# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2011 | May-Jun | (P1-9709/13) | Q#5

Hits: 467

Question

In the diagram, OABCDEFG is a rectangular block in which  cm and  cm. Unit vectors ,  and  are parallel to ,  and  respectively. The point P is the mid-point of DG, Q is the centre of the square face CBFG and R lies on AB such that AR = 4 cm.

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

ii.       Use a scalar product to calculate angle RPQ.

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 .

First, let’s find coordinates of point . Consider the diagram.

It is given that  is parallel to  and it is evident from the diagram that  is equal and parallel to  being opposite sides of rectangular block OABCDEFG. It is also given that  is the centre of the square face CBFG and  cm.  Since  is the centre of the square face CBFG, distance of point  along  from the origin is 3 units.

It is given that  is parallel to  and . It is evident from the diagram that  is equal and parallel to  being opposite sides of rectangular block OABCDEFG. It is evident from the diagram that distance of point  along  from the origin is 12 units.

It is given that  is parallel to  and . It is evident from the diagram that  is equal and parallel to  being opposite sides of rectangular block OABCDEFG. Since  is the centre of the square face CBFG, distance of point  along along  from the origin is 3 units..

Hence, coordinates of .

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

Now, let’s find coordinates of point . Consider the diagram.

It is given that  is parallel to  and it is evident from the diagram that distance of point  along  from the origin is ZERO.

It is given that  is parallel to  and . It is evident from the diagram that  is equal and parallel to  and  being opposite sides of rectangular block OABCDEFG. Since it is given point  is the mid-point of DG, distance of point  along  from the origin is 6 units.

It is given that  is parallel to  and . It is evident from the diagram that  is equal and parallel to  being opposite sides of rectangular block OABCDEFG. Therefore, distance of point  along  from the origin is 6 units.

Hence, coordinates of .

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

Now have;

Therefore, we can write the vector  as;

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 .  We already have found the coordinates and position vector of point .

So, let’s find coordinates of point . Consider the diagram.

It is given that  is parallel to  and 6.

I
t is evident from the diagram that distance of point  along  from the origin is 6 units.

It is given that  is parallel to  and it is evident from the diagram that  is parallel to  being
opposite sides of rectangular block
OABCDEFG. It is also given that AR = 4 cm. Hence, it is evident from the diagram that distance of point  along  from the origin is 4 units.

It is given that  is parallel to  and t is evident from the diagram that  distance of point  along along  from the origin is ZERO.

Hence, coordinates of .

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

Now have;

Therefore, we can write the vector  as;

ii.

We recognize that angle RQP is between  and . Therefore, we need the scalar/dot product of these two to find angle RQP.

From (i) we have;

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

Therefore, for the given case;

Therefore;

Hence;

Now we can equate the two equations of ;

Hence the angle RPQ is .