Vector Algebra Question 200
Question: A vector $ \mathbf{n} $ of magnitude 8 units is inclined to x-axis at $ 45^{o} $ , y-axis at $ 60^{o} $ and an acute angle with z-axis. If a plane passes through a point $ (\sqrt{2},,-1,,1) $ and is normal to $ \mathbf{n} $ , then its equation in vector form is
Options:
A) $ \mathbf{r}.(\sqrt{2}\mathbf{i}+\mathbf{j}+\mathbf{k})=4 $
B) $ \mathbf{r}.(\sqrt{2}\mathbf{i}+\mathbf{j}+\mathbf{k})=2 $
C) $ \mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=4 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Let $ \gamma $ be the angle made by $ \mathbf{n} $ with z-axis. Then direction cosines of $ \mathbf{n} $ are $ l=\cos 45^{o}=\frac{1}{\sqrt{2}}, $ $ m=\cos 60^{o}=\frac{1}{2} $ and $ n=\cos \gamma $ . \ $ l^{2}+m^{2}+n^{2}=1\Rightarrow {{( \frac{1}{\sqrt{2}} )}^{2}}+{{( \frac{1}{2} )}^{2}}+n^{2}=1 $
Þ $ n^{2}=\frac{1}{4}\Rightarrow n=\frac{1}{2} $ , [ $ \because $ $ \gamma $ is acute, \ $ n=\cos \gamma >0 $ ] We have $ |\mathbf{n}|=8 $ , \ $ \mathbf{n}=|\mathbf{n}|(l\mathbf{i}+m\mathbf{j}+n\mathbf{k}) $
$ \Rightarrow \mathbf{n}=8( \frac{1}{\sqrt{2}}\mathbf{i}+\frac{1}{2}\mathbf{j}+\frac{1}{2}\mathbf{k} ) $ $ =4\sqrt{2}\mathbf{i}+4\mathbf{j}+4\mathbf{k} $ The required plane passes through the point $ (\sqrt{2},-1,,1) $ having position vector $ \mathbf{a}=\sqrt{2}\mathbf{i}-\mathbf{j}+\mathbf{k} $ . So, its vector equation is $ (\mathbf{r}-\mathbf{a}).\mathbf{n}=0 $ or $ \mathbf{r}.,\mathbf{n}=\mathbf{a}.,\mathbf{n} $
Þ $ \mathbf{r}.(4\sqrt{2}\mathbf{i}+4\mathbf{j}+4\mathbf{k})=(\sqrt{2}\mathbf{i}-\mathbf{j}+\mathbf{k}).(4\sqrt{2}\mathbf{i}+4\mathbf{j}+4\mathbf{k}) $
Þ $ \mathbf{r}.(\sqrt{2}\mathbf{i}+\mathbf{j}+\mathbf{k})=2 $ .
BETA
