Find the equation of the plane passing throug | vedclass

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Question

(A) Any plane parallel to the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$ is of the form $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=\lambda$. Since

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$x+y+z=a+b+c$

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(A) Any plane parallel to the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$ is of the form $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=\lambda$. Since the plane passes through the point $(a, b, c)$,its position vector is $\vec{r}=a\hat{i}+b\hat{j}+c\hat{k}$. Substituting this into the equation,we get $(a\hat{i}+b\hat{j}+c\hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=\lambda$,which implies $a+b+c=\lambda$. Substituting $\lambda=a+b+c$ back into the equation,we get $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=a+b+c$. In Cartesian form,substituting $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$,we obtain $x+y+z=a+b+c$.

Find the equation of the plane passing through $(a, b, c)$ and parallel to the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$.

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