Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | Oct-Nov | (P1-9709/13) | Q#6

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Question

The diagram shows a solid figure OABCDEFG with a horizontal rectangular base OABC in which  OA = 8 units and AB = 6 units. The rectangle DEFG lies in a horizontal plane and is such that D is 7  units vertically above O and DE is parallel to OA. The sides DE and DG have lengths 4 units and 2  units respectively.

Unit vectors  and  and  are parallel to  and   and  respectively.

Use a scalar product to find angle OBF, giving your answer in the form , where  and are  integers.

Solution

We are required to find the angle OBF.

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, need to find and .

First, we find .

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  6 units (OC & AB are parallel).

·   It is given that is parallel to and we can see that distance of point along  from the origin is  0 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 .

A vector in the direction of  is;

For the given case;

Therefore, we need the position vectors of points  and  respectively.

Therefore, we need to find which is the position vector of point F.

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.  (DE is parallel to OA and DE=4 units) 

·   It is given that is parallel to and we can see that distance of point along  from the origin is  2 units (DG & OC are parallel & GD and EF are parallel and GD=2 units).

·   It is given that is parallel to and we can see that distance of point along  from the origin is  7 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 B.

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 ;

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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