Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2016  OctNov  (P19709/11)  Q#9
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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 = 90^{o}.
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 .
We have already found above;
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 ;
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