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The best place to find a helping hand is at the end of your own arm, Swedish Proverb.
Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.
Definition. The magnitude of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $\vert{}\vert{}\langle 3, 2, 1 \rangle\vert{}\vert{} = \sqrt{3^2+2^2+1^2} = \sqrt{14}$, $\vert{}\vert{}\langle 3, -4, 5 \rangle\vert{}\vert{} = \sqrt{3^2+(-4)^2+5^2} = \sqrt{50} = 5\sqrt{2}$, or $\vert{}\vert{}\langle 1, 0, 0 \rangle\vert{}\vert{} = \sqrt{1^2+0^2+0^2} = \sqrt{1} = 1$.
Vector Addition: The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k} = \langle a_1+b_1, a_2+b_2, a_3+b_3 \rangle$.
Vector Subtraction: The subtraction of two vectors is similar to addition and is also done component-wise, it is given by simply subtracting their corresponding components (x, y, and z in 3D), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k} = \langle a_1-b_1, a_2-b_2, a_3-b_3 \rangle$.
Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction (unless the scalar is negative, which reverses the direction). It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = \langle ca_1, ca_2, ca_3 \rangle$.
The dot or scalar product is a fundamental operation between two vectors. It produces a scalar quantity that represents the projection of one vector onto another. The dot product is the sum of the products of their corresponding components: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3.$, e.g. $\vec{A}·\vec{B} = \sum a_ib_i = \langle 2, 2, -1 \rangle \cdot \langle 5, -3, 2 \rangle = (2 \cdot 5) + (2 \cdot -3) + (-1 \cdot 2) = 10 - 6 - 2 = 2.$
Geometrically, it is the product of their magnitudes multiplied by the cosine of the angle $\theta$ between them, $\boxed{\vec{A} \cdot \vec{B} = \vert{}\vert{}\vec{A}\vert{}\vert{} \cdot \vert{}\vert{}\vec{B}\vert{}\vert{} \cos(\theta)}$
The projection of $\vec{a}$ onto $\vec{b}$ is $proj_{\vec{b}}\vec{a} = \frac{\vec{a}·\vec{b}}{||\vec{b}||^2}\vec{b}$.
The cross product, denoted by $\vec{A} \times \vec{B}$, is a binary operation on two vectors in three-dimensional space. It yields a new vector that is perpendicular (orthogonal, at a right angle) to both original vectors and has a magnitude equal to the area of the parallelogram formed by the two input vectors.
The cross product $\vec{A}x\vec{B}$ can be computed using the determinant of a $3 \times 3$ matrix formed by the standard basis vectors ($\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$) and the components of the two vectors: $\vec{A} \times \vec{B} = det(\begin{smallmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\ a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\\ b_1 & b_2\end{smallmatrix}|\vec{k}$.
Example: If $\vec{A} = \langle 1, 2, 3 \rangle$ and $\vec{B} = \langle 4, 5, 6 \rangle$: $\vec{A} \times \vec{B} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} = (2\cdot 6 - 3\cdot 5)\hat{\mathbf{i}} - (1\cdot 6 - 3\cdot 4)\hat{\mathbf{j}} + (1\cdot 5 - 2\cdot 4)\hat{\mathbf{k}} = -3\hat{\mathbf{i}} + 6\hat{\mathbf{j}} - 3\hat{\mathbf{k}} = \langle -3, 6, -3 \rangle$
Solution:
Finding the Velocity Function.
To find the velocity function $\vec{v}(t)$, we need to integrate the acceleration function $\vec{a}(t)$ with respect to time:
$\vec{v}(t) = \int \vec{a}(t)dt =\int \left( \vec{i}+2\vec{j}+6t\vec{k} \right)dt = t\vec{i}+2t\vec{j}+3t^2\vec{k} + C,$ where C is the constant of integration.
We determine C using the initial velocity, $\vec{v}(0)=\vec{j}-\vec{k}$: $\vec{v}(0) = 0 \cdot \vec{i}+2\cdot \vec{j}+3 \cdot 0^2\vec{k} + C$. Thus, $C = \vec{j}-\vec{k}$
Combining the components gives the exact velocity function: $\vec{v}(t) = t\vec{i}+2t\vec{j}+3t^2\vec{k} + \vec{j}-\vec{k} = t\vec{i}+(2t+1)\vec{j}+(3t^2-1)\vec{k}$
Finding the Position Function.
Analogously, we can calculate the position function by integrating the velocity function (i.e., position $\vec{r}(t)$ is the indefinite integral of the velocity vector $\vec{v}(t)$ with respect to time):
$\vec{r}(t) = \int \vec{v}(t)dt = \int \left(t\vec{i}+(2t+1)\vec{j}+(3t^2-1)\vec{k} \right) dt = \frac{t^2}{2}\vec{i}+(t^2+t)\vec{j} + (t^3-t)\vec{k} + C,$ where C is the constant of integration.
Using the initial condition $\vec{r}(0)=\vec{i}-2\vec{j}+3\vec{k}: \vec{r}(0) = \frac{0^2}{2}\vec{i}+(0^2+0)\vec{j} + (0^3-0)\vec{k} + C = C$. Thus, C = $\vec{i}-2\vec{j}+3\vec{k}$.
Grouping like component terms yields the final position vector function: $\vec{r}(t) = \left( \frac{t^2}{2} + 1 \right)\vec{i} + (t^2 + t - 2)\vec{j} + (t^3 - t + 3)\vec{k}$
Conclusion. This two-step vector integration process illustrates how to integrate acceleration to find velocity, and then integrate velocity to find position, using the problem’s given initial conditions to determine the constants of integration.
Compute Velocity.
The velocity vector is obtained by integrating the acceleration vector with respect to time: $\vec{v}(t) = \int (3\hat{\mathbf{i}} + 3\hat{\mathbf{k}}) dt = 3t\hat{\mathbf{i}} + 3t\hat{\mathbf{k}} + C$ where $C$ is the constant of integration.
To determine $\vec{C}$, use the initial condition for the velocity: $\vec{v}(0) = 8\hat{\mathbf{j}}$. This yields $C = 8\hat{\mathbf{j}}$.
The velocity function is: $\vec{v}(t) = 3t\hat{\mathbf{i}} + 8\hat{\mathbf{j}} + 3t\hat{\mathbf{k}}.$
Compute Position.
The position vector is obtained by integrating the velocity vector with respect to time t.
$\vec{r}(t) = \int (3t\hat{\mathbf{i}} + 8\hat{\mathbf{j}} + 3t\hat{\mathbf{k}}) dt = \frac{3t^2}{2}\hat{\mathbf{i}} + 8t\hat{\mathbf{j}} + \frac{3t^2}{2}\hat{\mathbf{k}} + C$ where $C$ is the constant of integration.
To determine C, use the initial condition for the position: $\vec{r}(0) = \vec{0} \implies C = 0$.
The final position function is: $\vec{r}(t) = \frac{3t^2}{2}\hat{\mathbf{i}} + 8t\hat{\mathbf{j}} + \frac{3t^2}{2}\hat{\mathbf{k}}.$
Proof that $\vec{r} \cdot \vec{v} = 0$ (i)
Since the position vector $\vec{r}(t)$, which describes the position of a point in space as a function of time t, is said to have a constant length or magnitude over time, the dot product of $\vec{r}(t)$ with itself is also constant: $\vec{r} \cdot \vec{r}$ remains constant: $\vec{r} \cdot \vec{r} = \vert{}\vec{r}(t)\vert{}^2 = R^2 \quad (\text{constant})$
Differentiating both sides with respect to t using the vector product rule: $\frac{d}{dt}(\vec{r} \cdot \vec{r}) = \frac{d}{dt}(\vec{r})\cdot \vec{r} + \vec{r} \cdot \frac{d}{dt}(\vec{r}) = \vec{v}\cdot \vec{r}+\vec{r}\cdot\vec{v}$.
Since the dot product is commutative ($\vec{v}\cdot \vec{r} = \vec{r}\cdot\vec{v}$): $2(\vec{r}\cdot\vec{v}) = \frac{d}{dt}(\vec{r} \cdot \vec{r}) = \frac{d}{dt}(R^2) = 0 \implies \vec{r}\cdot\vec{v} = 0$.
The position vector $\vec{r}$ and velocity vector $\vec{v}$ are orthogonal at all times.
To demonstrate that $\vec{r} \times \vec{v}$ is constant, we will prove that its time derivative is zero $\frac{d}{dt}(\vec{r} \times \vec{v}) = 0$
Using the product rule for cross products: $\frac{d}{dt}(\vec{r} \times \vec{v}) = \frac{d}{dt}(\vec{r}) \times \vec{v} + \vec{r}\times \frac{d}{dt}(\vec{v}) = \vec{v} \times \vec{v} + \vec{r} \times \vec{a}$
The reader should notice that:
(1 $\star$) The cross product of any vector with itself is zero: $\forall \vec{v}, \vec{v} \times \vec{v}=0$ (geometrically, the cross product of two vectors is the area of the parallelogram formed by both vectors, but as we are talking about the same vector, there is no area).
(2 $\star$) Our acceleration is purely radial. Substituting $\vec{a}(t) = c\vec{r}(t)$, we get: $\vec{r} \times \vec{a} = \vec{r}\times c\vec{r}(t) = c(\vec{r}\times \vec{r}) = c\vec{0} = \vec{0}$.
Therefore: $\frac{d}{dt}(\vec{r} \times \vec{v}) = \vec{0} + \vec{0} = \vec{0}$.
The cross product $\vec{r} \times \vec{v}$ remains constant throughout the motion.
An example of a position vector that satisfies these conditions is: $\vec{r}(t) = \langle \cos(t), \sin(t), 0 \rangle$ (Figure A).
This is the standard uniform circular motion in the xy-plane.
Let’s test all required conditions:
Constant Length:
$\vert{}\vec{r}(t)\vert{} = \sqrt{\cos^2(t) + \sin^2(t) + 0} = 1 \quad (\text{constant})$
Acceleration Requirement ($\vec{a} = c\vec{r}$):
$\vec{v}(t) = \frac{d\vec{r}}{dt} = \langle -\sin(t), \cos(t), 0 \rangle$
$\vec{a}(t) = \frac{d\vec{v}}{dt} = \langle -\cos(t), -\sin(t), 0 \rangle = -1 \cdot \vec{r}(t)$
In this particular example: $c = -1 \neq 0$, satisfying $\vec{a}(t) = c\vec{r}(t)$.
Orthogonality Check ($\vec{r} \cdot \vec{v} = 0$):
$\vec{r} \cdot \vec{v} = \cos(t)(-\sin(t)) + \sin(t)\cos(t) + 0 = 0$
Constant Cross Product Check ($\vec{r} \times \vec{v}$):
$\vec{r} \times \vec{v} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \cos(t) & \sin(t) & 0 \\ -\sin(t) & \cos(t) & 0 \end{vmatrix} = \left(\cos^2(t) + \sin^2(t)\right)\hat{\mathbf{k}} = \hat{\mathbf{k}}$
Since $\hat{\mathbf{k}} = \langle 0, 0, 1 \rangle$ is constant in both magnitude and direction, this confirms the proof.
Therefore, the example $\vec{r}(t) = \langle \cos(t), \sin(t), 0 \rangle$ satisfies both conditions $\vec{r} \cdot \vec{v} = 0$ and $\vec{r} \times \vec{v}$ is constant.
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