2026 RI Promo Q12

2026 RI Promo Q12

Junior College 1
8 marks

Relative to the origin \(O\), the points \(P\), \(U\) and \(V\) have position vectors given by \(\mathbf p\), \(\mathbf u\) and \(\mathbf v\) respectively. Vectors \(\mathbf u\) and \(\mathbf v\) are perpendicular unit vectors. The points \(O\), \(P\), \(U\) and \(V\) form a tetrahedron with four triangular faces.

The position vector \(\mathbf q\) of a general point on the plane \(\pi\) is defined by \(\mathbf q=\mathbf p+\lambda\mathbf u+\mu\mathbf v\) where \(\lambda\) and \(\mu\) are real parameters.

  1. Show by using definition of the scalar product that \(\mathbf q\cdot\mathbf q=|\mathbf q|^2\).[1]
  2. By completing the squares in \(\lambda\) and \(\mu\), show
    \[|\mathbf q|^2=(\lambda+A)^2+(\mu+B)^2+C-A^2-B^2,\]
    where real numbers \(A\), \(B\) and \(C\) are to be determined in terms of \(\mathbf p\), \(\mathbf u\) or \(\mathbf v\).
    Hence state the values of \(\lambda\) and \(\mu\) in terms of \(\mathbf p\), \(\mathbf u\) or \(\mathbf v\), corresponding to the minimum value of \(|\mathbf q|^2\).[4]

It is given that the origin \(O\) does not lie on \(\pi\). Point \(M\) lies on \(\pi\) and has position vector \(\mathbf m=\mathbf p+\lambda_1\mathbf u+\mu_1\mathbf v\) where \(\lambda_1\) and \(\mu_1\) are the values of \(\lambda\) and \(\mu\) found in part (b) respectively.

  1. State the geometrical interpretation of length \(OM\) in relation to \(\pi\) and explain why the volume of the tetrahedron \(OPUV\) is given by \(\dfrac16|\mathbf m|\).[3]

[The volume of a tetrahedron is \(\dfrac13\times\text{base area}\times\text{height}\).]

Solution:

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Answer:(a) \(\mathbf q\cdot\mathbf q=|\mathbf q|^2\). (b) \(A=\mathbf p\cdot\mathbf u,\ B=\mathbf p\cdot\mathbf v,\ C=|\mathbf p|^2\); \(\lambda=-\mathbf p\cdot\mathbf u,\ \mu=-\mathbf p\cdot\mathbf v\). (c) \(OM\) is the perpendicular distance to \(\pi\), volume \(=\dfrac16|\mathbf m|\).

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