Find the Cartesian equation of the plane pass | vedclass

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(A) The vector equation of a plane passing through a point with position vector $\vec{a}$ and normal to vector $\vec{n}$ is given by:$(\vec{r} - \vec{

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$x + 5y - 6z + 18 = 0$

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(A) The vector equation of a plane passing through a point with position vector $\vec{a}$ and normal to vector $\vec{n}$ is given by:$(\vec{r} - \vec{a}) \cdot \vec{n} = 0$ or $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \dots (i)$Given that the plane passes through the point $P(3, -3, 1)$,we have $\vec{a} = 3\hat{i} - 3\hat{j} + \hat{k}$.The plane is normal to the line joining $A(3, 4, -1)$ and $B(2, -1, 5)$. Thus,the normal vector $\vec{n}$ is the vector $\vec{AB}$:$\vec{n} = \vec{AB} = (2 - 3)\hat{i} + (-1 - 4)\hat{j} + (5 - (-1))\hat{k} = -\hat{i} - 5\hat{j} + 6\hat{k}$.Substituting $\vec{a}$ and $\vec{n}$ into equation $(i)$:$\vec{r} \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = (3\hat{i} - 3\hat{j} + \hat{k}) \cdot (-\hat{i} - 5\hat{j} + 6\hat{k})$$\vec{r} \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = (3)(-1) + (-3)(-5) + (1)(6)$$\vec{r} \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = -3 + 15 + 6 = 18$.To find the Cartesian equation,substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = 18$$-x - 5y + 6z = 18$$x + 5y - 6z + 18 = 0$.

Find the Cartesian equation of the plane passing through the point $(3, -3, 1)$ and normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$.

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