Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | Oct-Nov | (P1-9709/11) | Q#8

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Question

The diagram shows a solid figure OABCDEF having a horizontal rectangular base OABC with OA =  6 units and AB = 3 units. The vertical edges OF, AD and BE have lengths 6 units, 4 units and 4  units respectively. Unit vectors ,  and  are parallel to ,   and   respectively.

    i.      Find .

  ii.      Find the unit vector in the direction of .

 iii.       Use a scalar product to find angle EFD.

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 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 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.

·   It is given that is parallel to and we can see that distance of point along  from the origin is  6 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 D.

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  6 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  4 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 .


ii.
 

We are required to find unit vector in the direction of .

A unit vector in the direction of  is;

Therefore, for the given case;

Therefore, we need vector and its magnitude .

Let’s first find .

A vector in the direction of  is;

Therefore, for the given case;

We have found in (i) that;

We need to find .

Next, 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  6 units. 

·   It is given that is parallel to and we can see that distance of point along  from the origin is  3 units (as OC and AB are opposite/parallel sides of rectangle OABC).

·   It is given that is parallel to and we can see that distance of point along  from the origin is  4 units (as BE and OF are opposite/parallel verticals on vertex of rectangle OABC).

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 .

Hence;

Now we find magnitude of .

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

Therefore;

Now we can find unit vector in the direction of .

 iii.
 

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;

We have from (ii) 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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