When a parallelogram of two vectors is a rectangle

When a parallelogram of two vectors is a rectangle

IP 4
8 marks

It is given that \(\mathbf a=\begin{pmatrix}3\\ 4\end{pmatrix}\) and \(\mathbf b=\begin{pmatrix}k\\ 2\end{pmatrix}\), where \(k\) is a real constant. The vectors \(\mathbf a\) and \(\mathbf b\) are drawn from the same point \(O\) as adjacent sides of a parallelogram \(OACB\).

  1. Show that \(|\mathbf a+\mathbf b|\ge6\) for every value of \(k\), state the value of \(k\) for which equality holds, and give a geometric reason why \(|\mathbf a+\mathbf b|\) can never be less than \(6\).[3]
  2. Find the value of \(k\) for which the diagonals \(OC\) and \(AB\) of the parallelogram have equal length.[3]
  3. For this value of \(k\), show that \(OA\) is perpendicular to \(OB\) by comparing the gradients of \(\mathbf a\) and \(\mathbf b\), and explain what kind of parallelogram \(OACB\) is.[2]

Solution:

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Answer:(a) \(|\mathbf a+\mathbf b|^2=(3+k)^2+36\ge36\); equality at \(k=-3\); the vertical component is always \(6\) (b) \(k=-\dfrac83\) (c) gradients \(\dfrac43\) and \(-\dfrac34\) multiply to \(-1\); a rectangle

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