Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2013  OctNov  (P19709/13)  Q#4
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Question
The diagram shows a pyramid OABC in which the edge OC is vertical. The horizontal base OAB is a triangle, rightangled at O, and D is the midpoint of AB. The edges OA, OB and OC have lengths of 8 units, 6 units and 10 units respectively. Unit vectors , and are parallel to , and respectively.
i. Express each of the vectors and in terms of , and .
ii. Use a scalar product to find angle ODC.
Solution
i.
Tofind consider the diagram below. It is evident from the diagram that;
We are given that and D is the midpoint of AB. Therefore;
Hence;
Now we need vectors and .
First we 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 8 units.
· It is given that is parallel to and we can see that distance of point along from the origin is ZERO.
· It is given that is parallel to and we can see that distance of point along from the origin is ZERO.
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 find vector .
A vector in the direction of is;
For the given case;
Therefore, we need the position vectors of points and . We already have position vector of point .
To find the position vector of point .
A point has position vector from the origin . Then the position vector of is denoted by or .
Hence we need coordinates of the point .
Let’s find the 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 ZERO.
· It is given that is parallel to and we can see that distance of point along from the origin is 6 units.
· It is given that is parallel to and we can see that distance of point along from the origin is ZERO.
Hence, coordinates of .
Now we can represent the position vector of point as follows;
Now we have;
Therefore we can write the vector ;
Now we have;
Therefore, we can write;
Next we are required to find the vector .
A vector in the direction of is;
For the given case;
Therefore, we need the position vectors of points and . We already have position vector of point .
To find the position vector of point .
A point has position vector from the origin . Then the position vector of is denoted by or .
Hence we need coordinates of the point .
Let’s find the 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 ZERO.
· It is given that is parallel to and we can see that distance of point along from the origin is ZERO.
· It is given that is parallel to and we can see that distance of point along from the origin is 10 units.
Hence, coordinates of .
Now we can represent the position vector of point as follows;
Now we have;
Therefore we can write the vector ;
ii.
We recognize that is angle between and .
Hence we use scalar/dot product of and .
From (i) we have both and .
The scalar or dot product of two vectors and in component form is given as;


Since ;
Therefore for the given case;
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;
Hence the angle ODC is .
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