Math Problem Statement

Soit un repère orthonormé (O; i⃗, j⃗, k⃗). Soit les points A(3, -3, 7), B(-4, -6, 6) et C(24, y, 10). Donner la valeur de y pour que A, B et C soient alignés.

Solution

The problem asks to find the value of yy such that the points A(3,3,7)A(3, -3, 7), B(4,6,6)B(-4, -6, 6), and C(24,y,10)C(24, y, 10) are collinear (aligned in a straight line).

Step-by-step solution:

To determine if three points are collinear, the vectors AB\overrightarrow{AB} and AC\overrightarrow{AC} must be proportional. In other words, AB\overrightarrow{AB} and AC\overrightarrow{AC} must be linearly dependent, meaning their cross product must be the zero vector.

1. Find the vectors:

  • AB=BA=(43,6(3),67)=(7,3,1)\overrightarrow{AB} = B - A = (-4 - 3, -6 - (-3), 6 - 7) = (-7, -3, -1)
  • AC=CA=(243,y(3),107)=(21,y+3,3)\overrightarrow{AC} = C - A = (24 - 3, y - (-3), 10 - 7) = (21, y + 3, 3)

2. Take the cross product of AB\overrightarrow{AB} and AC\overrightarrow{AC}:

The cross product of two vectors u=(u1,u2,u3)\mathbf{u} = (u_1, u_2, u_3) and v=(v1,v2,v3)\mathbf{v} = (v_1, v_2, v_3) is given by: u×v=(u2v3u3v2,u3v1u1v3,u1v2u2v1)\mathbf{u} \times \mathbf{v} = (u_2 v_3 - u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2 - u_2 v_1)

For AB=(7,3,1)\overrightarrow{AB} = (-7, -3, -1) and AC=(21,y+3,3)\overrightarrow{AC} = (21, y + 3, 3): AB×AC=((3)(3)(1)(y+3),(1)(21)(7)(3),(7)(y+3)(3)(21))\overrightarrow{AB} \times \overrightarrow{AC} = \left( (-3)(3) - (-1)(y + 3), (-1)(21) - (-7)(3), (-7)(y + 3) - (-3)(21) \right)

This simplifies to: AB×AC=(9+y+3,21+21,7(y+3)+63)\overrightarrow{AB} \times \overrightarrow{AC} = (-9 + y + 3, -21 + 21, -7(y + 3) + 63) =(y6,0,7y21+63)= (y - 6, 0, -7y - 21 + 63) =(y6,0,7y+42)= (y - 6, 0, -7y + 42)

3. Set the cross product equal to the zero vector:

For AA, BB, and CC to be collinear, the cross product must equal (0,0,0)(0, 0, 0). Therefore, we have the following system of equations: y6=0and7y+42=0y - 6 = 0 \quad \text{and} \quad -7y + 42 = 0

From y6=0y - 6 = 0, we get: y=6y = 6

Conclusion:

The value of yy for which the points AA, BB, and CC are collinear is y=6y = 6.


Would you like further details or have any questions? Here are five related questions to expand on this concept:

  1. How can we verify if a set of three points in 2D are collinear?
  2. What does it mean for vectors to be linearly dependent in 3D space?
  3. Could the collinearity condition be solved using determinants?
  4. How does the cross product help in detecting collinearity of points?
  5. What is the geometric interpretation of the cross product?

Tip: The cross product is a powerful tool to detect the orientation and collinearity of vectors in 3D space.

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Math Problem Analysis

Mathematical Concepts

Analytic Geometry
Vectors
Collinearity
Linear Dependence

Formulas

Vector Subtraction: \( \overrightarrow{AB} = B - A \)
Cross Product: \( \mathbf{u} \times \mathbf{v} = (u_2 v_3 - u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2 - u_2 v_1) \)

Theorems

Vectors are collinear if and only if their cross product is the zero vector.
Linear Dependence Theorem

Suitable Grade Level

Grades 10-12