Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2018  MayJun  (P19709/12)  Q#5
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Question
The diagram shows a threedimensional shape. The base OAB is a horizontal triangle in which angle AOB is 90^{o}. The side OBCD is a rectangle and the side OAD lies in a vertical plane. Unit vectors and are parallel to OA and OB respectively and the unit vector is vertical.
The position vectors of A, B and D are given by , and
i. Express each of the vectors and in terms of , and .
ii. Use a scalar product to find angle CAD.
Solution
i.
We are required to workout .
A vector in the direction of is;
For the given case;
Therefore, we need the position vectors of points and .
We are given that position vectors of A, B and D are given as;
Therefore;
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 .
We are already given that;
Next we find .
To find consider the diagram below.
It is evident from the diagram that;
We are given that OBCD is a rectangle, therefore, its opposite sides are parallel and equal. Hence, and are equal and parallel and we can write above equation as;
We are given that;
Hence;
Now we can find ;
ii.
We are required to find the angle CAD.
It is evident from the diagram that angle CAD is between and .
Therefore, we use scalar/dot product of and to find angle CAD.
From (i) we have;
It can be written as;
We need to find .
To find consider the diagram below.
It is evident from the diagram that;
Hence;
We are given that;
Therefore;
It can be written as;
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 CAD is .
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