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Question
If the vectors `hati - 2xhatj + 3 yhatk and hati + 2xhatj - 3yhatk` are perpendicular, then the locus of (x, y) is ______.
Options
a circle
an ellipse
a hyperbola
None of these
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Solution
If the vectors `hati - 2xhatj + 3 yhatk and hati + 2xhatj - 3yhatk` are perpendicular, then the locus of (x, y) is an ellipse.
Explanation:
\[\text{ Let }, \vec{a} = \hat{ i } - 2x \hat{j} + 3y \hat{k} \text{ and } \vec{b} = \hat{i} + 2x \hat{j} - 3y \hat{k} \]
\[\text{ It is given that the vectors are perpendicular. So, their dot product is zero }.\]
\[ \vec{a} . \vec{b} = 0\]
\[ \Rightarrow \left( \hat{i} - 2x \hat{j} + 3y \hat{k} \right) . \left( \hat{i} + 2x \hat{j} - 3y \hat{k} \right) = 0\]
\[ \Rightarrow 1 - 4 x^2 - 9 y^2 = 0\]
\[ \Rightarrow 4 x^2 + 9 y^2 = 1\]
\[\text{ Dividing both sides by } 36, \text{ we get }\]
\[\frac{x^2}{9} + \frac{y^2}{4} = 1\]
\[\text{ This is an ellipse }.\]
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