Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2013 | Oct-Nov | (P1-9709/11) | Q#3

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Question

The diagram shows a pyramid OABCD in which the vertical edge OD is 3 units in length. The point E is the centre of the horizontal rectangular base OABC. The sides OA and AB have lengths of 6 units and 4 units respectively. Unit vectors ,  and  are parallel to ,  and  respectively.

    i.       Express each of the vectors  and  in terms of ,  and .

   ii.       Use a scalar product to find angle BDE.

Solution


i.
 

We are required to find vectors  and

First let’s find vector .

A vector in the direction of  is;

For the given case;

Therefore, we need the position vectors of points  and .

A point  has position vector from the origin . Then the position vector of  is denoted by  or .

Hence we need coordinates of the points  and .

Let’s find the 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  is equal and parallel to   being opposite sides of rectangular base OABC. Therefore, 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 ZERO.

Hence, coordinates of .

Now we can represent the position vector of point  as follows;

Now let’s find the 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 ZERO. 

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

·       It is given that  is parallel to  and we can see that distance of point  along  from the origin is 3 units.

Hence, coordinates of .

Now we can represent the position vector of point  as follows;

Now have;

Therefore, we can write the vector ;

Now we find the vector .

A vector in the direction of  is;

For the given case;

Therefore, we need the position vectors of points  and .

A point  has position vector from the origin . Then the position vector of  is denoted by  or .

Hence we need coordinates of the points  and .

Let’s find the coordinates of the point . Consider the diagram below.

We are given that point E is the centre of the horizontal rectangular base OABC.

·       It is given that  is parallel to  and we can see that distance of point   along  from the origin is half of  i.e. 3 units. 

·       It is given that  is parallel to  and we can see  is equal and parallel to   being opposite sides of rectangular base OABC. Therefore, distance of point   along  from the origin is half of  i.e. 2 units.

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

Hence, coordinates of .

Now we can represent the position vector of point  as follows;

We already have the position vector of the point . Therefore, we have;

Therefore, we can write the vector ;


ii.
 

We recognize that  is angle between  and  .
Hence we use
scalar/dot product of  and .

From (i) we have both    and .

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 BDE is .

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