IB Mathematics Applications & Interpretation Question Type 6: Finding the Area of a Parallelogram or Triangle Spanned by Two Vectors Exercises

Find the area of the triangle with vertices at the origin, $2\mathbf{u}$ and $3\mathbf{v}$, where $\mathbf{u}=\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$ and $\mathbf{v}=\left(\begin{array}{c}1\\ 2\\ 1\end{array}\right)$.

Using the scalar product, find the area of the parallelogram spanned by the vectors $\mathbf{u}=\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$ and $\mathbf{v}=\left(\begin{array}{c}1\\ 2\\ 1\end{array}\right)$ without computing the cross product.

Compute the area of the parallelogram defined by the vectors $3\mathbf{u}$ and $2\mathbf{v}$, where $\mathbf{u}=(2,1,2)$ and $\mathbf{v}=(1,2,1)$.

The points $A,B$ and $C$ have position vectors $\mathbf{u}=\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$, $\mathbf{v}=\left(\begin{array}{c}1\\ 2\\ 1\end{array}\right)$ and $\mathbf{u}+\mathbf{v}$ respectively.

Find the area of triangle $ABC$.

Find the area of the parallelogram spanned by $\mathbf{u}+\mathbf{v}$ and $\mathbf{u}-\mathbf{v}$, where $\mathbf{u}=(2,1,2)$ and $\mathbf{v}=(1,2,1)$.

Find a unit normal vector to the plane of the parallelogram spanned by $\mathbf{u}=(2,1,2)$ and $\mathbf{v}=(1,2,1)$.

Determine the angle $\theta$ between the vectors $\mathbf{u}=\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$ and $\mathbf{v}=\left(\begin{array}{c}1\\ 2\\ 1\end{array}\right)$.

Determine the area of the quadrilateral with vertices $O$, $\mathbf{u}$, $\mathbf{u}+2\mathbf{v}$ and $2\mathbf{v}$ in that order, where $\mathbf{u}=(2,1,2)$ and $\mathbf{v}=(1,2,1)$.

Calculate the area of the parallelogram defined by the vectors $\mathbf{u}=(2,1,2)$ and $\mathbf{v}=(1,2,1)$.

Find the height of the parallelogram spanned by $\mathbf{u}=(2,1,2)$ and $\mathbf{v}=(1,2,1)$, measured perpendicular to the base vector $\mathbf{u}$.

Find the area of the triangle with vertices at the origin, the point with position vector $\mathbf{u}=(2,1,2)$ and the point with position vector $\mathbf{v}=(1,2,1)$.