# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Oct-Nov | (P1-9709/11) | Q#9

Question

The diagram shows a cuboid OABCDEFG with a horizontal base OABC in which OA = 4 cm and  AB = 15 cm. The height OD of the cuboid is 2 cm. The point X on AB is such that AX = 5 cm and  the point P on DG is such that DP = p cm, where p is a constant. Unit vectors ,  and  are parallel to OA, OC and OD respectively.

i.       Find the possible values of p such that angle OPX = 90o.

ii.       For the case where p = 9, find the unit vector in the direction of .

iii.       A point Q lies on the face CBFG and is such that XQ is parallel to AG. Find .

Solution

i.

It is evident that angle OPX is between  and .

We are given that angle OPX is .

If  and  & , then  and  are perpendicular.

Therefore, we need scalar/dot product of  and ;

Let’s find  and .

First we workout .

To find  consider the diagram below.

·       It is given that  is parallel to  and we can see that distance of point  along  from the origin is  0 units.

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

·       It is given that  is vertically upwards (parallel to )and we can see that distance of point  along  from the origin is 2 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 .

Next we find out .

A vector in the direction of  is;

For the given case;

Therefore, we need the position vectors of points  and .

Let’s find position vectors of point .

To find  consider the diagram below.

·       It is given that  is parallel to  and we can see that 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  5 units.

·       It is given that  is vertically upwards (parallel to )and we can see that distance of point   along  from the origin is 0 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 .

For the given case;

We have found;

 and

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

Since ;

For the given case;

Now we have two options.

ii.

We are required to find unit vector in the direction of .

A unit vector in the direction of  is;

Therefore, we need vector  and its magnitude  to find unit vector in the direction of .

First we find the vector .

A vector in the direction of  is;

For the given case;

We have found in (i);

Here we are given that p=9.

Therefore;

Next we need magnitude of  to find unit vector in the direction of .

Hence;

iii.

We are required to find  which is parallel to .

Let’s first find .

A vector in the direction of  is;

For the given case;

Therefore, we need the position vectors of points  and .

Let’s find position vector of point

To find  consider the diagram below.

·       It is given that  is parallel to  and we can see that distance of point  along  from the origin is  0 units.

·       It is given that  is parallel to  and we can see that distance of point  along  from the origin is  15 units.

·       It is given that  is vertically upwards (parallel to ) and we can see that distance of point   along  from the origin is 2 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 .

Let’s now find position vector of point .

To find  consider the diagram below.

·       It is given that  is parallel to  and we can see that 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  0 units.

·       It is given that  is vertically upwards (parallel to )and we can see that distance of point   along  from the origin is 0 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 .

For the given case;

Next we need to find vector of .

A vector in the direction of  is;

For the given case;

Therefore, we need the position vectors of points  and

We have already found position vector of point  in (i);

Let’s find position vector of point .

To find  consider the diagram below. We are given that point Q lies on the face CBFG.

·       It is given that  is parallel to  and we can see that distance of point  along  from the origin is  x units (because it can be anywhere on the face CBFG.

·       It is given that  is parallel to  and we can see that distance of point  along  from the origin is  15 units (because it lies on the face CBFG which is 15 units from the origin along .)

·       It is given that  is vertically upwards (parallel to )and we can see that distance of point  along  from the origin is z 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 .

For the given case;

Finally;

From this we can write three following equations;

It is evident that   can be solved for .

Substituting  in other two equations we can also find  and .

Hence, we can get  by substituting these x and y values in following expression of ;