3D Geometry
Minimum distance from plane; Cauchy-Schwarz
Grade Class 12
Question:
The minimum value of $x^2+y^2+z^2$ if $ax+by+cz=p$ is
$\left(\dfrac{p}{a+b+c}\right)^2$
$\dfrac{p^2}{a^2+b^2+c^2}$
$\dfrac{a^2+b^2+c^2}{p^2}$
0
Step-by-Step Solution
Key Concept: By Cauchy-Schwarz: $(ax+by+cz)^2\leq(a^2+b^2+c^2)(x^2+y^2+z^2)$. So $x^2+y^2+z^2\geq p^2/(a^2+b^2+c^2)$.
Min $=p^2/(a^2+b^2+c^2)$.
Correct Answer: 2