Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2019 | May-Jun | (P1-9709/11) | Q#7

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Question

The diagram shows a three-dimensional 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 mid-point 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 mid-point 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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