Vectors

1. The points and D have coordinates and respectively. If what are the values of and ?

 Solutions

Compute the vectors

and

The condition gives the following equations:

Solving these,

2. Points A and B have coordinates and respectively. If point is such that , what are the coordinates of ?

 Solutions

Points and . Let . The vector condition is

Since and , we have

Rearrange to solve for :

With and ,

hence

A quick check: , so , as required.

Therefore,

3. The point P has coordinates ( ).
   The point Q is such that .
   The point R has coordinates .
   For which value of is PQR an equilateral triangle?

 Hint

 Solutions

The given points are

Hence

Compute the squared length of :

Thus every side of the equilateral triangle must have squared length .

Compute squared lengths to :

For to be equilateral we require

From the first equation:

From the second equation:

The only value common to both sets is .

Check: for ,

so all three sides have squared length .

Therefore,

4. Point has coordinates ( ). Point B has coordinates ( ). Find the point C so that and are in the same direction and .

 Solutions

Compute the direction vector:

Let with . Then

Compute the length of :

Hence

So

and therefore

Quick check:

Therefore,

5. Forces and , where and are scalars, act on a box. Prove that it is not possible for their resultant force to act in the direction of .

 Solutions

Let and , where are scalars. Their resultant is

For to act in the direction of (i.e., to be parallel to and nonzero) its - and -components must vanish:

Subtracting the second equation from the first gives

so . Substituting back yields . Hence .

This means the resultant force is zero vector, and contradicts the requirement that it acts in the direction of which requires a nonzero vector.

6. The points and R have position vectors and .
   (a) Show that the triangle PQR is isosceles.

 Solutions

The position vectors of are

Compute the side vectors:

Compute squared lengths to avoid unnecessary square roots:

Since , we have

so two sides of triangle are equal. Therefore is isosceles (with vertex at ).

(b) M is the midpoint of QR . Find the position vector of M .

 Solutions

Midpoint formula:

Calculate the area of the triangle PQR .

 Hint

The area of triangle PQR is given by half of magnitude of the cross product of two side vectors:

So try to Calculate the cross product first. Then find its magnitude. Finally divide by 2 to get the area.

 Solutions

Given

Their cross product is

so its magnitude is

The area of triangle PQR is half this magnitude:

7. The points and D have position vectors and .
   (a) Write down the vector .

 Solutions

(b) Calculate the length BC

 Solutions

The position vectors are

The vector is

The length of is

Therefore the length is (approximately ).

Prove that ABCD is a rhombus.

 Hint

Calculate the vectors and and and . Then show that opposite sides are parallel by showing that and . Next calculate the lengths of and sides and show they are equal.

 Solutions

Note that

so opposite sides are parallel.

Compute the side lengths:

and

similarly .

All four sides are equal in length and opposite sides are parallel; hence ABCD is a rhombus.