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Question: The vector equation of the plane through the point \(\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) and per...

The vector equation of the plane through the point i+2j−k\mathbf{i} + 2\mathbf{j} - \mathbf{k} and perpendicular to the line of intersection of the planes r.(3i−j+k)=1\mathbf{r}.(3\mathbf{i} - \mathbf{j} + \mathbf{k}) = 1 and i+4j−2k=2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k} = 2 is

A

r.(2i+7j−13k)=1\mathbf{r}.(2\mathbf{i} + 7\mathbf{j} - 13\mathbf{k}) = 1

B

r.(2i−7j−13k)=1\mathbf{r}.(2\mathbf{i} - 7\mathbf{j} - 13\mathbf{k}) = 1

C

r.(2i+7j+13k)=0\mathbf{r}.(2\mathbf{i} + 7\mathbf{j} + 13\mathbf{k}) = 0

D

None of these

Answer

r.(2i−7j−13k)=1\mathbf{r}.(2\mathbf{i} - 7\mathbf{j} - 13\mathbf{k}) = 1

Explanation

Solution

The line of intersection of the planes r.(3i−j+k)=1\mathbf{r}.(3\mathbf{i} - \mathbf{j} + \mathbf{k}) = 1 and r.(i+4j−2k)=2\mathbf{r}.(\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}) = 2 is common to both the planes. Therefore, it is perpendicular to normals to the two planes i.e., n1=3i−j+k\mathbf{n}_{1} = 3\mathbf{i} - \mathbf{j} + \mathbf{k} and n2=i+4j−2k\mathbf{n}_{2} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}. Hence it is parallel to the vector n1×n2=−2i+7j+13k.\mathbf{n}_{1} \times \mathbf{n}_{2} = - 2\mathbf{i} + 7\mathbf{j} + 13\mathbf{k}. Thus, we have to find the equation of the plane passing through a=i+2j−k\mathbf{a} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}and normal to the vector n=n1×n2\mathbf{n} = \mathbf{n}_{1} \times \mathbf{n}_{2}.The equation of the required plane is (r−a). 6mun=0(\mathbf{r} - \mathbf{a}).\mspace{6mu}\mathbf{n} = 0 or r.n=a.n\mathbf{r}.\mathbf{n} = \mathbf{a}.\mathbf{n}

or r.(−2i+7j+13k)\mathbf{r}.( - 2\mathbf{i} + 7\mathbf{j} + 13\mathbf{k}) =(i+2j−k).(−2i+7j+13k)(\mathbf{i} + 2\mathbf{j} - \mathbf{k}).( - 2\mathbf{i} + 7\mathbf{j} + 13\mathbf{k})

or r.(2i−7j−13k)=1\mathbf{r}.(2\mathbf{i} - 7\mathbf{j} - 13\mathbf{k}) = 1.