MCQ
If the vectors $\hat{\text{i}}-2\text{x}\hat{\text{j}}+3\text{y}\hat{\text{k}}$ and $\hat{\text{i}}+2\text{x}\hat{\text{j}}-3\text{y}\hat{\text{k}}$ are perpendicular, then the locus of $(x,y)$ is:
  • A
    A circle.
  • An ellipse.
  • C
    A hyperbola.
  • D
    None of these.

Answer

Correct option: B.
An ellipse.
Let, $\vec{\text{a}}=\hat{\text{i}}-2\text{x}\hat{\text{j}}+3\text{y}\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+2\text{x}\hat{\text{j}}-3\text{y}\hat{\text{k}}$
It is given that the vectors are perpendicular.
so, their dot product is zero.
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\big(\hat{\text{i}}-2\text{x}\hat{\text{j}}+3\text{y}\hat{\text{k}}\big).\big(\hat{\text{i}}+2\text{x}\hat{\text{j}}-3\text{y}\hat{\text{k}}\big)=0$
$\Rightarrow1-4\text{x}^2-9\text{y}^2=0$
$\Rightarrow4\text{x}^2+9\text{y}^2=1$
Dividing both sides by $36$, we get
$\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=1$
This is an ellipse.

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