A) a sphere
B) an ellipse
C) a circle
D) a plane
Correct Answer: A
Solution :
Let \[\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\] and \[\vec{a}=a\hat{i}+b\hat{j}+c\hat{k},\] where \[a,\,\,\,b,\,\,\,c\] are constant. Now, \[|\vec{r}{{|}^{2}}-2\vec{r}.\vec{a}+p=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2(ax+by+cz)+p=0\] Which represent a sphere, where radius \[=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-p}=+ve\] \[[\because \,\,|\vec{a}|>p|\]You need to login to perform this action.
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