# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2019 | Oct-Nov | (P1-9709/13) | Q#10

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Question

Relative to an origin O, the position vectors of points A, B and X are given by

 and

i.
Find and show that AXB is a straight line.

The position vector of point C is given by .

ii.
Show that CX is perpendicular to AX.

iii.
Find the area of triangle ABC.

Solution

i.

First, we are required to find .

A vector in the direction of  is;

For the given case;

We are given that;

Next, we are required to show that AXB is a straight line.

It is evident that if AXB is straight line then angle between AX and BX must be .

Let’s first find BX and then angle between AX and BX.

A vector in the direction of  is;

For the given case;

We are given that;

Now, to find the angle between  and we find scalar/dot product of these two vectors.

The scalar or dot product of two vectors and in component form is given as;

Since;

For the given case;

Scalar/Dot product is also defined as below.

The scalar or dot product of two vectors andis number or scalar, where is the  angle between the directions of and.

For;

Therefore, we need to find  and .

Expression for the length (magnitude) of a vector is;

Therefore;

Hence;

Equating both scalar/dot products found above;

Therefore;

Hence, AXB is a straight line.

ii.

We are required to show that CX is perpendicular to AX.

If and  & , then andare perpendicular.

We need to show that scalar/dot product of CX and AX is ZERO.

Let’s first find CX and then scalar/dot product of CX and AX.

A vector in the direction of  is;

For the given case;

We are given that;

Now, we find scalar/dot product of these two vectors.

The scalar or dot product of two vectors and in component form is given as;

Since;

For the given case;

Hence, the two vectors  are perpendicular to each other.

iii.

We are required to find the area of triangle ABC.

As per given conditions, AXB is a straight line and CX is perpendicular to AX, we can construct following triangle ABC.

We are required to find the area of this triangle.

Expression for the area of the triangle is;

For the given case;

We need to first find .

A vector in the direction of  is;

For the given case;

We are given that;

We havea lready found above that;

Therefore, we need to find  and .

Expression for the length (magnitude) of a vector is;

Therefore;

Hence;