A parallelogram is constructed on the vectors $ \overrightarrow{\mathbf{a}}=3 \vec{\alpha}-\vec{\beta}, \overrightarrow{\mathbf{b}}=\vec{\alpha}+3 \vec{\beta} $, if $ |\overrightarrow{\alpha \alpha}|=|\vec{\beta}|=2 $ and angle between $ \vec{\alpha} $ and $ \vec{\beta} $ is $ \frac{\pi}{3} $, then length of diagonal of the parallelogram is

Question

A parallelogram is constructed on the vectors $ \overrightarrow{\mathbf{a}}=3 \vec{\alpha}-\vec{\beta}, \overrightarrow{\mathbf{b}}=\vec{\alpha}+3 \vec{\beta} $, if $ |\overrightarrow{\alpha \alpha}|=|\vec{\beta}|=2 $ and angle between $ \vec{\alpha} $ and $ \vec{\beta} $ is $ \frac{\pi}{3} $, then length of diagonal of the parallelogram is, $ 4 \sqrt{7} $, $ 4 \sqrt{3} $, $ 4 \sqrt{17} $, None of these

MARKS App · Accepted Answer

Let A B → = a → = 3 α → - β → , B C → = b → = α → + 3 β → Diagonal A C → = A B → + B C → = a → + b → ⟹ A C → = a → + b → ⟹ A C → = 4 α → + 2 β → ⟹ A C → 2 = 16 α → 2 + 4 β → 2 + 16 α → ⋅ β → ⟹ A C → 2 = 64 + 16 + 16 α → β → c o s π 3 ⟹ A C → 2 = 80 + 16 × 4 × 1 2 = 112 ⟹ A C → = 4 7 Other diagonal is | B D → = | a → - b → | ⟹ B D → 2 = | 2 α → - 4 β → | 2 = 4 α → | 2 + 16 β → | 2 - 16 α → β → cos ⁡ π 3 = 64 + 16 - 16 × 4 × 1 2 = 48 ⟹ | B D → = 48 = 4 3

Search any question & find its solution

Looking for more such questions to practice?