Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | Feb-Mar | (P1-9709/12) | Q#7

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Question

Fig. 1 shows a rectangle with sides of 7 units and 3 units from which a triangular corner has been  removed, leaving a 5-sided polygon OABCD. The sides OA, AB, BC and DO have lengths of 7  units, 3 units, 3 units and 2 units respectively. Fig. 2 shows the polygon OABCD forming the  horizontal base of a pyramid in which the point E is 8 units vertically above D. Unit vectors ,  and are parallel to ,   and   respectively.

    i.       Find  and the length of CE.

   ii.       Use a scalar product to find angle ECA, giving your answer in the form where m and  n are integers.

Solution


i.
 

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 the position vector  of point E.

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  2 (length of OD in Fig 1 is given as 2 units).

·       It is given that is parallel to and we can see that distance of point along  from the origin  is 8 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 C.

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  3 (width of rectangle in Fig 1 is given as 3 units).

·       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 .

Now we can find .

Next we are required to find the length of .

Expression for the length (magnitude) of a vector is;


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;

Next, let us find the vector .

A vector in the direction of  is;

For the given case;

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

From (i) we already have ;

First we find the position vector of point A. 

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  7 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 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 .

Now we can find .

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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