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  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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