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Question:

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 vertex B to the side AD, then →r is given by

→r=−→q+3(→p.→q)(→p.→p)→p

→r=−→q+(→p.→q)(→p.→p)→p

→r=→q−(→p.→q)(→p.→p)→p

→r=3→q−(→p.→q)(→p.→p)→p

Solution:

Let E be the point between A and D such that BE ⊥ AD.
→AE = vector component of →q on →p
→AE = (→p ⋅ →q / →p ⋅ →p)→p
Since →r is the vector that coincides with the altitude directed from the vertex B to the side AD, →r = →q - →AE
→r = →q - (→p ⋅ →q / →p ⋅ →p)→p
Therefore, the correct option is →r = →q - (→p.→q)(→p.→p)→p