Let be the set of the three-dimensional real column vectors and be the set of the three-by-three real matrices. Let , , and be linearly independent unit-length vectors and be a unit-length vector not parallel to , , or . Let and be square matrices defined as
Here, and denote the transpose of a matrix and a vector , respectively. Answer the following questions.
(1) Find the condition for such that the rank of is three.
(2) In the three-dimensional Euclidean space , consider four planes is a real number, and that satisfy the following three conditions: (i) the rank of is three, (ii) is not the empty set, and (iii) there exists a sphere to which are tangent. The position vector of the center of is represented by using a vector . Express using .
(3) Show that is a positive definite symmetric matrix.
(4) Consider the point from which the sum of squared distances to four planes is a real number, and is minimized. The position vector of is represented by using a vector . Express using and .
(5) Let be a straight line through a point , the position vector of which is , parallel to in . Let be the orthogonal projection of an arbitrary point , the position vector of which is , onto . The position vector of is represented by using a matrix . The identity matrix is denoted by .
(a) Express using and .
(b) Show that .
() Consider a plane is a non-zero vector, and is a real number). Let be the point from which the sum of squared distances to , , and is minimized. When , , and are orthogonal to each other, the position vector of is represented by . using a vector which is independent of and . Express using and .
we need to determine the conditions on that ensure has a rank of three.
Let's assume that the third row of matrix can be written as a linear combination of the first two rows. Thus, we assume:
where and are some scalars. Substituting as a linear combination of , , and , we have:
Expanding and rearranging the equation, we get:
Grouping like terms:
For this equation to hold for arbitrary vectors , , and , the coefficients of each vector must match:
For :
For :
For :
Thus, we have the following system of equations:
Substituting into the last equation:
So, the conditions under which the third row of can be written as a linear combination of the first two rows (i.e., the matrix would not have full rank) are:
For the matrix to have full rank (rank 3), must be such that the above condition does not hold. Therefore, the condition for the rank of to be three is:
where , and are unit vectors, and and are real numbers. The center of a sphere tangent to all four planes is represented as , where is a 3x3 matrix, and is what we need to find.
First, we show that is symmetric. Since each term is symmetric (as the outer product of a vector with itself is symmetric), their sum is also symmetric.
Next, to prove that is positive definite, we need to show that for any non-zero vector , the quadratic form .
Since each term is nonnegative, we have
Now suppose
Then
so in particular
Because are linearly independent in , they form a basis of . Therefore the only vector orthogonal to all three is the zero vector, so , contradicting the assumption that .
The sum of squared distances from a point to the four planes is minimized when is the point of orthogonal projection of the origin onto these planes. The squared distance from a point with position vector to the plane is given by:
Since are unit vectors (), this simplifies to:
The sum of squared distances to all four planes is:
To minimize , we take the gradient with respect to and set it equal to zero:
This equation can be rearranged into the form:
The matrix is defined as:
which is a matrix. Therefore, the position vector that minimizes the sum of squared distances can be expressed as:
Given that is the point minimizing the sum of squared distances, its position vector is , where is defined by: