IB Mathematics HL – Vectors

1. Vectors operations – IB Maths HL

If $a= \begin{pmatrix} -3 \\ 5 \\ 11 \end{pmatrix}$ and

$b= \begin{pmatrix} -4 \\ 6 \\ 8 \end{pmatrix}$

How can we find the vectors $a-3b$ and $4a+3b$

Solution

Vectors operations, addition, subtraction, scalar multiplication – IB Maths HL

First of all, let me remind you the properties of vector addition

The vector addition (subtraction) is Commutative, Associative, there is an additive identity which is the zero vector ad there is an additive inverse which is the vector with equal magnitude and opposite direction. Also, there is the scalar multiplication which is when a vector is multiplied by a scalar.

So about this question, we have the following

$a-3b=\begin{pmatrix} -3 \\ 5 \\ 11 \end{pmatrix} - 3 \begin{pmatrix} -4 \\ 6 \\ 8 \end{pmatrix}$

$=\begin{pmatrix} -3 \\ 5 \\ 11 \end{pmatrix} -\begin{pmatrix} -12 \\ 18 \\ 24 \end{pmatrix}$

$=\begin{pmatrix} -3+12 \\ 5-18 \\ 11-24 \end{pmatrix}$

$=\begin{pmatrix} 9 \\ -13 \\ -13 \end{pmatrix}$ and

$4a+3b =4 \begin{pmatrix} -3 \\ 5 \\ 11 \end{pmatrix} + 3 \begin{pmatrix} -4 \\ 6 \\ 8 \end{pmatrix}$

$=\begin{pmatrix} -12 \\ 20 \\ 44 \end{pmatrix} +\begin{pmatrix} -12 \\ 18 \\ 24 \end{pmatrix}$

$=\begin{pmatrix} -12-12 \\ 20+18 \\44+24 \end{pmatrix}$

$=\begin{pmatrix} -24 \\ 38 \\ 68 \end{pmatrix}$

2. Vectors, Magnitude – IB Maths HL

If $a= \begin{pmatrix} -2 \\ 3 \\ 5 \end{pmatrix}$

How can we find the magnitude of vector $v$

Solution

Vectors, Magnitude – IB Maths HL

The Magnitude $|v|$ of a vector $v$ with coordinates $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ is given by the following formula:

$|v|= \sqrt{x^2 + y^2 +z^2}$

Therefore the magnitude of vector $v$ is:

$|v|= \sqrt{(-2)^2 +(3)^2+(5)^2}=\sqrt{38}$

3. Vectors, Unit Vector – IB Maths HL

If $v= \begin{pmatrix} -2 \\ 3 \\ 5 \end{pmatrix}$

How can we find the unit vector of vector $v$

Solution

IB Maths HL – Vectors, Unit Vector

The unit vector of a vector $v$ is defined as ffollows

$\hat{v}= \frac{v}{|v|}$

In this case,

$v= \begin{pmatrix} -2 \\ 3 \\ 5 \end{pmatrix}$ and

$|v|= \sqrt{(-2)^2 +(3)^2+(5)^2}=\sqrt{38}$

Therefore,

$\hat{v}= \frac{1}{\sqrt{38}} \begin{pmatrix} -2 \\ 3 \\ 5 \end{pmatrix}$

4. Vectors, Dot Product, scalar product – IB Maths HL

If $v= \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$

$w= \begin{pmatrix} 6 \\ -3 \\ 1 \end{pmatrix}$

How can we find the dot product $v \cdot w$ ?

Solution

Vectors, Dot Product, scalar product, inner product – IB Maths HL

The dot (scalar) product of two vectors is defined as follows

$v \cdot w=|v||w|cos \theta$ , where $\theta$ is the angle between the two vectors.

Another expression of scalar product is

$v \cdot w=v_{1} w_{1}+ v_{2} w_{2}+ v_{3} w_{3}$

where $v= \begin{pmatrix} v_{1} \\ v_{2} \\ v_{3} \end{pmatrix}$

$w= \begin{pmatrix} w_{1} \\ w_{2} \\ w_{3} \end{pmatrix}$

The result of the scalar (inner) product is a scalar (a number)

Thus, $a \cdot b=(1 \cdot 3) + (2 \cdot (-2)) + (4 \cdot 5) =19$

5. Vectors, Scalar Product, Perpendicular vectors – IB Maths HL

How can we find the value of the parameter $n$ for which the vectors

$v= \begin{pmatrix} 3n \\ -4n \\ 7 \end{pmatrix}$ and

$w= \begin{pmatrix} 6 \\ -8 \\ 4 \end{pmatrix}$ are perpendicular?

Solution

IB Maths HL – Vectors, dot Product, perpendicular vectors

We know that two vectors are perpendicular if and only their scalar product equals zero.

Therefore, $v \cdot w=(3n \cdot 6) + (-4n \cdot (-8)) + (7 \cdot 4) =18n+32n+28=0$

$\Rightarrow 50n= -28 \Rightarrow n=\frac{-28}{50}=\frac{-14}{25}$

6. Vectors, Vector equation of a line – IB Math HL

How can we find the vector equation of the line passing through the point $(2,-1)$ and is parallel to the vector $v= \begin{pmatrix} 3 \\ 5 \end{pmatrix}$ ?

Solution

Vectors, Vector equation of a line – IB Math HL

The vector equation of a line is given by the following formula

$r=a+ \lambda b$

where $a$ is a position vector of the line (i.e. a point on the line) , $b$ is a direction vector (i.e. a vector parallel to the line) and $\lambda$ is a real number parameter.

So the vector equation of the line is :

$r= \begin{pmatrix} 2 \\ -1 \end{pmatrix}+ \lambda \begin{pmatrix} 3 \\ 5 \end{pmatrix}$

7. IB Maths HL – Vectors, Parametric equation form of a line

How can we find the parametric form of the equations of the line passing through the point $(-2,6)$ and is parallel to the vector $v= \begin{pmatrix} 4 \\ 7 \end{pmatrix}$ ?

Solution

IB Maths HL – Vectors, the Parametric equation of a line

The Parametric equation form of a straight line is :

$x=a_{1} + \lambda b_{1}$

$y=a_{2} + \lambda b_{2}$

, where $a= \begin{pmatrix} a_{1} \\ a_{2} \end{pmatrix}$ is a position vector (a point on the line and $b= \begin{pmatrix} b_{1} \\ b_{2} \end{pmatrix}$ is a direction vector ( a vector parallel to the line)

$\lambda$ is a real number parameter.

Therefore, we have that the parametric equation form is

$x=-2 +4 \lambda$

$y=6 + 7\lambda$

8. IB Maths HL – Vectors, Cartesian equation form of a line

How can we find the Cartesian equation form of the line passing through the point $(-4,6)$ and is parallel to the vector $b= \begin{pmatrix} 6 \\ -8 \end{pmatrix}$ ?

Solution

IB Math HL – Cartesian equation form of a line

The Cartesian form for the equation of a straight line is :

$\frac{x- a_{1}}{ b_{1}}= \frac{y- a_{2}}{ b_{2}}$

, where $a= \begin{pmatrix} a_{1} \\ a_{2} \end{pmatrix}$ is a position vector (a point on the line) and $b= \begin{pmatrix} b_{1} \\ b_{2} \end{pmatrix}$ is a direction vector ( a vector parallel to the line)

Concerning this question we have that the Cartesian equation form is given by the following formula:

$\frac{x-(-4)}{6}= \frac{y-6}{-8}$

$\frac{x+4}{6}= \frac{y-6}{-8}$

9. Vectors, 3D Vector equation – IB Maths HL

How can we find the vector equation of the line passing through the point (-1,2,3) and is parallel to the vector $b= \begin{pmatrix} -1 \\ 5 \\ 7 \end{pmatrix}$ ?

Solution

Vectors, Vector equation of a line – IB Maths HL

The vector equation of a line is given by the following formula

$r=a+ \lambda b$ where a is a position vector of the line (i.e. a point on the line) , b is a direction vector (i.e. a vector parallel to the line) and $\lambda$ is a real number parameter.

So the vector equation of the line is given by the formula

$r= \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}+ \lambda \begin{pmatrix} -1 \\ 5 \\ 7 \end{pmatrix}$

10. Vectors, Parametric equation form of a 3D line – IB Maths HL

How can we find the parametric form of the equations of the line passing through the point $(3,-5,2)$ and is parallel to the vector $b= \begin{pmatrix} 1 \\ 7 \\ -6 \end{pmatrix}$ ?

Solution

Vectors, Parametric equation form of a line – IB Maths HL

The Parametric form for the equation of a straight line is :

$x=a_{1} + \lambda b_{1}$

$y=a_{2} + \lambda b_{2}$

$z=a_{3} + \lambda b_{3}$

, where $a= \begin{pmatrix} a_{1} \\ a_{2} \\ a_{3} \end{pmatrix}$ is a position vector (a point on the line) and $b= \begin{pmatrix} b_{1} \\ b_{2} \\ b_{3} \end{pmatrix}$ is a direction vector ( a vector parallel to the line)

$\lambda$ is a real number parameter.

Therefore, we have that the parametric equation form is

$x=3 + 1\lambda$

$y=-5 +7 \lambda$

$y=2 -6 \lambda$

11. Vectors, Cartesian equation form of a line 3D, IB Maths HL

How can we find the Cartesian equation form of the line passing through the point $(2,-1,8)$ and is parallel to the vector $b= \begin{pmatrix} 4 \\ -2 \\ 7 \end{pmatrix}$ ?

Solution

The Cartesian form for the equation of a straight line is :

$\frac{x- a_{1}}{ b_{1}}= \frac{y- a_{2}}{ b_{2}}= \frac{z- a_{3}}{ b_{3}}$

Concerning this question we have that the Cartesian equation form is given by the following formula:

$\frac{x-2}{4}= \frac{y-(-1)}{-2}= \frac{z-8}{7}$

12. The intersection of two lines, Vectors Lines – IB Maths HL

How can we find the point of intersection of the following lines

$\frac{x-3}{2}= \frac{y-5}{2}= \frac{z+1}{-4}$

$\frac{x-8}{5}= \frac{y-9}{3}= \frac{z+3}{6}$

Solution

Vectors Lines – IB Math HL

The parametric equations form of the lines are

For the first line

$\frac{x-3}{2}= \frac{y-5}{2}= \frac{z+1}{-4}= \lambda$

$x-3=2 \lambda \Leftrightarrow x=3+ 2 \lambda$

$y-5=2 \lambda \Leftrightarrow x=5+2 \lambda$

$z+1=-4 \lambda \Leftrightarrow x=-1- 4 \lambda$

For the second line

$\frac{x-8}{5}= \frac{y-9}{3}= \frac{z+2}{6}= \mu$

$x-8= 5 \mu \Leftrightarrow x=8+ 5 \mu$

$y-9= 3 \mu \Leftrightarrow y=9+3 \mu$

$z+3= 6 \mu \Leftrightarrow z=-3+ 6 \mu$

Then, for the lines to intersect, there must be a value of $\lambda$ and $\mu$ that will provide the same point (the point of intersection). Using the above equations we equate the coordinates and then solve the simultaneous equations in $\lambda$ and $\mu$ in order to find the unique solution.

$3+ 2 \lambda = 8+ 5 \mu$

$5+2 \lambda = 9+3 \mu$

$-1- 4 \lambda =-3+ 6 \mu$ and we obtain $\lambda = \frac{5}{4} \ and \ \mu =\frac{-1}{2}$

Therefore the two lines meet at the point

$\begin{pmatrix} 2 \\ 4 \\ 1 \end{pmatrix}$

13. Vectors, Coincident – IB Maths HL

How can we determine if the following lines intersect in one point, are parallel and distinct, are coincident, or are skew?

$L_{1}: \frac{x-2}{2}= \frac{1-y}{4}= \frac{z+1}{6}$ and $L_{2}:$

$x=4+6 \mu$

$y=-3-12 \mu$

$z=5+ 18 \mu$

Solution

IB Mathematics HL – Vectors, Parallel, coincident or skew lines

We first rewrite the Cartesian equation for $L_{1}:$

$L_{1}: \frac{x-2}{2}= \frac{y-1}{-4}= \frac{z-(-1)}{3}$

We observe that the lines $L_{1}, L_{2}$ have parallel direction vectors since

$\begin{pmatrix} 6 \\ -12 \\ 18 \end{pmatrix}=3 \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}$

Thus the lines $L_{1}, L_{2}$ are either parallel or coincident.

For example the point

$\begin{pmatrix} 4 \\ -3 \\ 5 \end{pmatrix}$ lies on $L_{2}$ for

$\mu =0$ . Then we check whether it is also on $L_{1}$

Set $\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} 4 \\ -3 \\ 5 \end{pmatrix}$ into the equation of $L_{1}$

and we have $\frac{4-2}{2}= \frac{-3-1}{-4}= \frac{5-(-1)}{3}=1$

Therefore, the point $\begin{pmatrix} 4 \\ -3 \\ 5 \end{pmatrix}$ lies on $L_{1}$ . As they share a common point, $L_{1}\ and\ L_{2}$ cannot be parallel, thus they are coincident.