MathematicsHard202×since 2002Q5809Let α→=3i^+j^\overrightarrow \alpha = 3\widehat i + \widehat jα=3i+j and β→=2i^−j^+3k^\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat kβ=2i−j+3k . If β→=β→1−β2→\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}}β=β1−β2, where β→1{\overrightarrow \beta _1}β1 is parallel to α→\overrightarrow \alphaα and β2→\overrightarrow {{\beta _2}}β2 is perpendicular to α→\overrightarrow \alphaα , then β→1×β2→{\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}}β1×β2 is equal toA3i^−9j^−5k^3\widehat i - 9\widehat j - 5\widehat k3i−9j−5kB12{1 \over 2}21(−3i^+9j^+5k^- 3\widehat i + 9\widehat j + 5\widehat k−3i+9j+5k)C−3i^+9j^+5k^- 3\widehat i + 9\widehat j + 5\widehat k−3i+9j+5kD12{1 \over 2}21(3i^−9j^+5k^3\widehat i - 9\widehat j + 5\widehat k3i−9j+5k)Check answerSkip