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(B) The equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\vec{r} \cdot (2\

Vedclass · Accepted Answer

$\vec{r} \cdot (3\hat{i} + 2\hat{j} + 0\hat{k}) = 8$

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(B) The equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$ is given by:$[\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) - 5] + \lambda [\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) - 3] = 0$$\Rightarrow \vec{r} \cdot [(1 + 2\lambda)\hat{i} + (3 - \lambda)\hat{j} + (-1 + \lambda)\hat{k}] = 5 + 3\lambda \quad (i)$Since the plane passes through the point $(2, 1, -2)$,the position vector is $\vec{r} = 2\hat{i} + \hat{j} - 2\hat{k}$. Substituting this into equation $(i)$:$(2\hat{i} + \hat{j} - 2\hat{k}) \cdot [(1 + 2\lambda)\hat{i} + (3 - \lambda)\hat{j} + (-1 + \lambda)\hat{k}] = 5 + 3\lambda$$2(1 + 2\lambda) + 1(3 - \lambda) - 2(-1 + \lambda) = 5 + 3\lambda$$2 + 4\lambda + 3 - \lambda + 2 - 2\lambda = 5 + 3\lambda$$7 + \lambda = 5 + 3\lambda$$2 = 2\lambda \Rightarrow \lambda = 1$Substituting $\lambda = 1$ in equation $(i)$:$\vec{r} \cdot [(1 + 2(1))\hat{i} + (3 - 1)\hat{j} + (-1 + 1)\hat{k}] = 5 + 3(1)$$\vec{r} \cdot (3\hat{i} + 2\hat{j} + 0\hat{k}) = 8$.

Find the equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$,and passing through the point $(2, 1, -2)$.

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