If a plane is at a distance of 6 units from t | vedclass

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(A) Given, distance from the origin to the plane $d = 6$ units. Normal vector $\vec{N} = 2 \hat{i} + 6 \hat{j} - 3 \hat{k}$. First, find the magnitude

Vedclass · Accepted Answer

$2 x + 6 y - 3 z - 42 = 0$

Vedclass · Answer

(A) Given, distance from the origin to the plane $d = 6$ units. Normal vector $\vec{N} = 2 \hat{i} + 6 \hat{j} - 3 \hat{k}$. First, find the magnitude of the normal vector: $|\vec{N}| = \sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$. The unit normal vector is $\hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{2 \hat{i} + 6 \hat{j} - 3 \hat{k}}{7}$. The equation of the plane in normal form is $\vec{r} \cdot \hat{n} = d$. Substituting the values, we get $\vec{r} \cdot \left( \frac{2 \hat{i} + 6 \hat{j} - 3 \hat{k}}{7} \right) = 6$. Multiplying by $7$, we get $\vec{r} \cdot (2 \hat{i} + 6 \hat{j} - 3 \hat{k}) = 42$. Substituting $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$, we get $(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (2 \hat{i} + 6 \hat{j} - 3 \hat{k}) = 42$. This simplifies to $2x + 6y - 3z = 42$, or $2x + 6y - 3z - 42 = 0$.

If a plane is at a distance of $6$ units from the origin and the vector $2 \hat{i} + 6 \hat{j} - 3 \hat{k}$ is its normal, then the equation of the plane in Cartesian form is

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