Vectors, Sheet #2

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Problem sheet

Skill Building Questions

Problem 1.

Consider the three vectors $\vec{a}= (2,1,0)$, $\vec{b}= (-1,2,3)$, and $\vec{c}= (1,2,1)$. Calculate the following:

(a) $\vec{a}+\vec{b}$

$\Rightarrow \quad \boxed{\vec{a}+\vec{b}=(1,3,3)}$








(b) $2\vec{a}-\vec{b}$

$2\vec{a}-\vec{b}=((2\times2+1),\space (2\times1-2),\space (2\times0-3)) = \boxed{ (5,0,-3)}$








(c) $\vec{a}\circ\vec{b}\circ \vec{c}$

$\vec{a}\circ\vec{b}\circ \vec{c}=(\vec{a}\circ\vec{b})\circ\vec{c}=(-2,2,0)(1,2,1)=\boxed{(-2,4,0)}$








(d) $| \vec{c} |$

$\| c \| = \sqrt{1^2+2^2+1^2}= \boxed{ \sqrt{6}}$








(e) Find the unit vector in direction of $\vec{c}$.

$\boxed{\hat{c} = \frac{\vec{c}}{|c|} = \left(\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}}\right)}$








(f) Find the volume of the parallelepiped described by the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ (hint: Use Triple scalar product).

$V=|\vec{a}\cdot(\vec{b}\times\vec{c})|=|(2,1,0)\cdot(-4,4,-4)|=|-4|=\boxed{4 \text{ volume units}}$








Problem 2.

Scalar product calculation.

(a) Let $\vec{a} = \hat{i}+2\hat{j} $ and $ \vec{b} =2\hat{i}+\hat{j}. $ Is $ | \vec{a} | = | \vec{b} |?$ Are the vectors $\vec{a}$ and $\vec{b}$ equal ?

We have $$|\vec{a}|=\sqrt{1^2+2^2}=\sqrt{5}$$ and $$|\vec{b}|=\sqrt{2^2+1^2}=\sqrt{5}$$ So, $$\boxed{|\vec{a}|=|\vec{b}|.}$$ But, the two vectors are not equal since their corresponding components are distinct.








(b) Find the vector from point $P(2, 3, 0)$ to $Q(-1, -2, -4)$:

Since the vector is to be directed from P to Q, clearly P is the initial point and Q is the terminal point. So, the required vector joining P and Q is the vector $\vec{PQ}$ given by $$\vec{PQ} = (-1-2)\hat{i}+(-2-3)\hat{j}+(-4-0)\hat{k}$$ $$\boxed{\vec{PQ} = -3\hat{i}-5\hat{j}-4\hat{k}.}$$








(c) Find the angle $\theta$ between the vectors $\vec{a} = \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ :

The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by $$\cos{\theta} = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$$ $$\vec{a}\cdot\vec{b} = (\hat{i}+\hat{j}-\hat{k})\cdot (\hat{i}-\hat{j}+\hat{k})=-1$$ $$|\vec{a}||\vec{b}|=\sqrt{1^2+1^2+(-1)^2}\sqrt{1^2+(-1)^2+1^2}=\sqrt{3}\times\sqrt{3}=3$$ Therefore, we have $$\cos\mathrm{\theta} = -\frac{1}{3}$$ Hence,the required angle is $$\boxed{ \theta =\cos^{-1}\left(-\frac{1}{3}\right) = 109.5^{\circ} .}$$








(d) If $\vec{a} = 5\hat{i}-\hat{j}-3\hat{k}$ and $\vec{b} = \hat{i}+3\hat{j}-5\hat{k}$, then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.

Two non-zero vectors are perpendicular if their scalar product is zero. $$\vec{a}+\vec{b} = (5\hat{i}-\hat{j}-3\hat{k})+(\hat{i}+3\hat{j}-5\hat{k})=6\hat{i}+2\hat{j}-8\hat{k}$$ $$\vec{a}-\vec{b} =(5\hat{i}-\hat{j}-3\hat{k})-(\hat{i}+3\hat{j}-5\hat{k})=4\hat{i}-4\hat{j}+2\hat{k}$$ Therefore, we have: $$\boxed{(\vec{a}-\vec{b})\cdot(\vec{a}+\vec{b}) =(6\hat{i}+2\hat{j}-8\hat{k})\cdot(4\hat{i}-4\hat{j}+2\hat{k})=24-8-16=0}$$








(e) Show the points $A(-2\vec{i}+3\vec{j}+5\vec{k})$, $B(\vec{i}+2\vec{j}+3\vec{k})$, and $C(7\vec{i}-\vec{k})$ are collinear (They lie on the same straight line).

We have $$\vec{AB}=(1+2)\hat{i}+(2-3)\hat{j}+(3-5)\hat{k}=3\hat{i}-\hat{j}-2\hat{k},$$ $$\vec{BC}=(7-1)\hat{i}+(0-2)\hat{j}+(-1-3)\hat{k}=6\hat{i}-2\hat{j}-4\hat{k},$$ $$\vec{CA}=(-2-7)\hat{i}+(3-0)\hat{j}+(5+1)\hat{k}=-9\hat{i}+3\hat{j}+6\hat{k},$$ $$|\vec{AB}|=\sqrt{14},\space |\vec{BC}|=2\sqrt{14}, $$ and $$|\vec{AC}|=3\sqrt{14},$$ Therefore, $$\boxed{|\vec{AC}|=|\vec{AB}|+|\vec{BC}|.}$$








Scalar Product Visualization

To help with intuition when answering these questions, here is a visualization built in Manim. You can play around with it here!

Problem 3.

Cross product calculation.

(a) Find $|\vec{a}\times\vec{b}|,$ if $\vec{a}=2\vec{i}+\vec{j}+3\vec{k}$ and $\vec{b}=3\vec{i}+5\vec{j}-2\vec{k}$

We have $$\vec{a}\times\vec{b}=$$ $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \\ \end{vmatrix}$$ Therefore, $$\vec{a}\times\vec{b}= \hat{i}(-2-15)-(-4-9)\hat{j}+(10-3)\hat{k}=-17\hat{i}+13\hat{j}+7\hat{k}$$ Hence, $$|\vec{a}\times\vec{b}|=\sqrt{(-17)^2+(13)^2+(7)^2}= \boxed{\sqrt{507}.}$$








(b) Find the area of a triangle having the points $A(1,1,1)$, $B(1,2,3)$ and $C(2,3,1)$ as its vertices

We have $$\vec{AB}=\hat{j}+2\hat{k} \text{ and } \vec{AC}=\hat{i}+2\hat{j}$$ The area of given triangle is $$\frac{1}{2}|\vec{AB}\times\vec{AC}|$$ Now $$\vec{AB}\times\vec{AC}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \\ \end{vmatrix}$$ Therefore $$|\vec{AB}\times\vec{AC}|= \sqrt{16+4+1}=\sqrt{21}$$ Thus, area of the triangle is $\boxed{\frac{1}{2}\sqrt{21} \text{ units}^2}$








(c) Find a unit vector perpendicular to each of the vectors $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$, where $\vec{a} = \hat{i}+\hat{j}+\hat{k} $, $\vec{b}= \hat{i}+2\hat{j}+3\hat{k}$.

We have $\vec{a}+\vec{b}=2\hat{i}+3\hat{j}+4\hat{k}$ and $\vec{a}-\vec{b}=-1\hat{j}-2\hat{k}$. Therefore, a vector is perpendicular to both $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ is given by $$(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \\ \end{vmatrix}=-2\hat{i}+4\hat{j}-2\hat{k}(=\vec{c}, \text{ say})$$ Now $$|\vec{c}| = \sqrt{4+16+4}=\sqrt{24}=2\sqrt{6}$$ Therefore, the required unit vector is $$\boxed{\frac{\vec{c}}{|\vec{c}|} =-\frac{1}{\sqrt{6}}\hat{i}+\frac{2}{\sqrt{6}}\hat{j}-\frac{1}{\sqrt{6}}\hat{k}.}$$








(d) Find the area of a parallelogram which adjacent sides are given by the vectors $\vec{a}= 3\hat{i}+\hat{j}+4\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The area of a parallelogram with $\vec{a}$ and $\vec{b}$ as its adjacent sides is given by $|\vec{a}\times\vec{b}|$ $$\vec{a}\times\vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \\ \end{vmatrix}=5\hat{i}+\hat{j}-4\hat{k}$$ Therefore $$|\vec{a}\times\vec{b}| = \sqrt{25+1+16} = \sqrt{42}$$ Hence, the required area is $\boxed{\sqrt{42} \text{ units}^2.}$








Cross Product Visualization

To help with intuition when answering these questions, here is a visualization built in Manim. You can play around with it here!

Problem 4.

Analytical geometry calculation.

(a) Find the equation for a plane through the point $(0,1,-7)$ which is perpendicular to the vector $(4,-1,6)$.

Let $A=(0,1,-7)$. Let $\vec{n}=(4,-1,6)$. Then, for $P=(x,y,z)$, the equation for the plane is $$\vec{n}.(P-A)= \cos(90^\circ)=0$$ This becomes $$(4,-1,6)\cdot(x-0,y-1,z+7)=0$$ Or $$4x-(y-1)+6(z+7)=0$$ Often, we prefer to write this as $\boxed{4x-y+6z+43=0.}$








(b) Find the equation for a plane through the points $A=(0,1,-7),B=(3,1,-9), $ and $C=(0,-5,-8)$.

Let $\vec{AB}=\vec{b}=(3,1,-9)-(0,1,-7)=(3,0,-2)$. Let $\vec{AC}=\vec{c}=(0,-5,-8)-(0,1,7)=(0,-6,-1)$. Then, we can find a normal vector by taking their cross product $$\vec{n}=\vec{b}\times\vec{c}=\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ -3&0&2\\ 0&6&1\\ \end{vmatrix}=\hat{i}(0-12)-\hat{j}(-3-0)+\hat{k}(-18-0)=(-12,3,-18)$$ We will pick the first point $A=(0,1,-7)$ and we assume a point $P$ is on the plane $P=(x,y,z)$. Let $\vec{a}=\vec{AP}=(x-0,y-1,z+7)$. The equation for the plane becomes $$\vec{n}\cdot\vec{a}=(-12,3,-18)\cdot(x,y-1,z+7)=0$$ Which we rewrite as $$-12x+3(y-1)-18(z+7)=0$$ or $$-12x+3y-18z-129=0$$ $$\boxed{4x-y+6z+43=0}$$








(c) Find the angle $\theta$ between two lines in the x, y-plane, given by the equations: $3x-4y+1=0$ and $2x+y-5=0$

Normal vector to the lines are, respectively, $\vec{n_1} = (3 , -4)$ and $\vec{n_2}=(2,1)$. Therefore, $$\cos\theta=\frac{n_1\cdot n_2}{|n_1|\cdot|n_2|}$$ $$\cos\theta=\frac{3\cdot2+(-4)\cdot1}{\sqrt{3^2+(-4)^2}\sqrt{2^2+1^2}}= \frac{2}{5\sqrt{5}}$$ $$\boxed{\theta=\cos^{-1}\left(\frac{2}{5\sqrt{5}}\right)=79.7^\circ}$$








(d) Let ABC be a triangle in the x, y-plane, with the vertices at the points $A = (2,-1)$, $B=(4,4)$ and $C=(9,7)$. Find the distance from point A normal to line BC:

First, we can begin by drawing a diagram of the points: vectors Find the equation of line BC: $$ \text{Gradient} = C - B = \begin{bmatrix} 5\\\\ 3 \end{bmatrix}$$ $$ \therefore BC = \begin{bmatrix} 4 \\\\ 4 \end{bmatrix} + \lambda \begin{bmatrix} 5 \\\\ 3 \end{bmatrix}$$ The point D is the perpendicular intersection between line BC and A, where D can be expressed as a point on the line BC with a value of lambda to be found: $$D = \begin{bmatrix} 4 + 5\lambda \\\\ 4 + 3\lambda \end{bmatrix}$$ Therefore, we can form an equation for line AD: $$ \text{Gradient} = A - D = \begin{bmatrix} -2 - 5\lambda\\\\ -5 - 3\lambda \end{bmatrix}$$ $$\therefore AD = \begin{bmatrix} 2 \\\\ -1 \end{bmatrix} + \mu \begin{bmatrix} -2 - 5\lambda\\\\ -5 - 3\lambda \end{bmatrix} $$ As we know that lines BC and AD are perpendicular, the dot product of their gradients will be equal to 0: $$(-10 - 25\lambda) + (-15 - 9\lambda) = 0$$ $$\lambda = \frac{-25}{34}$$ We can use the AD gradient with this value of lambda to find the distance between A and D, d: $$d^2 = (-2 - 5\lambda)^2 + (-5 - 3\lambda)^2$$ $$\boxed{ { d } ={\frac{19}{\sqrt{34}}}}$$








(e) Find the unit vector in the direction of the sum of the vectors, $\vec{a}=2\vec{i}+2\vec{j}-5\vec{k}$ and $\vec{b}=2\vec{i}+\vec{j}+3\vec{k}$.

The sum of given vector is: $$\vec{a}+\vec{b}(=\vec{c},\text{say})=4\vec{i}+3\vec{j}-2\vec{k}$$ And $$|\vec{c}|=\sqrt{4^2+3^2+(-2)^2}=\sqrt{29}$$ Thus, the required unit vector is $$\hat{c}=\frac{1}{|\vec{c}|}\vec{c}=\frac{1}{\sqrt{29}} (4\vec{i}+3\vec{j}-2\vec{k})=\boxed{\frac{4}{\sqrt{29}}\vec{i}+\frac{3}{\sqrt{29}}\vec{j}-\frac{2}{\sqrt{29}}\vec{k}}$$








Problem 5.

Show that the points $A = (2\hat{i},-\hat{j},\hat{k})$, $B = (\hat{i},-3\hat{j},-5\hat{k})$ and, $C = (3\hat{i},-4\hat{j},-4\hat{k})$ are the vertices of a right angled triangle.

We have $\vec{AB}=(1-2)\hat{i}+(-3+1)\hat{j}+(-5-1)\hat{k}=-\hat{i}-2\hat{j}-6\hat{k}$ $$\vec{BC}=(3-1)\hat{i}+(-4+3)\hat{j}+(-4+5)\hat{k}=2\hat{i}-\hat{j}+\hat{k}$$ $$\vec{CA}=(2-3)\hat{i}+(-1+4)\hat{j}+(1+4)\hat{k}=-\hat{i}+3\hat{j}+5\hat{k}$$ Further, note that $$|\vec{AB}|^2=41=6+35=|\vec{BC}|^2+|\vec{CA}|^2$$ Hence, $\boxed{\textrm{the triangle is a right angled triangle.}}$








Problem 6.

Use the scalar triple product to show that the vectors $a=2i+3j+k, b=i-j, c=7i+3j+2k$, are coplanar, that is, they lie in the same plane.

We have the volume of parallelepiped (remember the definition of the dot product and cross product)
Volume = $|a||\cos\phi||b\times c|=\vec{a}\cdot(\vec{b}\times \vec{c})$ Volume=$\vec{a}\cdot(\vec{b}\times\vec{c})=(2,3,1)\cdot \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 7 & 3 & 2 \\ \end{vmatrix}=(2,3,1)\big(\begin{vmatrix} -1&0\\ 3&2\\ \end{vmatrix}\hat{i}-\begin{vmatrix} 1&0\\ 7&2\\ \end{vmatrix}\hat{j}+\begin{vmatrix} 1&-1\\ 7&3\\ \end{vmatrix}\hat{k}\big)$
Volume=$(2,3,1)(-2,-2,10)=-4-6+10=0$ Therefore, the volume of the parallelepiped formed by vector a,b,c is 0. $\boxed{\textrm{This means } \vec{a},\vec{b}, \textrm{ and } \vec{c} \textrm{ are coplanar.}}$








Problem 7.

Given the basis vectors $\vec{a} = 2\hat{i} - 3\hat{j}$ and $\vec{b} = \hat{i} + 2\hat{j}$, what is the vector $(4, -9)$ equivalent to?

$(4, -9)$ is equivalent to $4\vec{a} - 9\vec{b}$
$4\vec{a} = 8\hat{i} - 12\hat{j}$
$-9\vec{b} = -9\hat{i} - 18\hat{j}$
$(4, -9) = (8-9)\hat{i} + (-12 - 18)\hat{j}$
$(4, -9) = \boxed{-\hat{i} -30\hat{j}}$








Problem 8.

The lines $l_1$ and $l_2$ have equations $\begin{pmatrix} 1, \
0, \
0 \
\end{pmatrix} + \lambda\begin{pmatrix} 0, \
1, \
1 \
\end{pmatrix} $ and $\begin{pmatrix} -1, \
3, \
-1 \
\end{pmatrix} + \mu\begin{pmatrix} 2, \
-1, \
-1 \
\end{pmatrix} $ respectively, where $\lambda$ and $\mu$ are scalars.
Find the shortest distance, $d$, between these two lines.

Method 1 (The long way):
Let $A$ and $B$ be general points on $l_1$ and $l_2$ respectively.
$A = \begin{pmatrix} 1 \\\\ \lambda \\\\ \lambda \\\\ \end{pmatrix}$ and $B = \begin{pmatrix} -1 + 2\mu \\\\ 3 - \mu \\\\ -1 - \mu \\\\ \end{pmatrix}$
$\vec{AB} = \begin{pmatrix} -1 + 2\mu \\\\ 3 - \mu \\\\ -1 - \mu \\\\ \end{pmatrix} - \begin{pmatrix} 1 \\\\ \lambda \\\\ \lambda \\\\ \end{pmatrix} = \begin{pmatrix} -2 + 2\mu \\\\ 3 - \mu -\lambda \\\\ -1 - \mu - \lambda \\\\ \end{pmatrix} $
As $\vec{AB}$ is perpendicular to both $l_1$ and $l_2$, the scalar product of the direction vectors of the lines is zero. This can be used to generate linear equations in $\lambda$ and $\mu$.
$\vec{AB}$ is perpendicular to $l_1$:
$\begin{pmatrix} -2 + 2\mu \\\\ 3 - \mu -\lambda \\\\ -1 - \mu - \lambda \\\\ \end{pmatrix}\cdot\begin{pmatrix} 0 \\\\ 1 \\\\ 1 - \mu - \lambda \\\\ \end{pmatrix} = 0$
$3 - \mu - \lambda - 1 - \mu - \lambda = 0 \Rightarrow\quad 2 - 2\mu - 2\lambda = 0\quad (1)$
$\vec{AB}$ is perpendicular to $l_2$:
$\begin{pmatrix} -2 + 2\mu \\\\ 3 - \mu -\lambda \\\\ -1 - \mu - \lambda \\\\ \end{pmatrix}\cdot\begin{pmatrix} 2 \\\\ -1 \\\\ -1 - \mu - \lambda \\\\ \end{pmatrix} = 0$
$ -4 + 4\mu - 3 + \mu + \lambda + 1 + \mu + \lambda = 0 \Rightarrow\quad -6 + 6\mu + 2\lambda = 0\quad (2)$
Rearranging equation $(1)$ gives $\lambda = 1 - \mu$. This can be substituted into equation $(2)$.
$ -6 + 6\mu +2(1 - \mu) = 0 \Rightarrow\quad -4 + 4\mu = 0 \Rightarrow\quad \mu = 1$
Therefore, $\lambda = 1 - 1 = 0 $
$\vec{AB} = \begin{pmatrix} -2 + 2(1) \\\\ 3 - 1 - 0 \\\\ -1 - 1 - 0 \\\\ \end{pmatrix} = \begin{pmatrix} 0 \\\\ 2 \\\\ -2 \\\\ \end{pmatrix} $
The length of $\vec{AB}$ is the shortest distance between the lines.
Shortest distance, $d = |\vec{AB}| = \sqrt{0^2 + 2^2 + (-2)^2} = \boxed{2\sqrt{2}}$
Method 2 (The quick way):
Method 1 can be replaced by a short-cut equation:
For two lines given in the form $l_1 = \vec{a_1} + \lambda\vec{b_1}$ and $l_2 = \vec{a_2} + \mu\vec{b_2}$, $ d = \frac{|(\vec{a_2} - \vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|(\vec{b_1}\times\vec{b_2})|}$
$\vec{a_2} - \vec{a_1} = \begin{pmatrix} -1 \\\\ 3 \\\\ -1 \\\\ \end{pmatrix} - \begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\\\ \end{pmatrix} = \begin{pmatrix} -2 \\\\ 3 \\\\ -1 \\\\ \end{pmatrix}$
$ \vec{b_1}\times\vec{b_2} = \begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\\\ \end{pmatrix}\times\begin{pmatrix} 2 \\\\ -1 \\\\ -1 \\\\ \end{pmatrix} = \begin{pmatrix} 0 \\\\ 2 \\\\ -2 \\\\ \end{pmatrix} $
$ |\vec{b_1}\times\vec{b_2}| = \sqrt{0^2 + 2^2 + (-2)^2} = 2\sqrt{2} $
$ (\vec{a_2} - \vec{a_1})\cdot(\vec{b_1}\times\vec{b_2}) = \begin{pmatrix} -2 \\\\ 3 \\\\ -1 \\\\ \end{pmatrix}\cdot\begin{pmatrix} 0 \\\\ 2 \\\\ -2 \\\\ \end{pmatrix} = 0 + 6 + 2 = 8 $
Finally, $d = \frac{8}{2\sqrt{2}} = \boxed{2\sqrt{2}}$








Problem 9.

The locations of a pair of long straight pipes are specified using Cartesian co-ordinates as follows:

Do the pipes need re-aligning to avoid intersection? (Assume the origin position O is (0,0,0)).

For pipe $A$, the vector form of equation is (the other two forms are Parametric form and Cartesian form): $$\vec{r_A} =\vec{OP}+\lambda'\vec{PQ}= [2,5,3]+\lambda'[5,5,5]=[2,5,3]+\lambda\frac{[1,1,1]}{\sqrt{3}}$$ $$\vec{r_B} =\vec{OR}+\mu'\vec{RS}= [0,6,3]+\mu'[12,-6,6]=[0,6,3]+\mu\frac{[-2,-1,1]}{\sqrt{6}}$$ (Non-unit) perpendicular to both their axes is $$ p =\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 1 & 1\\ -2 & -1 & 1\\ \end{vmatrix}=[2,-3,1]$$ The length of mutually perpendicular pipe $A$ and $B$ is $d$. $$d=\Big|\frac{\vec{RP}\cdot\vec{n}}{|n|}\Big|=\Big|[2,-1,0]\cdot \frac{[2,-3,1]}{\sqrt{14}}\Big|= \boxed{1.87 \dots}$$ Sum of the radii of the pipes is $0.4+0.5=\boxed{0.9.}$
Hence, $\boxed{\text{the pipes do NOT need re-aligning to avoid intersection.}}$








Exam Style Questions

Problem 10.

The diagram below shows a cuboid $OABCDEFG$ with coordinates as shown. The point P has coordinates $(4,2,0)$.

cuboid

(a) Find the length of the diagonal $AG$

$AG = \sqrt{4^2 + 3^2 +5^2} \\ = \sqrt{50} \\ = \boxed{5\sqrt{2} }$








(b) Show that the vector $\textbf{n} = 15\textbf{i} - 20\textbf{j} +4\textbf{k}$ is normal to the plane $DPF$. Hence find the cartesian equation of this plane.

$\overrightarrow{DP} = 4\textbf{i} +2\textbf{j} -5\textbf{k} \text{ or } \overrightarrow{PD} = -4\textbf{i} -2\textbf{j} +5\textbf{k}$
$\overrightarrow{DF} = 4\textbf{i} +3\textbf{j} \space \text{(or }\ \overrightarrow{PF} = \textbf{j} + 5\textbf{k})$
The scalar product should equal $0$ if the vector $\textbf{n}$ is perpendicular to the plane.
$\textbf{n} \cdot \overrightarrow{DP} = 15 \times 4 - 20 \times 2 + 4 \times (-5) = 0 $
$\textbf{n} \cdot \overrightarrow{DF} = 15 \times 4 - 20 \times 3 = 0 $
$\text{(or } \textbf{n} \cdot{PF} = -20 \times 1 + 4 \times 5 = 0)$
$\textbf{r} \cdot \textbf{n} = \textbf{a} \cdot \textbf{n} \Rightarrow\boxed{15x - 20 y + 4z = 20}$








(c) The diagonal $AG$ intersects the plane $DPF$ at $Q$. Write down a vector equation of the line $AG$. Hence find the coordinates of the point $Q$, and the ratio $AQ:QG$.

$\textbf{r} = 4\textbf{i} + \lambda(-4\textbf{i} + 3\textbf{j} + 5\textbf{k})$
$\textbf{r} = (4 - 4\lambda)\textbf{i} + 3\lambda\textbf{j} + 5\lambda\textbf{k}$
Substitute the components of the vector equation of the line AG into the equation for the plane $DPF$.
$15(4 - 4 \lambda) - 20 (3 \lambda) + 4 (5 \lambda) = 20 \Rightarrow 40 = 100 \lambda, \lambda = 0.4$
$Q$ is $(2.4, 1.2, 2)$
$\overrightarrow{AQ} = (-1.6, 1.2, 2)$ and $\overrightarrow{QG} = (-2.4, 1.8, 3)$
$\frac{-1.6}{-2.4} = \frac{1.2}{1.8} = \frac{2}{3} $
Therefore, $\boxed{AQ : QG = 2 : 3}$








(d) Find the acute angle between the line $AG$ and the plane $DPF$.

Angle between $(-4\textbf{i} + 3\textbf{j} + 5\textbf{k}) \text{ and } (15\textbf{i} -20\textbf{j} + 4\textbf{k})$ is $\theta$ where
$ cos\theta = \frac{(-4\times15) + (3\times-20)+(5\times4)}{\sqrt{50}\sqrt{641}}$
$\theta = 56.0$ or $124.0$
$\boxed{\text{Angle between line and plane } = 34.0^\circ}$








Problem 11.

A mahi-mahi fish, initially located at $(-4,3,-2)km$, begins to swim towards a coral reef with velocity $\begin{pmatrix} 1,
7,
-6 \end{pmatrix} \text{km/h}$.

(a) What is the speed of the fish in km/h to 2 decimal places?

$|\vec{v}|= \sqrt{1^2 + 7^2 + (-6)^2} = \sqrt{86} = \boxed{9.27 \text{km/h} }$








(b) A deep sea buoy floats on the surface of the water at location $(8,6,0)km$ and is attached by a chain to an anchor that sits on the sea bed (the buoy does not move from its given position at sea-level). The vector $\begin{pmatrix} 1,
-1,
-4 \end{pmatrix} $ describes the direction of the anchor from the buoy. The ocean is $10$km deep. How long is the anchor chain to 2 decimal places?

$\lambda \begin{pmatrix} 1\\-1\\-4 \end{pmatrix} = \begin{pmatrix} x\\y\\-10 \end{pmatrix} \Rightarrow{\lambda = \frac{-10}{-4}= 2.5}$

$\Rightarrow$ Chain length $= \left| \begin{pmatrix} 2.5\\-2.5\\-10 \end{pmatrix} \right| = \sqrt{2.5^2 + (-2.5)^2 + (-10)^2} = \boxed{10.61 km}$








(c) What is the minimum distance from the buoy to the mahi-mahi fish to 2 decimal places?

The position of the mahi-mahi fish at any given $t$ is represented by the vector $\vec{s}$.
$\vec{s} = \begin{pmatrix} -4\\3\\-2 \end{pmatrix} + t \begin{pmatrix} 1\\7\\-6 \end{pmatrix} = \begin{pmatrix} -4+t\\3+7t\\-2-6t \end{pmatrix} $
When the fish is closest to the buoy, the velocity vector of the fish $\vec{v}$ will be perpendicular to the vector from the fish to the buoy, which is represented by $(\vec{b} - \vec{s})$.
$\vec{v} \cdot (\vec{b} - \vec{s}) = 0 = \begin{pmatrix} 1\\7\\-6 \end{pmatrix} \cdotp \begin{pmatrix} 8 -(-4+t)\\6-(3+7t)\\0-(-2-6t) \end{pmatrix} = 1(12-t) + 7(3-7t) - 6(2+6t) \\ = 21 - 86t = 0 \Rightarrow{t = \frac{21}{86}}$
The minimum distance between the fish and the buoy is the length of the vector $(\vec{b} - \vec{s})$.
$ |(\vec{b}-\vec{s})| = \sqrt{(12-t)^2 + (3-7t)^2 + (2+6t)^2} = \boxed{12.32 km} $








Extension Questions

Problem 12.

Prove any given three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ satisfy the relationship \((\vec{a}+\vec{b})+\vec{c} = \vec{a} + (\vec{b} + \vec{c})\)

Let the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ be represented by $\vec{PQ}$, $\vec{QR}$ and $\vec{RS}$, respectively, as shown in the graph.
From the graph, $$\vec{a}+\vec{b}=\vec{PQ}+\vec{QR}=\vec{PR}$$ $$\vec{b}+\vec{c}=\vec{QR}+\vec{RS}=\vec{QS}$$ $$(\vec{a}+\vec{b})+\vec{c}=\vec{PR}+\vec{RS}=\vec{PS}$$ $$\vec{a}+(\vec{b}+\vec{c})=\vec{PQ}+\vec{QS}=\vec{PS}$$ Therefore $$(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$$








Problem 13.

The points $A$, $B$ and $C$ have position vectors $(-2, -3,0)$, $(-1, -1, 3)$ and $(1, 1, 1)$ respectively. Find the centre and radius of the circle that passes through all three points.

Let the centre of the circle, $X$ have position vector $\textbf{r} = \textbf{c} + \lambda(\textbf{c}-\textbf{b}) + \mu(\textbf{c}-\textbf{a})$, where $\textbf{a} = \begin{pmatrix} -2\\-3\\0 \end{pmatrix}, \textbf{b} = \begin{pmatrix} -1\\-1\\3 \end{pmatrix}, \textbf{c} = \begin{pmatrix} 1\\1\\1 \end{pmatrix} $
Therefore, $\textbf{r} = \begin{pmatrix} 1\\1\\1 \end{pmatrix} + \lambda\begin{pmatrix} 2\\2\\-2 \end{pmatrix} + \mu\begin{pmatrix} 3\\4\\1 \end{pmatrix}$
$\overrightarrow{AX}$, $\overrightarrow{BX}$ and $\overrightarrow{CX}$ are radii of the circle which passes through $A$, $B$ and $C$.
$|\overrightarrow{AX}|^2 = |\overrightarrow{CX}|^2$
$(3+(2\lambda+3\mu))^2 + (4+(2\lambda=4\mu))^2 + (1+(-2\lambda+\mu))^2 = (2\lambda +3\mu)^2 + (2\lambda + 4\mu)^2 + (-2\lambda+\mu)^2 $
$ 26 + 6(2\lambda + 3\mu) + 8(2\lambda + 4\mu) + 2(-2\lambda + \mu) = 0 $
$ 26 + 24\lambda + 52\mu = 0,\quad (1)$
$ |\overrightarrow{BX}|^2 = |\overrightarrow{CX}|^2$
$ (2 + (2\lambda+3\mu))^2 + (2 + (2\lambda + 4\mu))^2 + (-2 + (-2\lambda + \mu))^2 = (2\lambda +3\mu)^2 + (2\lambda + 4\mu)^2 + (-2\lambda+\mu)^2$
$ 12 + 4(2\lambda + 3\mu) + 4(2\lambda + 4\mu) - 4(-2\lambda + \mu) = 0$
$ 12 + 24\lambda + 24\mu = 0,\quad (2)$
Solving equations $(1)$ and $(2)$ simultaneously,
$(1) - (2): 14 + 28\mu = 0 \Rightarrow \mu = -\frac{1}{2}$
Substituting $\mu = -\frac{1}{2}$ into $(2)$:
$ 12 + 24\lambda + 24(-\frac{1}{2}) = 0$
$ 24\lambda = 0 \Rightarrow\quad \lambda = 0 $
Therefore, $\textbf{r} = \begin{pmatrix} 1\\1\\1 \end{pmatrix} + -\frac{1}{2}\begin{pmatrix} 3\\4\\1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{2}\\-1\\\frac{1}{2} \end{pmatrix}$
Therefore, the coordinates of the centre of the circle are $\boxed{(-\frac{1}{2}, -1, -\frac{1}{2})}$.
The point X is equidistant to each point $A$, $B$ and $C$ (since $A$, $B$ and $C$ lie on circumference of circle centre X).
Considering point $C$, the radius $r$ is given by:
$ r = |\overrightarrow{CX}| = \sqrt{(1-(-\frac{1}{2}))^2 + (1-(-1))^2 + (1 - \frac{1}{2})^2} = \boxed{\sqrt{\frac{13}{2}}} $








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Revision Questions

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