The vector equation of the plane which is at | vedclass

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Question

(A) The vector equation of a plane at a distance $ d $ from the origin with a unit normal vector $ \hat{n} $ is given by $ \vec{r} \cdot \hat{n} = d $

Vedclass · Accepted Answer

$ \vec{r} \cdot(2 \hat{i}-3 \hat{j}+\hat{k})=3 $

Vedclass · Answer

(A) The vector equation of a plane at a distance $ d $ from the origin with a unit normal vector $ \hat{n} $ is given by $ \vec{r} \cdot \hat{n} = d $. Given the normal vector $ \vec{N} = 2 \hat{i} - 3 \hat{j} + \hat{k} $. First,find the magnitude of $ \vec{N} $: $ |\vec{N}| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} $. The unit normal vector is $ \hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{2 \hat{i} - 3 \hat{j} + \hat{k}}{\sqrt{14}} $. The distance from the origin is $ d = \frac{3}{\sqrt{14}} $. Substituting these into the equation $ \vec{r} \cdot \hat{n} = d $: $ \vec{r} \cdot \left( \frac{2 \hat{i} - 3 \hat{j} + \hat{k}}{\sqrt{14}} \right) = \frac{3}{\sqrt{14}} $. Multiplying both sides by $ \sqrt{14} $,we get: $ \vec{r} \cdot (2 \hat{i} - 3 \hat{j} + \hat{k}) = 3 $.

The vector equation of the plane which is at a distance of $ \frac{3}{\sqrt{14}} $ from the origin and the normal vector from the origin is $ 2 \hat{i}-3 \hat{j}+\hat{k} $ is:

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