Question

If the line $\overrightarrow{r}=2\overrightarrow{i}-\overrightarrow{j}-3\overrightarrow{k}+\lambda \left(\overrightarrow{i}+\overrightarrow{j}+\sqrt{2}\hat{k}\right)$ makes angles $\alpha ,\beta ,\gamma$ with $xy,yz,zx$ planes respectively then which of the following are not possible?

• A. $si{n}^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =2$ & ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$
• B. ${\tan }^2\alpha +{\tan }^2\beta +{\tan }^2\gamma =7$ & ${\cot }^2\alpha +{\cot }^2\beta +{\cot }^2\gamma =5/3$
• C. ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =1$ & ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =2$
• D. ${\sec }^2\alpha +{\sec }^2\beta +{\sec }^2\gamma =14/3$ & $\cos e{c}^2\alpha +\cos e{c}^2\beta +\cos e{c}^2\gamma =10$

Answer 1

The correct option is A $si{n}^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =2$ & ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

$\overrightarrow{r}=2\hat{i}-\hat{j}-3\hat{k}+\lambda (\hat{i}+\hat{j}+\sqrt{2}\hat{k})$

direction ratios of $\overrightarrow{r}=\hat{i}+\hat{j}+\sqrt{2}\hat{k}$

$\alpha ,\beta ,\gamma$ are angles that the plane makes with $z=0,x=0,y=0$.

angle made by a plane and line is complement of angle made by planes normal and line.

$\cos (90-\alpha )=\frac{\sqrt{2}}{2}=\sin \alpha \Rightarrow \alpha ={45}^0$

$\cos (90-\beta )=\frac{1}{2}=\sin \beta \Rightarrow \beta ={30}^0$

$\cos (90-\gamma )=\frac{1}{2}=\sin \gamma \Rightarrow \gamma ={30}^0$

$\therefore \sin {}^2\alpha +\sin {}^2\beta +\sin {}^2\gamma =1$

$\therefore \cos {}^2\alpha +\cos {}^2\beta +\cos {}^2\gamma =2$

$\therefore {\cot }^2\alpha +{\cot }^2\beta +{\cot }^2\gamma =\frac{5}{3}$

$\therefore {\tan }^2\alpha +{\tan }^2\beta +{\tan }^2\gamma =7$

$\therefore {\csc }^2\alpha +{\csc }^2\beta +{\csc }^2\gamma =10$

$\therefore {\sec }^2\alpha +{\sec }^2\beta +{\sec }^2\gamma =\frac{14}{3}$