If the vectors vec{a} = hat{i} - 2xhat{j} - 3 | vedclass

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(A) Two vectors are orthogonal if their dot product is equal to $0$.Given vectors are $\vec{a} = \hat{i} - 2x\hat{j} - 3y\hat{k}$ and $\vec{b} = \hat{

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a circle

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(A) Two vectors are orthogonal if their dot product is equal to $0$.Given vectors are $\vec{a} = \hat{i} - 2x\hat{j} - 3y\hat{k}$ and $\vec{b} = \hat{i} + 3x\hat{j} + 2y\hat{k}$.Taking the dot product: $\vec{a} \cdot \vec{b} = (1)(1) + (-2x)(3x) + (-3y)(2y) = 0$.This simplifies to $1 - 6x^2 - 6y^2 = 0$.Rearranging the terms, we get $6x^2 + 6y^2 = 1$, which implies $x^2 + y^2 = \frac{1}{6}$.This is the equation of a circle with center at the origin $(0, 0)$ and radius $\frac{1}{\sqrt{6}}$.Thus, the locus of the point $(x, y)$ is a circle.

If the vectors $\vec{a} = \hat{i} - 2x\hat{j} - 3y\hat{k}$ and $\vec{b} = \hat{i} + 3x\hat{j} + 2y\hat{k}$ are orthogonal to each other, then the locus of the point $(x, y)$ is

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