The equation of the plane in Cartesian form, | vedclass

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Question

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

Vedclass · Accepted Answer

$2 x-3 y+4 z=6$

Vedclass · Answer

(A) The equation of a plane at a distance $d$ from the origin with a normal vector $\vec{n}$ is given by $\vec{r} \cdot \hat{n} = d$, where $\hat{n}$ is the unit normal vector.Given the normal vector $\vec{n} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$, the magnitude is $|\vec{n}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.The unit normal vector is $\hat{n} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{29}}$.The distance from the origin is $d = \frac{6}{\sqrt{29}}$.Substituting these into the equation $\vec{r} \cdot \hat{n} = d$:$(x \hat{i} + y \hat{j} + z \hat{k}) \cdot \left( \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{29}} \right) = \frac{6}{\sqrt{29}}$.Multiplying both sides by $\sqrt{29}$, we get $2x - 3y + 4z = 6$.

The equation of the plane in Cartesian form, which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector drawn from the origin being $2 \hat{i}-3 \hat{j}+4 \hat{k}$, is

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