Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2003 | Oct-Nov | (P1-9709/01) | Q#7
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Question
The diagram shows a triangular prism with a horizontal rectangular base , where units and units. The vertical ends and are isosceles triangles with units. The mid-points of and are and respectively. The origin is at the mid-point of .
Unit vectors , and are parallel to , and respectively.
i.Find the length of OB.
ii.Express each of the vectors and in terms of , and .
iii.Evaluate and hence find angle , giving your answer correct to the nearest degree.
Solution
i.
To find the length of , we utilize the fact that and are isosceles triangles. We consider .
We are given that;
Therefore;
We know that origin is at the mid-point of , therefore;
Applying Pythagoras theorem to , as shown below.
Pythagorean Theorem
For the given case;
Hence, length of is 4 units.
ii.
First we find .
It is evident from the diagram that;
Now we need vectors and .
Since these are position vectors of points and , we need coordinates of the points and .
First we find the coordinates of the point and its position vectors.
Consider the diagram below.
It is given that is parallel to and we can see that distance of point along from the origin is half the distance because origin is at the mid-point of . Since vertical ends and are isosceles triangles with and we are also given that units.
Therefore, distance of point along from the origin is 3 units.
It is given that is parallel to and we can see thatdistance of point along from the origin is ZERO.
It is given that is parallel to and we can see thatdistance 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 the coordinates of the point and its position vectors.
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 thatdistance of point along from the origin is half the distance because point is at the mid-point of . Since , distance of point along from the origin is 6 units.
It is given that is parallel to and we can see thatdistance of point along from the origin is . From (i), we have. Therefore, distance of point from origin along is 4 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 have;
Hence, we can write;
Now we find out .
It is evident from the diagram that;
Now we need vectors and . We have already found ;
We need to find . Since it is position vector of point , we need coordinates of the point and . Therefore, we find the coordinates of the point and its position vector. 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 thatdistance of point along from the origin is the distance and we are given that . Therefore, distance of point along from the origin is 12 units.
It is given that is parallel to and we can see thatdistance of point along from the origin is ZERO.
Hence, coordinates of .
A point has position vector from the origin . Then the position vector of is denoted by or .
Now we have;
Therefore, we can write;
iii.
We recognize that is a scalar/dot product of and .
We have found in (ii) that;
The scalar or dot product of two vectors and in component form is given as;
Since;
Therefore for the given case;
To find the angle .
The scalar or dot product of two vectors andis number or scalar, where is the angle between the directions of and.
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