Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2018  OctNov  (P19709/12)  Q#7
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Question
The diagram shows a solid cylinder standing on a horizontal circular base with centre O and radius 4 units. Points A, B and C lie on the circumference of the base such that AB is a diameter and angle BOC = 90^{o}. Points P, Q and R lie on the upper surface of the cylinder vertically above A, B and C respectively. The height of the cylinder is 12 units. The midpoint of CR is M and N lies on BQ with BN = 4 units.
Unit vectors and are parallel to and and unit vector is vertically upwards.
Evaluate and hence find angle MPN.
Solution
We are required to evaluate the dot product first.
The scalar or dot product of two vectors and in component form is given as;


Since ;
Therefore, we need to find and .
Let us first find .
A vector in the direction of is;
For the given case;
Therefore, we need the position vectors of points and respectively.
First, we find the position vector of point N.
Let us find the vector .
Since this is the position vector of point , we need coordinates of the point . Consider the diagram below.
· It is given that is parallel to and we can see that distance of point along from the origin is 4 units (OA & OB are radii of circle with center O).
· It is given that is parallel to and we can see that distance of point along from the origin is 0 units.
· It is given that is vertically upwards and we can see that distance of point along from the origin is 4 units.
Hence, coordinates of .
Now we can represent the position vector of point as follows;
A point has position vector from the origin .
Then the position vector of is denoted by or .
Next, we find the position vector of point P.
Let us find the vector . Since this is the position vector of point , we need coordinates of the point . Consider the diagram below.
· It is given that is parallel to and we can see that distance of point along from the origin is 4 units.
· It is given that is parallel to and we can see that distance of point along from the origin is 0 units.
· It is given that is vertically upwards and we can see that distance of point along from the origin is 12 units.
Hence, coordinates of .
Now we can represent the position vector of point as follows;
A point has position vector from the origin . Then the position vector of is denoted by or .
Now we can find .
Let us first find .
A vector in the direction of is;
For the given case;
Therefore, we need the position vectors of points and respectively.
First, we find the position vector of point M.
Let us find the vector .
Since this is the position vector of point , we need coordinates of the point . Consider the diagram below.
· It is given that is parallel to and we can see that distance of point along from the origin is 0 units.
· It is given that is parallel to and we can see that distance of point along from the origin is 4 units (OA & OB & OC are radii of circle with center O).
· It is given that is vertically upwards and we can see that distance of point along from the origin is 6 units (M is midpoint of CR which is 12 units).
Hence, coordinates of .
Now we can represent the position vector of point as follows;
A point has position vector from the origin . Then the position vector of is denoted by or .
We have already found position vector of point P.
Now we can find .
Finally, we are able to find dot product .
The scalar or dot product of two vectors and in component form is given as;


Since ;
Now we need to find the angle MPN.
We recognize that is angle between and .
Hence, we use scalar/dot product of and to find angle .
The scalar or dot product of two vectors and in component form is given as;


Since ;
We have from (i) that;
The scalar or dot product of two vectors and is number or scalar , where is the angle between the directions of and .
Where





For the given case;
Therefore;
Equating both scalar/dot products we get;
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