Question

The equation \(r^2 - 2\vec{r}\cdot\vec{c} + h = 0,\ |\vec{c}|>\sqrt{h}\), represents

• A. circle
• B. ellipse
• C. cone
• D. sphere

Hint:

Vector form of sphere: \(|\vec{r}-\vec{a}|^2=R^2\), where \(\vec{a}\) is centre and \(R\) is radius.

Answer 1

The Correct Option is D

Step 1: Rewrite given equation.
Given:
\[ r^2 - 2\vec{r}\cdot\vec{c} + h = 0 \]
Step 2: Use identity for completing square.
\[ r^2 - 2\vec{r}\cdot\vec{c} = |\vec{r}-\vec{c}|^2 - |\vec{c}|^2 \]
So equation becomes:
\[ |\vec{r}-\vec{c}|^2 - |\vec{c}|^2 + h = 0 \]
\[ |\vec{r}-\vec{c}|^2 = |\vec{c}|^2 - h \]
Step 3: Interpret.
This is equation of a sphere with centre \(\vec{c}\) and radius:
\[ R=\sqrt{|\vec{c}|^2-h} \]
Step 4: Condition given.
\(|\vec{c}|>\sqrt{h}\) ensures \( |\vec{c}|^2-h>0\), so radius is real.
Final Answer:
\[ \boxed{\text{Sphere}} \]