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What is the distance between the planes x – 2y + z – 1 = 0 and – 3x + 6y – 3z + 2 = 0?
- a)3 units
- b)1 unit
- c)0 unit
- d)2 units
Correct answer is option 'D'. Can you explain this answer?
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What is the distance between the planes x 2y + z 1 = 0 and 3x + 6y ...
Given planes are:
x + 2y + z + 1 = 0
3x + 6y + 3z + 2 = 0
To find the distance between two planes, we need to find the perpendicular distance between them.
Step 1: Find the normal vectors of both planes.
The normal vector of the first plane is (1, 2, 1).
The normal vector of the second plane is (3, 6, 3).
Step 2: Find the dot product of the normal vectors.
(1, 2, 1)·(3, 6, 3) = 3 + 12 + 3 = 18
Step 3: Find the magnitude of the normal vector of one of the planes.
| (1, 2, 1) | = √(1² + 2² + 1²) = √6
Step 4: Use the formula to find the distance between the planes.
Distance = | ax + by + cz + d | / √(a² + b² + c²)
Here, a = 1, b = 2, c = 1, and d = -1 (taking any point on the first plane)
So, the distance = | (1)(-1) + (2)(-1) + (1)(-1) + 1 | / √(1² + 2² + 1²) = 2/√6 units.
Therefore, the distance between the given planes is 2/√6 units, which is approximately equal to 0.82 units. So, the correct answer is option (D) 2 units.
x + 2y + z + 1 = 0
3x + 6y + 3z + 2 = 0
To find the distance between two planes, we need to find the perpendicular distance between them.
Step 1: Find the normal vectors of both planes.
The normal vector of the first plane is (1, 2, 1).
The normal vector of the second plane is (3, 6, 3).
Step 2: Find the dot product of the normal vectors.
(1, 2, 1)·(3, 6, 3) = 3 + 12 + 3 = 18
Step 3: Find the magnitude of the normal vector of one of the planes.
| (1, 2, 1) | = √(1² + 2² + 1²) = √6
Step 4: Use the formula to find the distance between the planes.
Distance = | ax + by + cz + d | / √(a² + b² + c²)
Here, a = 1, b = 2, c = 1, and d = -1 (taking any point on the first plane)
So, the distance = | (1)(-1) + (2)(-1) + (1)(-1) + 1 | / √(1² + 2² + 1²) = 2/√6 units.
Therefore, the distance between the given planes is 2/√6 units, which is approximately equal to 0.82 units. So, the correct answer is option (D) 2 units.
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What is the distance between the planes x 2y + z 1 = 0 and 3x + 6y 3z + 2 = 0?a)3 unitsb)1 unitc)0 unitd)2 unitsCorrect answer is option 'D'. Can you explain this answer?
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