QCAA Specialist Mathematics Vectors in two and three dimensions

This option is incorrect. Substituting $x = t+2$ into this equation yields $y = (t+2)^2 - 4 = t^2 + 4t$, which does not match the given component $y = t^2$.

This option is incorrect. Substituting $x = t+2$ into this equation yields $y = (t+2)^2 + 4 = t^2 + 4t + 8$, which does not match the given component $y = t^2$.

From the vector $\vec{r}$, we have $x = t+2$ and $y = t^2$. Solving for $t$ gives $t = x-2$. Substituting this into the $y$ equation yields $y = (x-2)^2 = x^2 - 4x + 4$.

This option corresponds to $y = (x+2)^2$. This implies an incorrect substitution where $t = x+2$ instead of the correct relationship $t = x-2$.

This option incorrectly uses the position vector of the point $\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$ as the normal vector. The dot product must be taken with the normal vector $\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$.

The vector equation of a plane is given by $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$, where $\mathbf{n}$ is the normal vector and $\mathbf{a}$ is a point on the plane. Substituting $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ and $\mathbf{a} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$ yields this result.

The equation of a plane is defined using the scalar (dot) product to establish orthogonality, not the vector (cross) product.

While this involves the correct vectors, the cross product is typically used for the equation of a line. The plane equation requires the condition $(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0$.