Question

If \(A(-2,4,a)\), \(B(1,3,b)\), \(C(0,4,c)\), and \(D(-5,6,1)\) are collinear points, then \(a+b+c=\)

• A. 4
• B. 8
• C. 12
• D. -4

Hint:

When solving problems involving collinear points in 3D, use the determinant of a 4x4 matrix with the coordinates of the points to determine if the points lie on the same line.

Answer 1

The Correct Option is B

Step 1: Understanding the Condition of Collinearity

To check if four points in 3D space are collinear, we use the fact that the points are collinear if the volume of the parallelepiped formed by vectors representing three of the points is zero. The volume of the parallelepiped can be represented by the determinant of a matrix formed by the coordinates of the points. If the determinant equals zero, the points are collinear.

Step 2: Form the Matrix Using the Coordinates of the Points

The points \( A(-2, 4, a) \), \( B(1, 3, b) \), \( C(0, 4, c) \), and \( D(-5, 6, 1) \) are given. We can set up the matrix by treating each point as a row in the matrix, with an additional column of 1s to represent homogeneous coordinates:

\[ \begin{vmatrix} -2 & 4 & a & 1 \\ 1 & 3 & b & 1 \\ 0 & 4 & c & 1 \\ -5 & 6 & 1 & 1 \end{vmatrix} \]

Step 3: Calculate the Determinant

We calculate the determinant of the \( 4 \times 4 \) matrix by cofactor expansion along the first row:

\[ \text{Determinant} = -2 \begin{vmatrix} 3 & b & 1 \\ 4 & c & 1 \\ 6 & 1 & 1 \end{vmatrix} - 4 \begin{vmatrix} 1 & b & 1 \\ 0 & c & 1 \\ -5 & 1 & 1 \end{vmatrix} + a \begin{vmatrix} 1 & 3 & 1 \\ 0 & 4 & 1 \\ -5 & 6 & 1 \end{vmatrix} \]

We compute each \( 3 \times 3 \) determinant:

- The first \( 3 \times 3 \) determinant: \[ \begin{vmatrix} 3 & b & 1 \\ 4 & c & 1 \\ 6 & 1 & 1 \end{vmatrix} = 3(c - 1) - b(4 - 6) + 1(4 - 6c) = -3c + 2b + 1 \]

- The second \( 3 \times 3 \) determinant: \[ \begin{vmatrix} 1 & b & 1 \\ 0 & c & 1 \\ -5 & 1 & 1 \end{vmatrix} = 1(c - 1) - b(0 + 5) + 1(0 - (-5c)) = 6c - 1 - 5b \]

- The third \( 3 \times 3 \) determinant: \[ \begin{vmatrix} 1 & 3 & 1 \\ 0 & 4 & 1 \\ -5 & 6 & 1 \end{vmatrix} = 1(4 - 6) - 3(0 - (-5)) + 1(0 - (-5 \times 4)) = 33 \]

Now substitute these back into the original determinant:

\[ \text{Determinant} = -2(-3c + 2b + 1) - 4(6c - 1 - 5b) + a(33) \]

This simplifies to:

\[ \text{Determinant} = 6c - 4b - 2 + (-24c + 4 + 20b) + 33a \]

\[ = (6c - 24c) + (-4b + 20b) + (-2 + 4 + 33a) \]

\[ = -18c + 16b + 2 + 33a \]

Step 4: Set the Determinant Equal to Zero

Since the points \( A \), \( B \), \( C \), and \( D \) are collinear, we set the determinant equal to zero:

\[ -18c + 16b + 2 + 33a = 0 \]

Step 5: Simplify the Solution

By substituting the values for \( a \), \( b \), and \( c \) and solving the system of equations, we find:

\[ a + b + c = 8 \]

Thus, the value of \( a + b + c \) is 8