JEE MAIN - Mathematics (2003 - No. 31)
A tetrahedron has vertices at $$O(0,0,0), A(1,2,1) B(2,1,3)$$ and $$C(-1,1,2).$$ Then the angle between the faces $$OAB$$ and $$ABC$$ will be :
$${90^ \circ }$$
$${\cos ^{ - 1}}\left( {{{19} \over {35}}} \right)$$
$${\cos ^{ - 1}}\left( {{{17} \over {31}}} \right)$$
$${30^ \circ }$$
Explanation
Vector perpendicular to the face $$OAB$$
$$ = \overrightarrow {OA} \times \overrightarrow {OB} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 1 \cr 2 & 1 & 3 \cr } } \right| = 5\widehat i - \widehat j - 3\widehat k$$
Vector perpendicular to the face $$ABC$$
$$ = \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & { - 1} & 2 \cr { - 2} & { - 1} & 1 \cr } } \right| = \widehat i - 5\widehat j - 3\widehat k$$
Angle between the faces $$=$$ angle between their normals
$$\cos \theta = \left| {{{5 + 5 + 9} \over {\sqrt {35} \sqrt {35} }}} \right| = {{19} \over {35}}$$
or $$\theta = {\cos ^{ - 1}}\left( {{{19} \over {35}}} \right)$$
$$ = \overrightarrow {OA} \times \overrightarrow {OB} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 1 \cr 2 & 1 & 3 \cr } } \right| = 5\widehat i - \widehat j - 3\widehat k$$
Vector perpendicular to the face $$ABC$$
$$ = \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & { - 1} & 2 \cr { - 2} & { - 1} & 1 \cr } } \right| = \widehat i - 5\widehat j - 3\widehat k$$
Angle between the faces $$=$$ angle between their normals
$$\cos \theta = \left| {{{5 + 5 + 9} \over {\sqrt {35} \sqrt {35} }}} \right| = {{19} \over {35}}$$
or $$\theta = {\cos ^{ - 1}}\left( {{{19} \over {35}}} \right)$$
Comments (0)
