JEE MAIN - Mathematics (2004 - No. 48)
A line with direction cosines proportional to $$2,1,2$$ meets each of the lines $$x=y+a=z$$ and $$x+a=2y=2z$$ . The co-ordinates of each of the points of intersection are given by :
$$\left( {2a,3a,3a} \right),\left( {2a,a,a} \right)$$
$$\left( {3a,2a,3a} \right),\left( {a,a,a} \right)$$
$$\left( {3a,2a,3a} \right),\left( {a,a,2a} \right)$$
$$\left( {3a,3a,3a} \right),\left( {a,a,a} \right)$$
Explanation
Let a point on the line
$$x = y + a = z$$ is $$\left( {\lambda ,\lambda - a,\lambda } \right)$$
and a point on the line
$$x + a = 2y = 2z$$ is $$\left( {\mu - a,{\mu \over 2},{\mu \over 2}} \right),$$
then direction ratio of the line joining these points are
$$\lambda - \mu + a,\,\,\lambda - a - {\mu \over 2},\,\,\lambda - {\mu \over 2}$$
If it represents the required line, then
$${{\lambda - \mu + a} \over 2}$$
$$ = {{\lambda - a - {\mu \over 2}} \over 1}$$
$$ = {{\lambda - {\mu \over 2}} \over 2}$$
on solving we get $$\lambda = 3a,\,\mu = 2a$$
$$\therefore$$ The required points of intersection are
$$\left( {3a,3a - a,3a} \right)$$ and $$\left( {2a - a,{{2a} \over 2},{{2a} \over 2}} \right)$$
or $$\left( {3a,2a,3a} \right)$$ and $$\left( {a,a,a} \right)$$
$$x = y + a = z$$ is $$\left( {\lambda ,\lambda - a,\lambda } \right)$$
and a point on the line
$$x + a = 2y = 2z$$ is $$\left( {\mu - a,{\mu \over 2},{\mu \over 2}} \right),$$
then direction ratio of the line joining these points are
$$\lambda - \mu + a,\,\,\lambda - a - {\mu \over 2},\,\,\lambda - {\mu \over 2}$$
If it represents the required line, then
$${{\lambda - \mu + a} \over 2}$$
$$ = {{\lambda - a - {\mu \over 2}} \over 1}$$
$$ = {{\lambda - {\mu \over 2}} \over 2}$$
on solving we get $$\lambda = 3a,\,\mu = 2a$$
$$\therefore$$ The required points of intersection are
$$\left( {3a,3a - a,3a} \right)$$ and $$\left( {2a - a,{{2a} \over 2},{{2a} \over 2}} \right)$$
or $$\left( {3a,2a,3a} \right)$$ and $$\left( {a,a,a} \right)$$
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