IB Mathematics HL – Vectors
1. Vectors operations – IB Maths HL
If and
How can we find the vectors and
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
and
2. Vectors, Magnitude – IB Maths HL
If
How can we find the magnitude of vector
Solution
Vectors, Magnitude – IB Maths HL
The Magnitude of a vector
with coordinates
is given by the following formula:
Therefore the magnitude of vector is:
3. Vectors, Unit Vector – IB Maths HL
If
How can we find the unit vector of vector
Solution
IB Maths HL – Vectors, Unit Vector
The unit vector of a vector is defined as ffollows
In this case,
and
Therefore,
4. Vectors, Dot Product, scalar product – IB Maths HL
If
How can we find the dot product ?
Solution
Vectors, Dot Product, scalar product, inner product – IB Maths HL
The dot (scalar) product of two vectors is defined as follows
, where
is the angle between the two vectors.
Another expression of scalar product is
where
The result of the scalar (inner) product is a scalar (a number)
Thus,
5. Vectors, Scalar Product, Perpendicular vectors – IB Maths HL
How can we find the value of the parameter for which the vectors
and
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,
6. Vectors, Vector equation of a line – IB Math HL
How can we find the vector equation of the line passing through the point and is parallel to the vector
?
Solution
Vectors, Vector equation of a line – IB Math HL
The vector equation of a line is given by the following formula
where is a position vector of the line (i.e. a point on the line) ,
is a direction vector (i.e. a vector parallel to the line) and
is a real number parameter.
So the vector equation of the line is :
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 and is parallel to the vector
?
Solution
IB Maths HL – Vectors, the Parametric equation of a line
The Parametric equation form of a straight line is :
, where is a position vector (a point on the line and
is a direction vector ( a vector parallel to the line)
is a real number parameter.
Therefore, we have that the parametric equation form is
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 and is parallel to the vector
?
Solution
IB Math HL – Cartesian equation form of a line
The Cartesian form for the equation of a straight line is :
, where is a position vector (a point on the line) and
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:
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 ?
Solution
Vectors, Vector equation of a line – IB Maths HL
The vector equation of a line is given by the following formula
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
is a real number parameter.
So the vector equation of the line is given by the formula
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 and is parallel to the vector
?
Solution
Vectors, Parametric equation form of a line – IB Maths HL
The Parametric form for the equation of a straight line is :
, where is a position vector (a point on the line) and
is a direction vector ( a vector parallel to the line)
is a real number parameter.
Therefore, we have that the parametric equation form is
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 and is parallel to the vector
?
Solution
The Cartesian form for the equation of a straight line is :
, where is a position vector (a point on the line) and
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:
12. The intersection of two lines, Vectors Lines – IB Maths HL
How can we find the point of intersection of the following lines
Solution
Vectors Lines – IB Math HL
The parametric equations form of the lines are
For the first line
For the second line
Then, for the lines to intersect, there must be a value of and
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
and
in order to find the unique solution.
and we obtain
Therefore the two lines meet at the point
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?
and
Solution
IB Mathematics HL – Vectors, Parallel, coincident or skew lines
We first rewrite the Cartesian equation for
We observe that the lines have parallel direction vectors since
Thus the lines are either parallel or coincident.
For example the point
lies on
for
. Then we check whether it is also on
Set into the equation of
and we have
Therefore, the point lies on
. As they share a common point,
cannot be parallel, thus they are coincident.