Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2013 | Oct-Nov | (P1-9709/13) | Q#4

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Question

The diagram shows a pyramid OABC in which the edge OC is vertical. The horizontal base OAB is a triangle, right-angled at O, and D is the mid-point of AB. The edges OA, OB and OC have lengths of 8 units, 6 units and 10 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 ODC.

Solution


i.
 

Tofind  consider the diagram below. It is evident from the diagram that;

We are given that and D is the mid-point of AB. Therefore;

Hence;

Now we need vectors  and .

First we 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 8 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 vector .

A vector in the direction of  is;

For the given case;

Therefore, we need the position vectors of points  and . We already have position vector of point .

To find the position vector of point .

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

Hence we need coordinates of the point .

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 6 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 we have;

Therefore we can write the vector ;

Now we have;

Therefore, we can write;

Next we are required to find the vector .

A vector in the direction of  is;

For the given case;

Therefore, we need the position vectors of points  and . We already have position vector of point .

 

To find the position vector of point .

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

Hence we need coordinates of the point .

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 10 units.

Hence, coordinates of .

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

Now 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 ODC is .

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