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Question: The vector equation of the plane through the point (2, 1, –1) and passing through the line of inters...

The vector equation of the plane through the point (2, 1, –1) and passing through the line of intersection of the plane r.(i+3j−k)=0\mathbf{r}.(\mathbf{i} + 3\mathbf{j} - \mathbf{k}) = 0 and r.(j+2k)=0\mathbf{r}.(\mathbf{j} + 2\mathbf{k}) = 0 is

A

r.(i+9j+11k)=0\mathbf{r}.(\mathbf{i} + 9\mathbf{j} + 11\mathbf{k}) = 0

B

r.(i+9j+11k)=6\mathbf{r}.(\mathbf{i} + 9\mathbf{j} + 11\mathbf{k}) = 6

C

r.(i−3j−13k)=0\mathbf{r}.(\mathbf{i} - 3\mathbf{j} - 13\mathbf{k}) = 0

D

None of these

Answer

r.(i+9j+11k)=0\mathbf{r}.(\mathbf{i} + 9\mathbf{j} + 11\mathbf{k}) = 0

Explanation

Solution

The vector equation of a plane through the line of intersection of the planes r.(i+3j−k)=0\mathbf{r}.(\mathbf{i} + 3\mathbf{j} - \mathbf{k}) = 0 and r.(j+2k)\mathbf{r}.(\mathbf{j} + 2\mathbf{k}) =0 can be written as

(r.(i+3j−k))+λ(r.(j+2k))=0(\mathbf{r}.(\mathbf{i} + 3\mathbf{j} - \mathbf{k})) + \lambda(\mathbf{r}.(\mathbf{j} + 2\mathbf{k})) = 0 .....(i)

This passes through 2i+j−k2\mathbf{i} + \mathbf{j} - \mathbf{k}

∴ (2i+j−k).(i+3j−k)+λ(2i+j−k).(j+2k)=0(2\mathbf{i} + \mathbf{j} - \mathbf{k}).(\mathbf{i} + 3\mathbf{j} - \mathbf{k}) + \lambda(2\mathbf{i} + \mathbf{j} - \mathbf{k}).(\mathbf{j} + 2\mathbf{k}) = 0

or (2+3+1)+λ(0+1−2)=0⇒λ=6(2 + 3 + 1) + \lambda(0 + 1 - 2) = 0 \Rightarrow \lambda = 6

Put the value of λ\lambda in (i) we get

r.(i+9j+11k)=0\mathbf{r}.(\mathbf{i} + 9\mathbf{j} + 11\mathbf{k}) = 0, which is the required plane.