Vectors in a Plane and  Space
      The vector product and the mixed product use, examples
The vector product and the mixed product use examples
Example:   Given are vectors, a = i - 2 k and  b = - i + 3 j + k, determine the vector  ca ´ b and the area of a parallelogram formed by vectors, a and b.
Solution:
   
The area of the parallelogram,  
or
Example:   Vertices of a triangle are, A(-1, 0, 1), B(3, -2, 0) and C(4, 1, -2), find the length of the altitude hb.
Solution:  Using the area of a triangle,
as the area of the triangle where
 
then, the area of the triangle,  
and the length of the side b,  
thus, the length of the altitude,
 
Example:  Find the angles that the unit vector, which is orthogonal to the plane formed by vectors, 
a = -i + 2 j + k  and  b = 3i - 2 j + 4k, makes with the coordinate axes.
Solution:  The unit vector which is orthogonal to the plane, formed by the vectors, a and b, is the direction vector of a vector c, such that
 
Angles, which the unit vector c°  forms with coordinate axes, we find by using the formula,
Example:  Given are vectors, a = -2i - 3 jb = -i - 2 j + 3k and c = -i + 2 j + k, find the projection (the scalar component) of the vector a onto vector d = b ´ c.
Solution:  Let first find the vector d,
Example:  Given are vectors, a = -3i + 4 j - l kb = 2i -  j + k  and  c = i - 4 j -3l k, determine the 
parameter
l such that vectors to be coplanar.
Solution:  Vectors lie on the same plane if their scalar triple product is zero, i.e., V = 0, therefore vectors’
coordinates must satisfy the condition,
Example:  Examine if vectors, a = 4i + 2 + kb = 3i + 3 j  - 2k  and  c = - 5i -  j - 4k, are coplanar and if so, prove their linear dependence.
Solution:  Check for coplanarity,
Since vectors are coplanar each of them can be represented as linear combination of other two.
Example:  Prove that points, A(-2, 1, 0), B(3, -2, -1), C(1, -1, 2) and D(-3, 2, -5), all lie on the same plane.
Solution:  If all given points belong to the same plane then vectors, AB, AC and AD are coplanar, therefore the scalar triple product (AB ´ AC) · AD = 0.
thus, all given points lie on the same plane.
Pre-calculus contents I
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