Question

If the vector $2\hat{i}-2\hat{j}+7\hat{k}$ is inclined at angle $\alpha ,\beta ,{\rm Y}$ with the coordinate axes then

• A. $3\cos \alpha =\frac{2}{\sqrt{57}}$
• B. $2\cos \beta =-\frac{3}{\sqrt{57}}$
• C. $\cos {\rm Y}=\frac{7}{\sqrt{57}}$
• D. $\cos {\rm Y}=\frac{8}{57}$

Answer 1

Answer: (C)

$\cos {\rm Y}=\frac{7}{\sqrt{57}}$

$\overrightarrow{r}=2\hat{i}-2\hat{j}+7\hat{k}$

$\hat{r}=\frac{2\hat{i}-2\hat{j}+7\hat{k}}{|\overrightarrow{r}|}=\frac{2\hat{i}-2\hat{j}+7\hat{k}}{\sqrt{2^2+2^2+7^2}}=\frac{2\hat{i}-2\hat{j}+7\hat{k}}{\sqrt{57}}$

$\hat{r}=\cos \alpha \hat{i}+\cos \beta \hat{j}+\cos \gamma \hat{k}$

$\cos \alpha =\frac{2}{\sqrt{57}}$ , $\cos \beta =-\frac{2}{\sqrt{57}}$, $\cos \gamma =\frac{7}{\sqrt{57}}$