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Properties of Parallelogram

Let ABCD be a...

Question

Let ABCD be a parallelogram such that AB = q , AB = p, and $∠ B A D$ be an acute angle. If r is the vector that coincides with the altitude directed from the vertex B to the side AD, then r is given by

\to r = 3 \to q - \frac{3 (\to p \cdot \to q)}{(\to p \cdot \to q)} \to p

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\to r = - \to q + \frac{(\to p \cdot \to q)}{(\to p \cdot \to q)} \to p

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\to r = - \to - q + \frac{(\to p \cdot \to q)}{(\to p \cdot \to p)} \to p

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\to r = - 3 \to q - \frac{3 (\to p \cdot \to q)}{(\to p \cdot \to q)} \to p

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Solution

The correct option is C $\to r = - \to - q + \frac{(\to p \cdot \to q)}{(\to p \cdot \to p)} \to p$

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Similar questions

A B C D

- - \to A B = \to q, - - \to A D = \to p

∠ B A D

\to r

B

A D

\to r

\to A B = q, \to A D = \to p

∠

\to r

\to r

\to p, \to q, \to r

\to p \times \to q = \to r

\to a

[\to a \to q \to r] = 1, [\to a \to r \to p] = 2

[\to a \to p \to q] = 3

\to a

p, q, r

x^{3} + 2 x^{2} + 3 x + 3 = 0

Let PQRS be a kite such that PQ > PS. Prove that ∠PQR > ∠PSR. (Hint: Join QS).

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