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Question: The cartesian equation of the plane \(\mathbf{r = (1 + \lambda}\mathbf{-}\mathbf{\mu)i + (2}\mathbf...

The cartesian equation of the plane

r=(1+λ−μ)i+(2−λ)j+(3−2λ+2μ)k\mathbf{r = (1 + \lambda}\mathbf{-}\mathbf{\mu)i + (2}\mathbf{-}\mathbf{\lambda)j + (3}\mathbf{-}\mathbf{2\lambda + 2\mu)k} is

A

2x+y=52x + y = 5

B

2x−y=52x - y = 5

C

2x+z=52x + z = 5

D

2x−z=52x - z = 5

Answer

2x+z=52x + z = 5

Explanation

Solution

We have r=(1+λ−μ)i+(2−λ)j+(3−2λ+2μ)k\mathbf{r} = (1 + \lambda - \mu)\mathbf{i} + (2 - \lambda)\mathbf{j} + (3 - 2\lambda + 2\mu)\mathbf{k}

⇒ r=(i+2j+3k)+λ(i−j−2k)+μ(−i+2k)\mathbf{r} = (\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) + \lambda(\mathbf{i} - \mathbf{j} - 2\mathbf{k}) + \mu( - \mathbf{i} + 2\mathbf{k}),

Which is a plane passing through a=i+2j+3k\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}and parallel to the vectors b=i−j−2kb = \mathbf{i} - \mathbf{j} - 2\mathbf{k} and c=−i+2k\mathbf{c} = - \mathbf{i} + 2\mathbf{k}

Therefore, it is perpendicular to the vector n=b×c=−2i−k\mathbf{n} = \mathbf{b} \times \mathbf{c} = - 2\mathbf{i} - \mathbf{k}

Hence, its vector equation is (r−a).n=0(\mathbf{r} - \mathbf{a}).\mathbf{n} = 0

⇒ r.n=a.n⇒r.(−2i−k)=−2−3⇒r.(2i+k)=5\mathbf{r}.\mathbf{n} = \mathbf{a}.\mathbf{n} \Rightarrow \mathbf{r}.( - 2\mathbf{i} - \mathbf{k}) = - 2 - 3 \Rightarrow \mathbf{r}.(2\mathbf{i} + \mathbf{k}) = 5

So, the Cartesian equation is (xi+yj+zk).(2i+k)(x\mathbf{i} + y\mathbf{j} + z\mathbf{k}).(2\mathbf{i} + \mathbf{k}) =5

or 2x+z=52x + z = 5.