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Question: The vector equation of the plane containing the lines \(\mathbf{r} = (\mathbf{i} + \mathbf{j}) + \l...

The vector equation of the plane containing the lines

r=(i+j)+λ(i+2j−k)\mathbf{r} = (\mathbf{i} + \mathbf{j}) + \lambda(\mathbf{i} + 2\mathbf{j} - \mathbf{k}) and r=(i+j)+μ(−i+j−2k)\mathbf{r} = (\mathbf{i} + \mathbf{j}) + \mu( - \mathbf{i} + \mathbf{j} - 2\mathbf{k}) is

A

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

B

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

C

r.(i+j+k)=3\mathbf{r}.(\mathbf{i} + \mathbf{j} + \mathbf{k}) = 3

D

None of these

Answer

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

Explanation

Solution

Given two lines r=(i+j)+λ(i+2j−k)\mathbf{r} = (\mathbf{i} + \mathbf{j}) + \lambda(\mathbf{i} + 2\mathbf{j} - \mathbf{k}) and

r=(i+j)+μ(−i+j−2k)\mathbf{r} = (\mathbf{i} + \mathbf{j}) + \mu( - \mathbf{i} + \mathbf{j} - 2\mathbf{k}) pass through a=i+j\mathbf{a} = \mathbf{i} + \mathbf{j} and are parallel to the vectors b=i+2j−k\mathbf{b} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}and c=−i+j−2k\mathbf{c} = - \mathbf{i} + \mathbf{j} - 2\mathbf{k} respectively. Therefore the plane containing them passes through a=i+j\mathbf{a} = \mathbf{i} + \mathbf{j} and is perpendicular to

n=b×c=(i+2j−k)×(−i+j−2k)=−3i+3j+3k\mathbf{n} = \mathbf{b} \times \mathbf{c} = (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) \times ( - \mathbf{i} + \mathbf{j} - 2\mathbf{k}) = - 3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}.

Hence, the equation of the plane is

(r−a).n=0⇒r.n=a.n⇒r.(i−j−k)=0(\mathbf{r} - \mathbf{a}).\mathbf{n} = 0 \Rightarrow \mathbf{r}.\mathbf{n} = \mathbf{a}.\mathbf{n} \Rightarrow \mathbf{r}.(\mathbf{i} - \mathbf{j} - \mathbf{k}) = 0.