Vector Algebra Question 358
Question: The vector equation of a plane, which is at a distance of 8 unit from the origin and which is normal to the vector $ 2\mathbf{i}+\mathbf{j}+2\mathbf{k}, $ is
Options:
A) $ \mathbf{r}.(2\mathbf{i}+\mathbf{j}+\mathbf{k})=24 $
B) $ \mathbf{r}.(2\mathbf{i}+\mathbf{j}+2\mathbf{k})=24 $
C) $ \mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=24 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Here $ d=8 $ and $ \mathbf{n}=2\mathbf{i}+\mathbf{j}+2\mathbf{k}) $ \ $ \mathbf{\hat{n}}=\frac{\mathbf{n}}{|\mathbf{n}|}=\frac{2\mathbf{i}+\mathbf{j}+2\mathbf{k}}{\sqrt{4+1+4}}=\frac{2}{3}\mathbf{i}+\frac{1}{3}\mathbf{j}+\frac{2}{3}\mathbf{k} $ Hence, the required equation of the plane is $ \mathbf{r}.( \frac{2}{3}\mathbf{i}+\frac{1}{3}\mathbf{j}+\frac{2}{3}\mathbf{k} )=8 $ or $ \mathbf{r}.(2\mathbf{i}+\mathbf{j}+2\mathbf{k})=24 $ .
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