2021 RI P2 Q4

2021 RI P2 Q4

10 marks
Prelims
    1. Referred to the origin \(O\), points \(A\), \(B\) and \(C\) have position vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) respectively. The three points lie on a circle with centre \(O\) and diameter \(AB\) (see diagram).
      Using a suitable scalar product, show that the angle \(ACB\) is \({{90}^{\circ }}\).[4]
    2. The variable vector \(\mathbf{r}\) satisfies the equation \(\left( \mathbf{r}-\mathbf{i} \right)\cdot \left( \mathbf{r}-\mathbf{k} \right)=0\). Describe the set of vectors \(\mathbf{r}\) geometrically.[2]
    1. The plane \(\pi \) passes through the points with position vectors \(x\mathbf{i}\), \(y\mathbf{j}\) and \(z\mathbf{k}\) where \(x\), \(y\), and \(z\) are non-zero constants. It is given that \(d\) is the perpendicular distance from the origin to \(\pi \). Show, by finding the normal of \(\pi \), or otherwise, that \(\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=\frac{1}{{{d}^{2}}}\).[4]

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs tuition enrolment.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.

Create a free account to practise similar questions.

Answer:(a)(i) Angle ACB = \(90°\) (a)(ii) Variable factors \(\mathbf{r}\) are points on the plane with normal direction of \(\left( 1,0,1 \right)\) and shortest distance to origin of \(\frac{{{\left| \mathbf{r} \right|}^{2}}}{\sqrt{2}}\).

Need help? Join our JC Math tuition classes.

Learn more