Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Feb-Mar | (P1-9709/12) | 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  OC 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 is 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  and .

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 andis 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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