Askiitians Tutor Team
Last Activity: 4 Months ago
To find the values of a, b, and c in the equation of the line that passes through the point M(1, 1, 1) and is perpendicular to the line of intersection of the given planes, we need to follow a few logical steps. Let's break this down step by step.
Step 1: Determine the Normal Vectors of the Planes
The equations of the planes are:
- Plane 1: x + 2y - 4z = 0
- Plane 2: 2x - y + 2z = 0
The normal vector of a plane given by the equation Ax + By + Cz = 0 is (A, B, C). Therefore:
- Normal vector of Plane 1: N1 = (1, 2, -4)
- Normal vector of Plane 2: N2 = (2, -1, 2)
Step 2: Find the Direction Vector of the Line of Intersection
The direction vector of the line of intersection of the two planes can be found using the cross product of their normal vectors:
Let N1 = (1, 2, -4) and N2 = (2, -1, 2). The cross product is calculated as follows:
- N1 × N2 = |i j k|
- |1 2 -4|
- |2 -1 2|
Calculating this determinant gives:
- i(2*2 - (-4)(-1)) - j(1*2 - (-4)(2)) + k(1*(-1) - 2*2)
- i(4 - 4) - j(2 + 8) + k(-1 - 4)
- 0i - 10j - 5k
Thus, the direction vector of the line of intersection is (0, -10, -5).
Step 3: Find the Direction Vector of the Required Line
The line we are looking for must be perpendicular to the line of intersection. Therefore, its direction vector must be orthogonal to (0, -10, -5). A vector (a, b, c) is orthogonal to (0, -10, -5) if:
0*a + (-10)*b + (-5)*c = 0, which simplifies to:
-10b - 5c = 0. This can be rearranged to:
2b + c = 0, or c = -2b.
Step 4: Express the Direction Vector
We can express the direction vector of the line in terms of b:
(a, b, c) = (a, b, -2b). To maintain the ratio a:b:c, we can set b = 1 for simplicity:
(a, 1, -2).
Step 5: Find the Ratios
Now we can express the ratios:
a : b : c = a : 1 : -2. Since we don't have a specific value for a, we can denote it as k, where k is any non-zero scalar. Thus:
For the simplest case, if we let a = 1, we have:
1 : 1 : -2.
Final Result
Therefore, the ratio a:b:c equals 1 : 1 : -2.