -4/3)$.

Thus, $Q$ intersects the cone.

**Answer**: the statement is false.

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