Q42 / 120 · VITEEE 2008 · Mathematics

The equation $\overrightarrow{\mathrm{r}}^{2}-2 \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{c}}+\mathrm{h}=0,|\overrightarrow{\mathrm{c}}|>\sqrt{\mathrm{h}}$, represents

  • A
    circle
  • B
    ellipse
  • C
    cone
  • Correct answer
    D
    sphere

Solution

Vector equation of a sphere in central form with centre having position vector $\overrightarrow{\mathrm{c}}$ and radius $R$ is $\begin{array}{l} |\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}|=\mathrm{R} \Rightarrow|\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}|^{2}=\mathrm{R}^{2} \\ \Rightarrow(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}})=\mathrm{R}^{2} \\ \Rightarrow \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{r}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=\mathrm{R}^{2} \\ \Rightarrow|\overrightarrow{\mathrm{r}}|^{2}-2(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{c}})+|\overrightarrow{\mathrm{c}}|^{2}=\mathrm{R}^{2} \\ \Rightarrow|\overrightarrow{\mathrm{r}}|^{2}-2(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{c}})+|\overrightarrow{\mathrm{c}}|^{2}-\mathrm{R}^{2}=0 \end{array}$ If $\mathrm{h}=|\overrightarrow{\mathrm{c}}|^{2}-\mathrm{R}^{2}$ i.e. $|\overrightarrow{\mathrm{c}}|>\sqrt{\mathrm{h}}$, then $|\overrightarrow{\mathrm{r}}|^{2}-2(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{c}})+\mathrm{h}=0$ Hence, the given equation represents a sphere.

Hence, the correct answer is option: 'D'.

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