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Question: The position vectors of two points P and Q are \(3\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\mat...

The position vectors of two points P and Q are 3i+j+2k3\mathbf{i} + \mathbf{j} + 2\mathbf{k} and i2j4k\mathbf{i} - 2\mathbf{j} - 4\mathbf{k} respectively. The equation of the plane through Q and perpendicular to PQ is

A

r.(2i+3j+6k)=28\mathbf{r}.(2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) = 28

B

r.(2i+3j+6k)=32\mathbf{r}.(2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) = 32

C

r.(2i+3j+6k)+28=0\mathbf{r}.(2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) + 28 = 0

D

None of these

Answer

r.(2i+3j+6k)+28=0\mathbf{r}.(2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) + 28 = 0

Explanation

Solution

The required plane is {r(i2j4k)}.PQ=0\{\mathbf{r} - (\mathbf{i} - 2\mathbf{j} - 4\mathbf{k})\}.\overset{\rightarrow}{PQ} = 0.