A line AB in three-dimensional space makes angles 45∘ and 120∘ with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle θ with the positive z-axis, then θ equals
Solutions · 1
Official solution
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)
\[ \Rightarrow 0 = {\pi \over 3}\]