Answer :
To find the points of intersection of the line joining the points [tex]\((2, 4, 5)\)[/tex] and [tex]\((3, 5, -4)\)[/tex] with the specified planes, we need to examine the line in parametric form and solve for the parameter [tex]\( t \)[/tex] where the line intersects each plane.
Step-by-step solution:
1. First, determine the direction vector of the line:
[tex]\[ \text{Direction vector} = (3 - 2, 5 - 4, -4 - 5) = (1, 1, -9) \][/tex]
2. Express the line in parametric form:
[tex]\[ (x, y, z) = (2, 4, 5) + t \cdot (1, 1, -9) \][/tex]
So,
[tex]\[ x = 2 + t, \quad y = 4 + t, \quad z = 5 - 9t \][/tex]
### (a) Intersection with the xy-plane (where [tex]\( z = 0 \)[/tex])
To find the intersection with the xy-plane, set [tex]\( z = 0 \)[/tex]:
[tex]\[ 5 - 9t = 0 \implies t = \frac{5}{9} \][/tex]
Substitute [tex]\( t = \frac{5}{9} \)[/tex] back into the parametric equations for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = 2 + \frac{5}{9} = \frac{18}{9} + \frac{5}{9} = \frac{23}{9} \approx 2.56 \][/tex]
[tex]\[ y = 4 + \frac{5}{9} = \frac{36}{9} + \frac{5}{9} = \frac{41}{9} \approx 4.56 \][/tex]
Thus, the point of intersection with the xy-plane is approximately:
[tex]\[ \left(\frac{23}{9}, \frac{41}{9}, 0\right) \approx (2.56, 4.56, 0) \][/tex]
### (b) Intersection with the yz-plane (where [tex]\( x = 0 \)[/tex])
To find the intersection with the yz-plane, set [tex]\( x = 0 \)[/tex]:
[tex]\[ 2 + t = 0 \implies t = -2 \][/tex]
Substitute [tex]\( t = -2 \)[/tex] back into the parametric equations for [tex]\( y \)[/tex] and [tex]\( z \)[/tex]:
[tex]\[ y = 4 + (-2) = 2 \][/tex]
[tex]\[ z = 5 - 9(-2) = 5 + 18 = 23 \][/tex]
Thus, the point of intersection with the yz-plane is:
[tex]\[ (0, 2, 23) \][/tex]
### (c) Intersection with the zx-plane (where [tex]\( y = 0 \)[/tex])
To find the intersection with the zx-plane, set [tex]\( y = 0 \)[/tex]:
[tex]\[ 4 + t = 0 \implies t = -4 \][/tex]
Substitute [tex]\( t = -4 \)[/tex] back into the parametric equations for [tex]\( x \)[/tex] and [tex]\( z \)[/tex]:
[tex]\[ x = 2 + (-4) = -2 \][/tex]
[tex]\[ z = 5 - 9(-4) = 5 + 36 = 41 \][/tex]
Thus, the point of intersection with the zx-plane is:
[tex]\[ (-2, 0, 41) \][/tex]
### Summary of Intersection Points
1. The intersection with the xy-plane is:
[tex]\[ \left(\frac{23}{9}, \frac{41}{9}, 0\right) \approx (2.56, 4.56, 0) \][/tex]
2. The intersection with the yz-plane is:
[tex]\[ (0, 2, 23) \][/tex]
3. The intersection with the zx-plane is:
[tex]\[ (-2, 0, 41) \][/tex]
Step-by-step solution:
1. First, determine the direction vector of the line:
[tex]\[ \text{Direction vector} = (3 - 2, 5 - 4, -4 - 5) = (1, 1, -9) \][/tex]
2. Express the line in parametric form:
[tex]\[ (x, y, z) = (2, 4, 5) + t \cdot (1, 1, -9) \][/tex]
So,
[tex]\[ x = 2 + t, \quad y = 4 + t, \quad z = 5 - 9t \][/tex]
### (a) Intersection with the xy-plane (where [tex]\( z = 0 \)[/tex])
To find the intersection with the xy-plane, set [tex]\( z = 0 \)[/tex]:
[tex]\[ 5 - 9t = 0 \implies t = \frac{5}{9} \][/tex]
Substitute [tex]\( t = \frac{5}{9} \)[/tex] back into the parametric equations for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = 2 + \frac{5}{9} = \frac{18}{9} + \frac{5}{9} = \frac{23}{9} \approx 2.56 \][/tex]
[tex]\[ y = 4 + \frac{5}{9} = \frac{36}{9} + \frac{5}{9} = \frac{41}{9} \approx 4.56 \][/tex]
Thus, the point of intersection with the xy-plane is approximately:
[tex]\[ \left(\frac{23}{9}, \frac{41}{9}, 0\right) \approx (2.56, 4.56, 0) \][/tex]
### (b) Intersection with the yz-plane (where [tex]\( x = 0 \)[/tex])
To find the intersection with the yz-plane, set [tex]\( x = 0 \)[/tex]:
[tex]\[ 2 + t = 0 \implies t = -2 \][/tex]
Substitute [tex]\( t = -2 \)[/tex] back into the parametric equations for [tex]\( y \)[/tex] and [tex]\( z \)[/tex]:
[tex]\[ y = 4 + (-2) = 2 \][/tex]
[tex]\[ z = 5 - 9(-2) = 5 + 18 = 23 \][/tex]
Thus, the point of intersection with the yz-plane is:
[tex]\[ (0, 2, 23) \][/tex]
### (c) Intersection with the zx-plane (where [tex]\( y = 0 \)[/tex])
To find the intersection with the zx-plane, set [tex]\( y = 0 \)[/tex]:
[tex]\[ 4 + t = 0 \implies t = -4 \][/tex]
Substitute [tex]\( t = -4 \)[/tex] back into the parametric equations for [tex]\( x \)[/tex] and [tex]\( z \)[/tex]:
[tex]\[ x = 2 + (-4) = -2 \][/tex]
[tex]\[ z = 5 - 9(-4) = 5 + 36 = 41 \][/tex]
Thus, the point of intersection with the zx-plane is:
[tex]\[ (-2, 0, 41) \][/tex]
### Summary of Intersection Points
1. The intersection with the xy-plane is:
[tex]\[ \left(\frac{23}{9}, \frac{41}{9}, 0\right) \approx (2.56, 4.56, 0) \][/tex]
2. The intersection with the yz-plane is:
[tex]\[ (0, 2, 23) \][/tex]
3. The intersection with the zx-plane is:
[tex]\[ (-2, 0, 41) \][/tex]
