Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2016  FebMar  (P19709/12)  Q#7
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Question
The diagram shows a pyramid OABC with a horizontal triangular base OAB and vertical height OC. Angles AOB, BOC and AOC are each right angles. Unit vectors , and are parallel to OA, OB and OC respectively, with OA = 4 units, OB = 2.4 units and OC = 3 units. The point P on CA is such that CP = 3 units.
i. Show that .
ii. Express and in terms of , and .
iii. Use a scalar product to find angle BPC.
Solution
i.
It is evident from the diagram that from the side view we can see a right triangle AOC with angle AOC given as right angle.
Pythagorean Theorem
For right triangle AOC;
It is evident from the diagram that;
We can also write it as;
Therefore;
Now we need to workout .
A vector in the direction of is;
For the given case;
Therefore, we need the position vectors of points and .
A vector in the direction of is;
For the given case;
Therefore, we need the position vectors of points and .
To find consider the diagram below.
Now we need vector . Since this is the position vector of point , we need coordinates of the point .
· 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 .
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 0 units.
· It is given that is vertically upwards (parallel to )and we can see that distance of point along from the origin is 3 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 can find .
As we have seen above;
Therefore;
ii.
We are required to find and .
Let’s first find . Consider the diagram below.
It is evident from the diagram that;
We have seen in (i) that;
It is evident from the diagram that;
Therefore;
We can also write it as;
Therefore;
Hence;
We already have found in (i) that;
Hence;
Next we are required to find . Consider the diagram below.
It s evident from the diagram that;
We have found above that;
Let’s find .
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 2.4 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 .
Hence;
iii.
We are required to find the angle BPC.
It is evident from the diagram that angle BPC is between
Therefore, we use scalar/dot product of and to find angle BPC.
From (i) and (ii) 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 BPC is .
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