IB Mathematics Analysis and Approaches Topic AHL…

Let $\mathbf{a}=(2,-1,3),\qquad \mathbf{b}=(k,4,-2),\qquad \mathbf{c}=(1,2,1),$ where $k\in\mathbb{R}$ . Define the scalar triple product by $[\mathbf{a},\mathbf{b},\mathbf{c}]=\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$ .

Find the value of $k$ for which $\mathbf{a}$ and $\mathbf{b}$ are perpendicular.

For this value of $k$ , find the exact angle between and .

The diagram shows a parallelogram $ABCD$ with $A(1,2,3)$ , and .

Find the vector from $A$ to the point dividing $\overrightarrow{AB}$ in the ratio (start at ) for and .

In the study of kinematics, the position of a moving particle can be described by a position vector that depends on time. If a particle moves through three-dimensional space with constant velocity, its position at time $t$ seconds is given by a vector equation of the form

\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}

Line $L_1$ : $r = (\begin{array}{c} 1 \\ 2 \\ 3 \end{array}) + λ (\begin{array}{c} \end{array}$ and plane : .

For $k=5$ , show that the three vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ are not coplanar.

Find a unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ (for $k=5$ ). Give your answer in exact surd form.

Prove that the angle between $\mathbf{a}\times\mathbf{b}$ and $\mathbf{c}$ satisfies
$\cos\phi=\frac{[\mathbf{a},\mathbf{b},\mathbf{c}]}{|\mathbf{a}\times\mathbf{b}|\,|\mathbf{c}|}.$

Find $\overrightarrow{AB}$ .

Find $\overrightarrow{AD}$ .

Hence find $\overrightarrow{AC}$ .

Find the coordinates of point $C$ .

Find the position of each particle at $t = 1$ and at $t = 2$ .

Find the distance between $P$ and $Q$ at $t = 1$ and at $t = 2$ .

Show that the displacement vector from $P$ to $Q$ at time $t$ is given by

$d (t) = r_{Q} (t) - r_{P} (t) = (\begin{array}{c} - 2 + 3 t \\ 4 - t \end{array}$

The distance between $P$ and $Q$ at time $t$ is denoted $D(t) = |\mathbf{d}(t)|$ .

Since $D(t) \geq 0$ for all $t$ , the value of $t$ that minimises $D(t)$ is the same as the value of that minimises . This is a standard technique in optimization: working with the squared distance avoids the complication of differentiating a square root.

Hence find the value of $t$ at which $P$ and $Q$ are closest together, and this minimum distance.

The velocity vectors of $P$ and $Q$ are $\mathbf{v}_P = \begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix}$ and respectively. The relative velocity of with respect to is defined as .

Find the displacement vector $\mathbf{d}(1)$ at the moment of closest approach.

Hence show that $\mathbf{v}_{rel}$ is perpendicular to $\mathbf{d}(1)$ .

Consider now two particles with general position vectors

$\mathbf{r}_1(t) = \mathbf{a} + t\mathbf{u}, \qquad \mathbf{r}_2(t) = \mathbf{b} + t\mathbf{w}$

Hence show that at the time $t^*$ when the particles are closest, the relative velocity $\mathbf{q}$ is perpendicular to the displacement vector $\mathbf{d}(t^*)$ .

geometrically why the displacement vector between two particles must be perpendicular to their relative velocity at the instant of closest approach.

Find the point of intersection of the line $L_1$ and the plane $\pi$ .

if the line $L_1$ is parallel, perpendicular, or neither to the normal of the plane $\pi$ .

IB Mathematics Analysis and Approaches AHL 3.12—vector Definitions Questionbank

IB Mathematics Analysis and Approaches AHL 3.12—vector Definitions Questionbank

IB Mathematics Analysis and Approaches AHL 3.12—vector Definitions Questionbank

IB Mathematics Analysis and Approaches AHL 3.12—vector Definitions Questionbank

IB Mathematics Analysis and Approaches (AA) Topic AHL 3.12—vector Definitions Questionbank