JEE MAIN - Mathematics (2010 - No. 22)
A line $$AB$$ in three-dimensional space makes angles $${45^ \circ }$$ and $${120^ \circ }$$ with the positive $$x$$-axis and the positive $$y$$-axis respectively. If $$AB$$ makes an acute angle $$\theta $$ with the positive $$z$$-axis, then $$\theta $$ equals :
$${45^ \circ }$$
$${60^ \circ }$$
$${75^ \circ }$$
$${30^ \circ }$$
Explanation
Direction cosines of the line :
$$\ell = \cos {45^ \circ } = {1 \over {\sqrt 2 }},m = \cos {120^ \circ } = {{ - 1} \over 2},\pi = \cos \theta $$
where $$\theta $$ is the angle, which line makes with positive $$z$$-axis.
Now $${\ell ^2} + {m^2} + {n^2} = 1$$
$$ \Rightarrow {1 \over 2} + {1 \over 4} + {\cos ^2}\theta = 1,\,\,{\cos ^2}\theta = {1 \over 4}$$
$$ \Rightarrow \cos \theta = {1 \over 2}\,\,\,\,\left( \theta \right.$$ being acute)
$$ \Rightarrow 0 = {\pi \over 3}$$
$$\ell = \cos {45^ \circ } = {1 \over {\sqrt 2 }},m = \cos {120^ \circ } = {{ - 1} \over 2},\pi = \cos \theta $$
where $$\theta $$ is the angle, which line makes with positive $$z$$-axis.
Now $${\ell ^2} + {m^2} + {n^2} = 1$$
$$ \Rightarrow {1 \over 2} + {1 \over 4} + {\cos ^2}\theta = 1,\,\,{\cos ^2}\theta = {1 \over 4}$$
$$ \Rightarrow \cos \theta = {1 \over 2}\,\,\,\,\left( \theta \right.$$ being acute)
$$ \Rightarrow 0 = {\pi \over 3}$$
Comments (0)
