The vector equation of any plane passing thro | vedclass

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Question

(C) The equation of a plane passing through the line of intersection of the planes $\vec{r} \cdot \vec{n}_1 = q_1$ and $\vec{r} \cdot \vec{n}_2 = q_2$

Vedclass · Accepted Answer

$\vec{r} \cdot (4 \hat{i}-\hat{j}+\hat{k})=12$

Vedclass · Answer

(C) The equation of a plane passing through the line of intersection of the planes $\vec{r} \cdot \vec{n}_1 = q_1$ and $\vec{r} \cdot \vec{n}_2 = q_2$ is given by $\vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2) = q_1 + \lambda q_2$. Given planes are $\vec{r} \cdot (\hat{i}-2 \hat{j}+3 \hat{k})=5$ and $\vec{r} \cdot (3 \hat{i}+\hat{j}-2 \hat{k})=7$. The equation of the required plane is $\vec{r} \cdot [(\hat{i}-2 \hat{j}+3 \hat{k}) + \lambda (3 \hat{i}+\hat{j}-2 \hat{k})] = 5 + 7\lambda$. This simplifies to $\vec{r} \cdot [(1+3\lambda)\hat{i} + (-2+\lambda)\hat{j} + (3-2\lambda)\hat{k}] = 5 + 7\lambda$. Since the plane passes through the point $\vec{r} = 2\hat{i}-3\hat{j}+\hat{k}$,we substitute these coordinates: $2(1+3\lambda) - 3(-2+\lambda) + 1(3-2\lambda) = 5 + 7\lambda$. $2 + 6\lambda + 6 - 3\lambda + 3 - 2\lambda = 5 + 7\lambda$. $11 + \lambda = 5 + 7\lambda$. $6 = 6\lambda$,which gives $\lambda = 1$. Substituting $\lambda = 1$ into the equation: $\vec{r} \cdot [(1+3(1))\hat{i} + (-2+1)\hat{j} + (3-2(1))\hat{k}] = 5 + 7(1)$. $\vec{r} \cdot (4\hat{i} - \hat{j} + \hat{k}) = 12$.

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