Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2003 | Oct-Nov | (P1-9709/01) | Q#7

Hits: 870

Question

The diagram shows a triangular prism with a horizontal rectangular base , where  units and  units. The vertical ends  and  are isosceles triangles with  units. The mid-points of  and  are  and  respectively. The origin  is at the mid-point of .

Unit vectors ,  and  are parallel to ,  and  respectively.


i.       
Find the length of OB.

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

  iii.        Evaluate  and hence find angle , giving your answer correct to the nearest degree.

Solution


i.
 

To find the length of , we utilize the fact that  and  are isosceles triangles. We consider .

We are given that;

Therefore;

We know that origin  is at the mid-point of , therefore;

Applying Pythagoras theorem to , as shown below.

Pythagorean Theorem

For the given case;

Hence, length of  is 4 units.


ii.
 

First we find .

It is evident from the diagram that;

Now we need vectors  and .

Since these are position vectors of points  and , we need  coordinates of the points  and .

First we find the coordinates of the point  and its position vectors.

Consider the diagram below.


It is given that  is parallel to  and we can see that distance of point  along  from the origin is half the distance  because origin  is at the mid-point of . Since vertical ends  and  are isosceles triangles with  and we are also given that  units.
Therefore,
distance of point  along  from the origin is 3 units. 

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.

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 find the coordinates of the point  and its position vectors.


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 half the distance  because point  is at the mid-point of . Since , 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 . From (i), we have . Therefore, distance of point   from origin along  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 have;

Hence, we can write;

Now we find out .

It is evident from the diagram that;

Now we need vectors  and . We have already found ;

We need to find . Since it is position vector of point , we need  coordinates of the point   and . Therefore, we find the coordinates of the point  and its position vector. 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 the distance  and we are given that . Therefore, distance of point  along  from the origin is 12 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 .

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

Now we have;

Therefore, we can write;


iii.
 

We recognize that  is a scalar/dot product of and .

We have found in (ii) that;

The scalar or dot product of two vectors  and  in component form is given as;

Since ;

Therefore for the given case;

To find the angle .

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;

Hence the angle CMN is .

Comments