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(D) The equation of a plane passing through the intersection of two planes $P_1: \overline{r} \cdot \overline{n}_1 = d_1$ and $P_2: \overline{r} \cdot

Vedclass · Accepted Answer

$\overline{r} \cdot(\overline{i}+8 \overline{j}+2 \overline{k})=23$

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(D) The equation of a plane passing through the intersection of two planes $P_1: \overline{r} \cdot \overline{n}_1 = d_1$ and $P_2: \overline{r} \cdot \overline{n}_2 = d_2$ is given by $(\overline{r} \cdot \overline{n}_1 - d_1) + \lambda(\overline{r} \cdot \overline{n}_2 - d_2) = 0$. Given planes are $\overline{r} \cdot(\overline{i}-2 \overline{k}) - 3 = 0$ and $\overline{r} \cdot(2 \overline{j}+\overline{k}) - 5 = 0$. The equation of the required plane is $(\overline{r} \cdot(\overline{i}-2 \overline{k}) - 3) + \lambda(\overline{r} \cdot(2 \overline{j}+\overline{k}) - 5) = 0$. This plane passes through the point $\overline{a} = \overline{i}+2 \overline{j}+3 \overline{k}$. Substituting $\overline{r} = \overline{i}+2 \overline{j}+3 \overline{k}$ into the equation: $((\overline{i}+2 \overline{j}+3 \overline{k}) \cdot(\overline{i}-2 \overline{k}) - 3) + \lambda((\overline{i}+2 \overline{j}+3 \overline{k}) \cdot(2 \overline{j}+\overline{k}) - 5) = 0$. Calculating the dot products: $(1 - 6 - 3) + \lambda(4 + 3 - 5) = 0$ $-8 + \lambda(2) = 0 \implies 2\lambda = 8 \implies \lambda = 4$. Substituting $\lambda = 4$ back into the plane equation: $\overline{r} \cdot(\overline{i}-2 \overline{k}) - 3 + 4(\overline{r} \cdot(2 \overline{j}+\overline{k}) - 5) = 0$ $\overline{r} \cdot(\overline{i} + 8 \overline{j} - 2 \overline{k} + 4 \overline{k}) = 3 + 20$ $\overline{r} \cdot(\overline{i} + 8 \overline{j} + 2 \overline{k}) = 23$.

The vector equation of a plane passing through the line of intersection of the planes $\overline{r} \cdot(\overline{i}-2 \overline{k})=3$ and $\overline{r} \cdot(2 \overline{j}+\overline{k})=5$,and passing through the point $\overline{i}+2 \overline{j}+3 \overline{k}$,is:

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