Im struggling to understand the marking scheme's answer to this particular question.

The points A, B and C have position vectors, relative to the origin O, given by

$\\\vec{OA} = \mathbf{i +2j +3k}$

$\vec{OB} = \mathbf{4j +k}$

$\vec{OC} = \mathbf{2i+5j -k}$

The plane $p$ is parallel to $OA$ and the line $BC$ lies in $p$. Find the equation of $p$, giving your answer in the form $ax + by + cz = d.$

The marking scheme gave 4 possible methods to solve the question.


  1. Obtain two equations $a+2b+3c=0$ or $2a+b-2c=0$ and solve for the ratio of $a$,$b$, and $c$ and substitute a relevant point in the plane equation. Or,

  2. Calculate the vector product of relevant vectors and substitute a relevant point in the plane equation.

    • Form a 2-parameter equation with relevant vectors

    • State a correct equation. e.g $\mathbf{r=2i+5j-k} + \lambda(\mathbf{i +2j+3k}) + \mu(\mathbf{2i +j -2k})$

    • State 3 equations in $x$, $y$, $z$, $\lambda$, and $\mu$

    • Eliminate $\lambda$ and $\mu$

    • Obtain answer $-7x+8y-3z=29$ or equivalent

‏‏‎ ‎

    • Using a relevant point and relevant direction vectors, form a determinant equation for the plane

    • State a correct equation, e.g $$\begin{bmatrix} x-2 & y-5 & z-1 \\ 1 & 2 & 3 \\ 2 & 1 & -2 \end{bmatrix}=0$$

    • Attempt to expand the determinant

    • Obtain correct values of two cofactors

    • Obtain answer $-7x+8y-3z=29$ or equivalent

‏‏‎ ‎

I understand the first two methods, but im struggling to understand the third and fourth method. i know that $\mathbf{r=2i+5j-k} + \lambda(\mathbf{i +2j +3k}) + \mu(\mathbf{2i +j -2k})$ is a vector equation but i dont know how its obtained and how its related to the plane.

It will be really helpful if the fourth method is explained as well, but I have no idea what are determinant equations and cofactors

the original question (number 9) and the marking scheme


1 Answer 1


For the third method, we can think of a plane by starting at an initial point.$$\mathbf{r}=2\mathbf{i}+5\mathbf{j}−\mathbf{k}+\lambda(\mathbf{i}+2\mathbf{j}+3\mathbf{k})+\mu(2\mathbf{i}+\mathbf{j}−2\mathbf{k}),$$ Imagine you're a bug, starting at the point $(2,5,-1)$, which is on the plane. You can first travel as far as you want in the direction of $\langle 1,2,3\rangle$ or opposite this direction, which produces a line. Then you can also travel as much as you want in the direction of $\langle 2,1,-2\rangle$. Now you have a plane.

If you take the parametric equations that this equation gives, you'd get $$x=2+\lambda+2\mu,\quad y=5+2\lambda+\mu,\text{ and } z=-1+3\lambda-2\mu,$$ and you can eliminate $\lambda$ and $\mu$ to get a single equation involving just $x$, $y$, and $z$.

For the fourth method, the determinant of a matrix is zero precisely when one vector can be written as a combination of the others. So this is just giving a slick way to find when $\langle x-2,y-5,z-1\rangle$ is in the span of $\langle 1,2,3\rangle$ and $\langle 2,1,-2\rangle$.


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