A vector \[\vec{a}\] is parallel to the line of intersection of the plane determined by the vectors \[\hat{i}, \hat{i}+\hat{j}\] and the plane determined by the vectors \[\hat{i}-\hat{j}, \hat{i}+\hat{k}\]. The obtuse angle between \[\vec{a}\] and the vector \[\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}\] is :
Solutions · 1
Official solution
If \[{\overrightarrow n _1}\] is a vector normal to the plane determined by \[\widehat i\] and \[\widehat i + \widehat j\] then
\[{\overrightarrow n _1} = \left| {i11j01k00 } \right| = \widehat k\]
If \[{\overrightarrow n _2}\] is a vector normal to the plane determined by \[\widehat i - \widehat j\] and \[\widehat i + \widehat k\] then
\[{\overline n _2} = \left| {i11j−10k01 } \right| = - \widehat i - \widehat j + \widehat k\]
Vector \[\overrightarrow a \] is parallel to \[{\overrightarrow n _1} \times {\overrightarrow n _2}\]
i.e. \[\overrightarrow a \] is parallel to \[\left| {i0−1j0−1k11 } \right| = \widehat i - \widehat j\]
Given \[\overrightarrow b = \widehat i - 2\widehat j + 2\widehat k\]
Cosine of acute angle between
\[\overrightarrow a \] and \[\overrightarrow b = \left| {{{\overrightarrow a \,.\,\overrightarrow b } \over {|\overrightarrow a |\,.\,|\overrightarrow b |}}} \right| = {1 \over {\sqrt 2 }}\]
Obtuse angle between \[\overrightarrow a \] and \[\overrightarrow b = {{3\pi } \over 4}\]