Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2019  MayJun  (P19709/11)  Q#7
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Question
The diagram shows a threedimensional shape in which the base OABC and the upper surface DEFG are identical horizontal squares. The parallelograms OAED and CBFG both lie in vertical planes. The point M is the midpoint of AF.
Unit vectors and are parallel to OA and OC respectively and the unit vector is vertically upwards. The position vectors of A and D are given by
i. Express each of the vectors and in terms of , and .
ii. Use a scalar product to find angle GMA correct to the nearest degree.
Solution
i.
We are required to find vectors and .
First, we find vector .
We are given that point M is the midpoint of AF. Therefore;
Now we need to find .
A vector in the direction of is;
For the given case;
Therefore, we need the position vectors of points and , respectively.
We are given the position vector of point A.
Next, we find the position vector of point F.
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 . It is evident that distance of point along from the origin is 8 units. However, since , the distance of point along from the origin is (8+3=) 11 units.
· It is given that is parallel to and we can see that distance of point along from the origin is 8 units. (AB is parallel to OC being opposite sides of square OABC with each length 8 units)
· It is given that is vertically upwards. It is evident that distance of point along from the origin is 10 units. (BF is parallel to OD and ) 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 .
Finally, we can find .
Next, we are required to 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 .
We have already found above that;
A vector in the direction of is;
Therefore;
We are given that;
Hence;
Next, we find the position vector of point G.
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 . It is evident that distance of point along from the origin is 3 units. Since, , the distance of point along from the origin is 3 units and points and are on the same side of the square DEFG.
· It is given that is parallel to and we can see that distance of point along from the origin is 8 units. (OA and OC being adjacent sides of square OABC with each length 8 units)
· It is given that is vertically upwards. It is evident that distance of point along from the origin is 10 units. (CG is parallel to OD and ) 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 .
ii.
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 ;
Therefore, we need to find and .
We have from (i) that;
Now we find the scalar product of and .
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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