N1993 P1 Q14

N1993 P1 Q14

12 marks
Past Year National Exams
Free

The equation of a straight line \(l\) is \(\mathbf{r}=\left( \begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} \right)+t\left( \begin{matrix} -1 \\ 1 \\ 1 \\ \end{matrix} \right)\), where \(t\) is a parameter. The point \(A\)on \(l\) is given by \(t=0\), and the origin of position vectors is \(O\).

  1. Calculate the acute angle between \(OA\) and \(\ell \), giving your answer correct to the nearest degree.[3]
  2. Find the position vector of the point \(P\) on \(\ell \) such that \(OP\) is perpendicular to \(\ell \).[3]
  3. A point \(Q\) on \(\ell \) is such that the length of \(OQ\) is \(5\) units. Find the two possible position vectors of \(Q\).[3]
  4. The points \(R\) and \(S\) on \(\ell \) are given by \(t\,=\,\lambda \) and \(t\,=\,2\lambda \) respectively. Show that there is no value of \(\lambda \) for which \(OR\) and \(OS\) are perpendicular.[3]

Video Solution:

Default solution

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Answer:(i) \(52^\circ\) (ii) \(\overrightarrow{OP}=\dfrac13(7,2,5)\) (iii) \(\overrightarrow{OQ}=(0,3,4)\) or \(\dfrac13(14,-5,-2)\) (iv) \(\overrightarrow{OR}\cdot\overrightarrow{OS}=6(\lambda+1)^2+8>0\), so no real \(\lambda\)

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