IB Mathematics Applications & Interpretation Question Type 6: Finding the Distance Between Two Position Vectors Exercises

Given the vectors $\mathbf{a}=\left(\begin{array}{c}1\\ 1\\ 3\end{array}\right)$, $\mathbf{b}=\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)$, $\mathbf{c}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$ and $\mathbf{d}=\left(\begin{array}{c}0\\ 2\\ 2\end{array}\right)$, calculate the resultant vector $\mathbf{R}=\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}$ without drawing individual tip-to-tail diagrams.

Let $\mathbf{R}$ be the resultant vector defined by $\mathbf{R}=\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}$.

Find the magnitude of $\mathbf{R}$ and compare this value with the sum of the magnitudes of the individual vectors: $\Vert \mathbf{a}\Vert +\Vert \mathbf{b}\Vert +\Vert \mathbf{c}\Vert +\Vert \mathbf{d}\Vert$.

Three vectors in the $xy$-plane are $\mathbf{p}=\left(\begin{array}{c}4\\ 0\\ 0\end{array}\right)$, $\mathbf{q}=\left(\begin{array}{c}0\\ 3\\ 0\end{array}\right)$ and $\mathbf{r}=\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)$. Without drawing separate tip-to-tail diagrams, find the resultant and its magnitude.

Given $\mathbf{p}=(3,0,4)$, $\mathbf{q}=(0,5,-3)$ and $\mathbf{r}=(-2,1,1)$, find the unit vector in the direction of $\mathbf{p}+\mathbf{q}+\mathbf{r}$.

Let $\mathbf{a}=\left(\begin{array}{c}1\\ 1\\ 3\end{array}\right)$, $\mathbf{b}=\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)$, $\mathbf{c}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$ and $\mathbf{d}=\left(\begin{array}{c}0\\ 2\\ 2\end{array}\right)$. Determine the angle $\theta$ between $\mathbf{R}=\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}$ and $\mathbf{a}$.

For the vectors $\mathbf{u}=(2,-1,0)$, $\mathbf{v}=(-1,2,3)$ and $\mathbf{w}=(0,1,-2)$, find $\mathbf{S}=\mathbf{u}+\mathbf{v}+\mathbf{w}$ and compute its magnitude $\Vert \mathbf{S}\Vert .$

Given vectors $\mathbf{u}=2\mathbf{i}-\mathbf{j}+\mathbf{k}$, $\mathbf{v}=-\mathbf{i}+3\mathbf{j}+2\mathbf{k}$, $\mathbf{w}=\mathbf{i}+\mathbf{j}-\mathbf{k}$, find the missing vector $\mathbf{x}$ such that $\mathbf{u}+\mathbf{v}+\mathbf{w}+\mathbf{x}=0$.

Compute the scalar projection.

In the $xy$-plane, vectors $\mathbf{p}=\left(\begin{array}{c}5\\ 2\\ 0\end{array}\right)$, $\mathbf{q}=\left(\begin{array}{c}1\\ -3\\ 0\end{array}\right)$, $\mathbf{r}=\left(\begin{array}{c}-4\\ 1\\ 0\end{array}\right)$ and $\mathbf{s}=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$ are given. Use a single tip-to-tail chain to find the resultant and verify whether it equals the zero vector.

Express the vector sum $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}$ in unit vector notation ($\mathbf{i},\mathbf{j},\mathbf{k}$ form).

Vectors $\mathbf{u},\mathbf{v},\mathbf{w},\mathbf{x}$ satisfy $\mathbf{u}+\mathbf{v}+\mathbf{w}=\mathbf{d}$ and $\mathbf{x}=\mathbf{d}-(\mathbf{u}+\mathbf{v}+\mathbf{w})$. If $\mathbf{u}=(1,2,3)$, $\mathbf{v}=(0,1,-1)$ and $\mathbf{d}=(3,3,4)$, find $\mathbf{w}$ and $\mathbf{x}$.