What is the angle between two lines whose direction ratios satisfy following equations?

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What is the angle between two lines whose direction ratios satisfy following equations?

Find the angle between two lines whose direction ratios satisfy $2 l - m + 2 n = 0$ $2l-m+2n=0$ and $m n + n l + l m = 0$ $mn+nl+lm=0$

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Answer 1 · 2017-03-12T09:33:26

The two lines are perpendicular to each other.

Explanation:

Let the direction cosines of the two lines be $(l_{1}, m_{1}, n_{1})$ $(l_1,m_1,n_1)$ and $(l_{2}, m_{2}, n_{2})$ $(l_2,m_2,n_2)$ and as they satisfy the conditions we have

$2 l - m + 2 n = 0$ $2l-m+2n=0$ .............................(1) and

$m n + n l + l m = 0$ $mn+nl+lm=0$ .............................(2)

Note that if $(a_{1}, b_{1}, c_{1})$ $(a_1,b_1,c_1)$ and $(a_{2}, b_{2}, c_{2})$ $(a_2,b_2,c_2)$ are direction ratios of two lines, we have $\frac{l_{1}}{a_{1}} = \frac{m_{1}}{b_{1}} = \frac{n_{1}}{c_{1}}$ $l_1/a_1=m_1/b_1=n_1/c_1$ and $\frac{l_{2}}{a_{2}} = \frac{m_{2}}{b_{2}} = \frac{n_{2}}{c_{2}}$ $l_2/a_2=m_2/b_2=n_2/c_2$

and the angle $θ$ $theta$ between them is given by

$cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$ $costheta=(a_1a_2+b_1b_2+c_1c_2)/(sqrt(a_1^2+b_1^2+c_1^2)sqrt(a_2^2+b_2^2+c_2^2))$

From (1), we get $m = 2 (l + n)$ $m=2(l+n)$ and substituting in (2) we get

$2 n (l + n) + n l + 2 l (l + n) = 0$ $2n(l+n)+nl+2l(l+n)=0$ or $2 l^{2} + 5 ln + 2 n^{2} = 0$ $2l^2+5ln+2n^2=0$

or $(l + 2 n) (2 l + n) = 0$ $(l+2n)(2l+n)=0$

i.e either $l + 2 n = 0$ $l+2n=0$ or $2 l + n = 0$ $2l+n=0$

if $l + 2 n = 0$ $l+2n=0$ , $l = - 2 n$ $l=-2n$ and $m = 2 (- 2 n + n) = - 2 n$ $m=2(-2n+n)=-2n$ and we have

$\frac{l_{1}}{- 2} = \frac{m_{1}}{- 2} = \frac{n_{1}}{1}$ $l_1/(-2)=m_1/(-2)=n_1/1$

and if $2 l + n = 0$ $2l+n=0$ m $n = - 2 l$ $n=-2l$ and $m = 2 (l - 2 l) = - 2 l$ $m=2(l-2l)=-2l$ and we have

$\frac{l_{2}}{1} = \frac{m_{2}}{- 2} = \frac{n_{2}}{- 2}$ $l_2/(1)=m_2/(-2)=n_2/(-2)$

and $cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$ $costheta=(a_1a_2+b_1b_2+c_1c_2)/(sqrt(a_1^2+b_1^2+c_1^2)sqrt(a_2^2+b_2^2+c_2^2))$

= $\frac{(- 2) \times 1 + (- 2) \times (- 2) + 1 \times (- 2)}{\sqrt{4 + 4 + 1} \sqrt{1 + 4 + 4}}$ $((-2)xx1+(-2)xx(-2)+1xx(-2))/(sqrt(4+4+1)sqrt(1+4+4))$

= $\frac{- 2 + 4 - 2}{9} = 0$ $(-2+4-2)/9=0$

Hence the two lines are perpendicular to each other.

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What is the angle between two lines whose direction ratios satisfy following equations?

Find the angle between two lines whose direction ratios satisfy $2 l - m + 2 n = 0$ $2l-m+2n=0$ and $m n + n l + l m = 0$ $mn+nl+lm=0$

1 Answer

Explanation:

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