Mathematics I · Unit 9 · 4 hrs
Plane and Space Vectors
Exam-focused notes for Plane and Space Vectors (Mathematics I, MTH117): what the TU syllabus asks and how it has actually been tested, with 11 solved past questions from this unit.
What this unit covers
- Introduction
- Applications
- Dot product and cross Product
- Equations of lines and Planes
- Derivative and integrals of vector functions
- Arc length and curvature
- Normal and binormal vectors
- Motion in space
Normal and binormal vectors
Find the unit normal and binormal vectors for the circular helix $\mathbf{r}(t) = \cos t \hat{i} + \sin t \hat{j} + t \hat{k}$. [5]
$$\mathbf{r}(t) = \cos t\,\hat{i} + \sin t\,\hat{j} + t\,\hat{k}$$ All data present and readable. --- $$\mathbf{r}'(t) = -\sin t\,\hat{i} + \cos t\,\hat{j} + \hat{k}$$ $$\mathbf{r}'(t) = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}$$ $$\mathbf{T}(t) = \frac{1}...
Full solved answer →Motion in space
The position vector of an object moving in a plane is given by $\mathbf{r}(t) = t^2 \hat{i} + t^2 \hat{j}$. Find its velocity, speed, and acceleration when $t = 1$ and illustrate geometrically. [5]
$$\vec{r}(t) = t^2\hat{i} + t^2\hat{j}, \qquad t = 1$$ --- $$\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(t^2)\hat{i} + \frac{d}{dt}(t^2)\hat{j} = 2t\hat{i} + 2t\hat{j}$$ At $t = 1$: $$\vec{v}(1) = 2\hat{i} + 2\hat{j}$$ --- $$\vec{v}(t) = \sqrt{(2t)^2 + ...
Full solved answer →Introduction
If $\vec{a}=(4,0,3)$ and $\vec{b}=(-2,1,5)$, find $|\vec{a}|$, $3\vec{b}$, $\vec{a}+\vec{b}$ and $2\vec{a}+5\vec{b}$.
Estimate the value of $\lim_{x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2}$ [5+5]
Given data: - $\vec{a} = (4, 0, 3)$ - $\vec{b} = (-2, 1, 5)$ - Limit: $\displaystyle \lim{x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2}$ Required: $\vec{a}$, $3\vec{b}$, $\vec{a}+\vec{b}$, $2\vec{a}+5\vec{b}$, and the limit value. All data present. --- (a) Vector...
Full solved answer →Derivative and integrals of vector functions
Find the derivative of $\mathbf{r}(t) = t^2\mathbf{i} - te^{-t}\mathbf{j} + \sin(2t)\mathbf{k}$ and find the unit tangent vector at $t = 0$. [5]
Vector function: $$\mathbf{r}(t) = t^2\,\mathbf{i} - te^{-t}\,\mathbf{j} + \sin(2t)\,\mathbf{k}$$ Point of evaluation: $t = 0$ Required: (a) $\mathbf{r}'(t)$, (b) unit tangent vector $\mathbf{T}(0)$. All data present. --- i-component: $\dfrac{d}{dt}(t^2) = ...
Full solved answer →Find the derivative of $r(t) = (1 + t^2)\hat{i} - te^{-t}\hat{j} + \sin 2t\hat{k}$ and find the unit tangent vector at $t=0$. [5]
$$\mathbf{r}(t) = (1+t^2)\,\hat{i} - te^{-t}\,\hat{j} + \sin 2t\,\hat{k}$$ Evaluation point: $t = 0$. --- i-component: $\dfrac{d}{dt}(1+t^2) = 2t$ j-component: $\dfrac{d}{dt}(-te^{-t})$ Product rule: $\dfrac{d}{dt}(te^{-t}) = e^{-t} + t(-e^{-t}) = e^{-t}(1-...
Full solved answer →Find the derivatives of $r(t) = (1 + t^2)\hat{i} - te^t\hat{j} + \sin 2t\hat{k}$ and find the unit tangent vector at $t=0$. [5]
$$\mathbf{r}(t) = (1+t^2)\,\hat{i} - te^t\,\hat{j} + \sin 2t\,\hat{k}$$ Evaluation point: $t = 0$. All required data present. --- Differentiate each component: $\hat{i}$ component: $$\frac{d}{dt}(1+t^2) = 2t$$ $\hat{j}$ component (product rule on $te^t$): $...
Full solved answer →Dot product and cross Product
Find the angle between the vectors $a = (2, 2, -1)$ and $b = (1, 3, 2)$. [5]
$$\mathbf{a} = (2, 2, -1), \qquad \mathbf{b} = (1, 3, 2)$$ Formula for the angle $\theta$: $$\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a}\,\mathbf{b}}$$ --- $$\mathbf{a} \cdot \mathbf{b} = (2)(1) + (2)(3) + (-1)(2) = 2 + 6 - 2 = 6$$ $$\mathbf{...
Full solved answer →Find a vector perpendicular to the plane that passes through the points: $P(1, 4, 6)$, $Q(-2, 5, -1)$ and $R(1, -1, 1)$. [5]
- $P(1, 4, 6)$ - $Q(-2, 5, -1)$ - $R(1, -1, 1)$ A normal vector to the plane is the cross product of two vectors lying in the plane: $$\vec{n} = \overrightarrow{PQ} \times \overrightarrow{PR}$$ $$\overrightarrow{PQ} = Q - P = (-2-1,\ 5-4,\ -1-6) = (-3,\ 1,\...
Full solved answer →Find a vector perpendicular to the plane that passes through the points: $P(1, 4, 6)$, $Q(-2, 5, -1)$ and $R(1, -1, 1)$. [5]
- $P(1, 4, 6)$ - $Q(-2, 5, -1)$ - $R(1, -1, 1)$ A normal vector to the plane is the cross product of two vectors lying in the plane: $$\vec{n} = \overrightarrow{PQ} \times \overrightarrow{PR}$$ $$\overrightarrow{PQ} = Q - P = (-2-1,\ 5-4,\ -1-6) = (-3,\ 1,\...
Full solved answer →If $a = (4, 0, 3)$ and $b = (-2, 1, 5)$, find $|a|$, the vector $a - b$ and $2a + b$. [5]
$$\vec{a} = (4, 0, 3) \qquad \vec{b} = (-2, 1, 5)$$ --- $$\vec{a} = \sqrt{4^2 + 0^2 + 3^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5$$ $$\boxed{\vec{a} = 5}$$ --- $$\vec{a} - \vec{b} = (4-(-2),\ 0-1,\ 3-5) = (6,\ -1,\ -2)$$ $$\boxed{\vec{a} - \vec{b} = (6,\ -1,\ ...
Full solved answer →Define cross product of two vectors. If $\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = 2\hat{i} + 7\hat{j} - 5\hat{k}$, find the vector $\vec{a} \times \vec{b}$ and $\vec{b} \times \vec{a}$. [5]
$$\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k} \quad\Rightarrow\quad (a1, a2, a3) = (1, 3, 4)$$ $$\vec{b} = 2\hat{i} + 7\hat{j} - 5\hat{k} \quad\Rightarrow\quad (b1, b2, b3) = (2, 7, -5)$$ The cross product (vector product) of two vectors $\vec{a}$ and $\vec{b}$...
Full solved answer →Make Unit 9 stick
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