NJC Tutorial 1B Vectors II Q7

NJC Tutorial 1B Vectors II Q7

Tutorial

The planes \({p_1}\) and \({p_2}\) have equations \({\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 3 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}}\) and \({\mathbf{r} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = 0}\) respectively, and meet in a line \({l}\).

  1. Find a vector equation for \({l}\).
  2. The plane \({p_3}\) has equation \({y + 5z + 11 - 13x + k(y + 6z + 11 - 11x) = 0}\). Explain why \({l}\) lies in \({p_3}\) for any constant \({k}\). Hence, or otherwise, find a Cartesian equation of the plane in which both \({l}\) and the point \({(1,\ 2,\ 3)}\) lie.

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs a course plan.

Solution:

Solution locked

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

Answer:(i) \(\mathbf{r}=\left( \begin{matrix} 0 \\ -11 \\ 0 \\ \end{matrix} \right)+\mu \left( \begin{matrix} 1 \\ 23 \\ -2 \\ \end{matrix} \right),\mu \in \mathbb{R}\) (ii) \(-19x+y+2z=-11\)

Need help? Join our JC Math tuition classes.

Learn more