AHL 3.17—Vector equations of a plane

Use $P$ as the fixed point: $\mathbf{a}=\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)$. The direction vectors should be $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.

$\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=\left(\begin{array}{c}3-2\\ 1-0\\ 2-1\end{array}\right)=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$

$\overrightarrow{PR}=\mathbf{r}-\mathbf{p}=\left(\begin{array}{c}1-2\\ 2-0\\ 0-1\end{array}\right)=\left(\begin{array}{c}-1\\ 2\\ -1\end{array}\right)$

Is $\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)=k\left(\begin{array}{c}-1\\ 2\\ -1\end{array}\right)$? This requires $k=-1$ from the first component but $k=1/2$ from the second — impossible. The vectors are non-parallel. ✓

The parametric equation is $\mathbf{r}=\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)+\lambda \left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)+\mu \left(\begin{array}{c}-1\\ 2\\ -1\end{array}\right)$, which matches option A.

A valid equation uses a point on the plane and two non-parallel direction vectors that lie in the plane. The direction vectors must be differences of the given position vectors.

Option B uses direction vectors $\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$ and $\left(\begin{array}{c}2\\ 2\\ 2\end{array}\right)$. Since $\left(\begin{array}{c}2\\ 2\\ 2\end{array}\right)=2\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$, these are parallel — they only define a line, not a plane. Eliminated.

Option C uses $\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$ as the fixed point, which is not any of $P$, $Q$, or $R$. While a plane through $P$, $Q$, $R$ could be written starting from a different point, the direction vectors $\mathbf{p}$ and $\mathbf{r}$ (position vectors, not difference vectors) are not guaranteed to lie in the plane correctly. This form is structurally incorrect.

Option D uses $\mathbf{q}=\left(\begin{array}{c}3\\ 1\\ 2\end{array}\right)$ and $\mathbf{r}=\left(\begin{array}{c}1\\ 2\\ 0\end{array}\right)$ as direction vectors — these are position vectors of the points, not displacement vectors between them. Using position vectors as direction vectors is incorrect.

Option A uses $P$ as the fixed point, $\overrightarrow{PQ}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$ and $\overrightarrow{PR}=\left(\begin{array}{c}-1\\ 2\\ -1\end{array}\right)$ as non-parallel direction vectors — this is the correct construction.

The Cartesian equation $ax+by+cz=d$ has its coefficients $(a,b,c)$ directly equal to the components of the normal vector $\mathbf{n}$.

From $3x-y+2z=12$, we have $a=3$, $b=-1$, $c=2$, so $\mathbf{n}=\left(\begin{array}{c}3\\ -1\\ 2\end{array}\right)$.

The normal vector is not unique — any scalar multiple such as $\left(\begin{array}{c}-3\\ 1\\ -2\end{array}\right)$ or $\left(\begin{array}{c}6\\ -2\\ 4\end{array}\right)$ also qualifies. However, the most natural/standard choice is to read the coefficients directly.

Option A, $\left(\begin{array}{c}3\\ -1\\ 2\end{array}\right)$, directly matches the coefficients of $x$, $y$, $z$ in the equation.

We need the normal vector, which is perpendicular to the plane. Its components equal the coefficients of $x$, $y$, $z$ in the Cartesian equation.

The vector $\left(\begin{array}{c}12\\ 0\\ 0\end{array}\right)$ corresponds to $12x=12$, i.e. $x=1$, which is a different plane entirely. The value $d=12$ is the constant, not part of the normal vector.

While $\left(\begin{array}{c}-3\\ 1\\ -2\end{array}\right)$ is a valid normal vector (it is $-1$ times the standard one), the word 'only' makes this incorrect — the normal is not unique, and $\left(\begin{array}{c}3\\ -1\\ 2\end{array}\right)$ is equally valid.

The vector $\left(\begin{array}{c}4\\ 0\\ 0\end{array}\right)$ has no relation to the coefficients $(3,-1,2)$. It would be the normal to the plane $x=$ const, not our plane.

Option A, $\left(\begin{array}{c}3\\ -1\\ 2\end{array}\right)$, correctly reads the coefficients of $x$, $y$, $z$ from $3x-y+2z=12$.