Find the angle between the lines whose direction cosines are give… - Vidyadip

Question

Find the angle between the lines whose direction cosines are given by the equations:
$2l + 2m - n = 0, mn + ln + lm = 0$

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Answer

The given equation are,
$2l + 2m - n = 0 .....(1)$
$mn + ln + lm = 0 .....(2)$
From (1), We get n = 2l + 2m.
Putting $n = 2l + 2m$ in $(2)$, We get
$m(2l + 2m) + l(2l + 2m) + lm = 0$
$2lm + 2m^2 + 2l^2 + 2ml + lm = 0$
$2ml^2 + 5lm + 2l^2 = 0$
$2m^2 + 4lm + lm + 2l^2 = 0$
$(2m + l) (m + 2l) = 0$
$\Rightarrow\text{m}=-\frac{1}{2}$ or $\text{m}=-2\text{l}$
By putting $\text{m}=-\frac{\text{l}}{2}$ in (1) we get n = l
By putting m = 2l in (i) we get n = -2l
So, vector parallel to these lines are
$\vec{\text{a}}=\hat{\text{i}}-\frac{1}{2}\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$ respectively.
If $\theta$ is the angle between the lines, then $\theta$ is also the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$
then,
$\cos\theta=\frac{\vec{\text{a}}\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{1+1-2}{\sqrt{1+\frac{1}{4}+1}\sqrt{1+4+9}}=0$
$\theta=\cos^{-1}(0)=\frac{\pi}{2}$.

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