Find the vector equation of the plane passing | vedclass

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(D) The equation of a plane parallel to the plane $\vec{r} \cdot \vec{n} = d$ is given by $\vec{r} \cdot \vec{n} = d_1$. Given the plane $\vec{r} \cdo

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(D) The equation of a plane parallel to the plane $\vec{r} \cdot \vec{n} = d$ is given by $\vec{r} \cdot \vec{n} = d_1$. Given the plane $\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) = 7$,the normal vector is $\vec{n} = 4\hat{i} - 12\hat{j} - 3\hat{k}$. The equation of the required plane is $\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) = d_1$. Since the plane passes through the point $\vec{a} = 2\hat{i} + \hat{j} - 4\hat{k}$,we substitute $\vec{r} = \vec{a}$ into the equation: $d_1 = (2\hat{i} + \hat{j} - 4\hat{k}) \cdot (4\hat{i} - 12\hat{j} - 3\hat{k})$ $d_1 = (2)(4) + (1)(-12) + (-4)(-3)$ $d_1 = 8 - 12 + 12 = 8$. Thus,the equation of the plane is $\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) = 8$.

Find the vector equation of the plane passing through the point $2\hat{i} + \hat{j} - 4\hat{k}$ and parallel to the plane $\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) - 7 = 0$.

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