The angle between two straight lines is α. One line has direction cosines 12,12,12 and the other line has direction ratios 0,1,2. Then tanα=

# The angle between two straight lines is $\alpha$. One line has direction cosines $\left(\frac{1}{\sqrt{2}},\frac{1}{2},\frac{1}{2}\right)$ and the other line has direction ratios $\left(0,1,2\right)$. Then $\mathrm{tan}\alpha =$

1. A

$\frac{3}{2\sqrt{5}}$

2. B

$\frac{\sqrt{11}}{2\sqrt{5}}$

3. C

$\frac{\sqrt{11}}{3}$

4. D

$\frac{3}{4\sqrt{11}}$

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### Solution:

If $\theta$ is the acute angle between two lines having direction ratios $〈{a}_{1},{b}_{1},{c}_{1}〉$and $〈{a}_{2},{b}_{2},{c}_{2}〉$

then$\mathrm{cos}\theta =\left|\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}\cdot \sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}}\right|$

Given $\alpha$ is angle between two lines having direction ratios $\left(\frac{1}{\sqrt{2}},\frac{1}{2},\frac{1}{2}\right)$ and$\left(0,1,2\right)$

Hence,

$\begin{array}{c}\mathrm{cos}\alpha =\frac{\frac{1}{\sqrt{2}}\left(0\right)+\frac{1}{2}\left(1\right)+\frac{1}{2}\left(2\right)}{1\cdot \sqrt{{0}^{2}+{1}^{2}+{2}^{2}}}\\ =\frac{3}{2\sqrt{5}}\end{array}$

Use right angled triangle, to get the value of $\mathrm{tan}\alpha$

Therefore,$\mathrm{tan}\alpha =\overline{)\frac{\sqrt{11}}{3}}$

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