Find the vector equation of the plane passing | vedclass

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Question

(A) The equation of a plane passing through the intersection of two planes $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$ is give

Vedclass · Accepted Answer

$\vec{r} \cdot(20 \hat{i}+23 \hat{j}+26 \hat{k})=69$

Vedclass · Answer

(A) The equation of a plane passing through the intersection of two planes $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$ is given by $\vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2) = d_1 + \lambda d_2.$Given planes are $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5.$Thus,the equation of the required plane is $\vec{r} \cdot [(\hat{i}+\hat{j}+\hat{k}) + \lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})] = 6 - 5\lambda.$This can be written as $\vec{r} \cdot [(1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1+4\lambda)\hat{k}] = 6 - 5\lambda.$Since the plane passes through the point $(1,1,1),$ we substitute $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ into the equation:$(1+2\lambda)(1) + (1+3\lambda)(1) + (1+4\lambda)(1) = 6 - 5\lambda.$$1 + 2\lambda + 1 + 3\lambda + 1 + 4\lambda = 6 - 5\lambda.$$3 + 9\lambda = 6 - 5\lambda.$$14\lambda = 3 \implies \lambda = \frac{3}{14}.$Substituting $\lambda = \frac{3}{14}$ back into the equation:$\vec{r} \cdot [(1 + 2(\frac{3}{14}))\hat{i} + (1 + 3(\frac{3}{14}))\hat{j} + (1 + 4(\frac{3}{14}))\hat{k}] = 6 - 5(\frac{3}{14}).$$\vec{r} \cdot [(1 + \frac{3}{7})\hat{i} + (1 + \frac{9}{14})\hat{j} + (1 + \frac{6}{7})\hat{k}] = \frac{84 - 15}{14}.$$\vec{r} \cdot [\frac{10}{7}\hat{i} + \frac{23}{14}\hat{j} + \frac{13}{7}\hat{k}] = \frac{69}{14}.$Multiplying by $14,$ we get $\vec{r} \cdot (20\hat{i} + 23\hat{j} + 26\hat{k}) = 69.$

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5,$ and the point $(1,1,1).$

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