Question:
In quadrilateral ABCD, $\vec{AB} = \vec{a},\ \vec{BC} = \vec{b},\ \vec{DA} = \vec{a} - \vec{b}$. M is midpoint of BC and X lies on DM such that $\vec{DX} = \dfrac{4}{5}\vec{DM}$. Then the points A, X and C:
In quadrilateral ABCD, $\vec{AB} = \vec{a},\ \vec{BC} = \vec{b},\ \vec{DA} = \vec{a} - \vec{b}$. M is midpoint of BC and X lies on DM such that $\vec{DX} = \dfrac{4}{5}\vec{DM}$. Then the points A, X and C:
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Use section formula or linear dependence of vectors to verify collinearity.
Updated On: May 18, 2025
- form an equilateral triangle
- are collinear
- form an isosceles triangle
- form a right angled triangle
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The Correct Option is B
Solution and Explanation
Given vectors and placement imply point X divides line DM internally in 4:1 ratio. If A, X and C lie on the same straight line vectorially, their vectors are linearly dependent, proving collinearity.
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