Let ABCD be a parallelogram such that AB = q , AD = p and BAD be an acute angle. If r is the vector that coincides with the altitude directed from the - Mathematics MCQ AIEEE 2012 2026 | ExamTest.live

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Let A BC D be a parallelogram such that A B = q, A D = 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 A D, then r is given by:

Solution & Step-by-step Explanation

Let E be the foot of the altitude from B to A D . A E is the projection of A B onto A D . A E = (\frac{q \cdot p}{∣ p ∣ ^{2}}) p = \frac{p \cdot q}{p \cdot p} p .In Δ A BE, A B + BE = A E is not the correct path because r is directed from B to E . r = BE = A E - A B . r = \frac{p \cdot q}{p \cdot p} p - q .Rearranging: r = - q + \frac{p \cdot q}{p \cdot p} p .

Try it yourself before checking the explanation above.

Let A BC D be a parallelogram such that A B = q, A D = 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 A D, then r is given by:

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

r = - q + \frac{p \cdot q}{p \cdot p} p

r = q - \frac{p \cdot q}{p \cdot p} p

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

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