Find the vector and Cartesian equations of th | vedclass

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(A) The position vector of the point $(1, 4, 6)$ is $\vec{a} = \hat{i} + 4\hat{j} + 6\hat{k}$.The normal vector $\vec{N}$ perpendicular to the plane i

Vedclass · Accepted Answer

Vector: $(\vec{r} - (\hat{i} + 4\hat{j} + 6\hat{k})) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 0$,Cartesian: $x - 2y + z + 1 = 0$

Vedclass · Answer

(A) The position vector of the point $(1, 4, 6)$ is $\vec{a} = \hat{i} + 4\hat{j} + 6\hat{k}$.The normal vector $\vec{N}$ perpendicular to the plane is $\vec{N} = \hat{i} - 2\hat{j} + \hat{k}$.The vector equation of a plane passing through a point with position vector $\vec{a}$ and normal vector $\vec{N}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{N} = 0$.Substituting the values,we get the vector equation:$(\vec{r} - (\hat{i} + 4\hat{j} + 6\hat{k})) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 0$.To find the Cartesian equation,let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.Substituting this into the vector equation:$((x - 1)\hat{i} + (y - 4)\hat{j} + (z - 6)\hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 0$.Taking the dot product:$1(x - 1) - 2(y - 4) + 1(z - 6) = 0$.Simplifying the expression:$x - 1 - 2y + 8 + z - 6 = 0$.$x - 2y + z + 1 = 0$.Thus,the Cartesian equation is $x - 2y + z + 1 = 0$.

Find the vector and Cartesian equations of the plane that passes through the point $(1, 4, 6)$ and the normal vector to the plane is $\hat{i} - 2\hat{j} + \hat{k}$.

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