# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | May-Jun | (P1-9709/13) | Q#9

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Question

The diagram shows a pyramid OABCD with a horizontal rectangular base OABC.  The sides OA and AB have lengths of 8 units and 6 units respectively. The point E on  OB is such that OE = 2 units. The point D of the pyramid is 7 units vertically  above E. Unit vectors ,  and  are parallel to ,   and   respectively.

i.
Show that .

ii.    Use a scalar product to find angle BDO.

Solution

i.

We are given length / magnitude of  as;

We are required to find  as given.

A unit vector in the direction of  is;

Therefore, we can write that a vector with given length / magnitude and the  direction of a given vector / unit vector as;

Hence, for ;

We are already given  but we need to find .

To find consider the diagram below.

It is evident from the diagram that  is parallel to .

Therefore, unit vector of  and  is same.

Hence, first we find unit vector of .

A unit vector in the direction of  is;

Therefore, to find the unit vector of we need length /  magnitude and vector  .

First we find and then its length /  magnitude.

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 8 units.  (OABC being a rectangular base)

·       It is given that is parallel to ( and hence, , OABC being a rectangular  base) and we can see that distance of point along  from the origin is 6. Hence,  distance of of point along  from the origin is also 6.

·       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 need to find the unit vector of .

Hence, unit vector of can be found as;

We know that same is the unit vector of .

Finally, we can find the vector .

ii.

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

First let us find the vector . Consider the diagram below.

It is evident from the diagram that;

From (i), we have;

Next we find .

We are given that from point  to , the distance is covered only vertically, along and is 7 units, and no distance is covered along  and .

Hence;

Therefore;

Next we need to find .

Consider the diagram below.

From (i), we have;

Therefore;

We have also found above that;

Hence;

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 BDO is .