2025 VJC P2 Q3

2025 VJC P2 Q3

13 marks

With reference to the point \(O\) as the origin and the \(x\)-\(y\) plane as a horizontal plane, the pyramid \(OPQRV\) has a parallelogram base \(OPQR\) and height \(OV\). The position vectors of the points \(P\) and \(R\) are \(-3\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\) and \(5\mathbf{i} - 2\mathbf{j} - \mathbf{k}\) respectively.

  1. Find the coordinates of the point \(S\) that lies on the line \(PR\) such that the distance from \(O\) to \(S\) is a minimum.[3]
  2. Find the cartesian equations of the planes such that the perpendicular distance from each plane to the base \(OPQR\) is \(2\sqrt{86}\) units.[3]
  3. Find the acute angle between \(OV\) and the vertical.[2]
  4. Given that \(\overrightarrow{QV}\) is parallel to the vector \(-5\mathbf{j} + 8\mathbf{k}\), find the position vector of the point \(V\). Hence find the exact volume of the pyramid \(OPQRV\).[5]

\(\text{[Volume of a pyramid} = \frac{1}{3} \times \text{base area} \times \text{height]}\)

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Answer:(a) \(S\left(\dfrac{33}{29},\dfrac{26}{29},\dfrac{27}{29}\right)\) (b) \(x+6y-7z=\pm172\) (c) \(41.0^\circ\) (d) \(\overrightarrow{OV}=2\mathbf i+12\mathbf j-14\mathbf k\); volume \(\dfrac{344}{3}\text{ units}^3\)

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