The acute angle between the two lines whose direction ratios are connected by l+m−n=0  and  l2+m2−n2=0 is

# The acute angle between the two lines whose direction ratios are connected by $l+m-n=0\text{\hspace{0.17em}\hspace{0.17em}}and\text{\hspace{0.17em}\hspace{0.17em}}{l}^{2}+{m}^{2}-{n}^{2}=0$ is

1. A

$\frac{\pi }{3}$

2. B

0

3. C

$\frac{\pi }{6}$

4. D

$\frac{\pi }{4}$

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### Solution:

Suppose that $n=1$

Hence the equations become $l+m=1$and ${l}^{2}+{m}^{2}=1$

Substitute $m=1-l$ in the equation  ${l}^{2}+{m}^{2}=1$and then solve for $l$

It implies that $l=1,0$ and $m=0,1$

Therefore, the direction ratios are $〈1,0,1〉,〈0,1,1〉$

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

Suppose that $\theta$ be the angle between the lines whose direction ratios are $〈1,0,1〉,〈0,1,1〉$

Then $\mathrm{cos}\theta =\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$

Therefore, the measure of angle $\theta$ is $\overline{)\frac{\pi }{3}}$

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