The distance between the planes 3x – 2y + 6z + 21 = 0 and – 6x + 4y – 12z + 35 = 0 is:

a)
1/2

b)
2

c)
3/2

d)
11/2

Correct answer is option 'D'. Can you explain this answer?

Question

The distance between the planes 3x &ndash; 2y + 6z + 21 = 0 and &ndash; 6x + 4y &ndash; 12z + 35 = 0 is:a)1/2b)2c)3/2d)11/2Correct answer is option 'D'. Can you explain this answer?

Answer

Answer

To find the distance between two planes, we need to find a line that is perpendicular to both planes. This line will pass through the point of intersection of the two planes, and the distance between the planes will be the distance between this point and either of the planes.

Step 1: Find the normal vectors of the planes.
The normal vector of a plane is the coefficients of x, y, and z in the equation of the plane. In this case, the normal vectors of the planes are (3, 2, 6) and (6, 4, 12), respectively.

Step 2: Find the direction vector of the line.
Since the line is perpendicular to both planes, its direction vector will be the cross product of the normal vectors of the planes. The cross product of two vectors is found by taking the determinant of a matrix formed by the vectors and then finding the vector formed by the cofactors of the matrix.

So, the direction vector of the line is found by taking the cross product of (3, 2, 6) and (6, 4, 12):

(3, 2, 6) × (6, 4, 12)
= [(2*12 - 4*6), (6*6 - 12*3), (3*4 - 2*6)]
= (24 - 24, 36 - 36, 12 - 12)
= (0, 0, 0)

Since the direction vector of the line is (0, 0, 0), it means that the two planes are parallel. In this case, the distance between the planes is 0.

Therefore, the correct answer is option 'D' (0).

Answer

To find the distance between two parallel planes, we can use the formula:

Distance = |D2 - D1| / √(A² + B² + C²)

where D1 and D2 are the constants in the equations of the planes, and A, B, and C are the coefficients of x, y, and z in the equations respectively.

Given planes:
Plane 1: 3x + 2y + 6z + 21 = 0
Plane 2: 6x + 4y + 12z + 35 = 0

Let's calculate the distance between these two planes step by step:

Step 1: Write the equations of the planes in the standard form Ax + By + Cz + D = 0.
Plane 1: 3x + 2y + 6z + 21 = 0
=> 3x + 2y + 6z = -21
=> 3x + 2y + 6z + 21 = 0 (No change)
Plane 2: 6x + 4y + 12z + 35 = 0
=> 6x + 4y + 12z = -35
=> 3x + 2y + 6z = -35/2

Step 2: Identify the coefficients A, B, C, and the constants D1 and D2.
For Plane 1: A = 3, B = 2, C = 6, D1 = -21
For Plane 2: A = 3, B = 2, C = 6, D2 = -35/2

Step 3: Calculate the distance using the formula.
Distance = |D2 - D1| / √(A² + B² + C²)
= |-35/2 - (-21)| / √(3² + 2² + 6²)
= |-35/2 + 42/2| / √(9 + 4 + 36)
= |-35 + 42| / √49
= 7 / 7
= 1

Therefore, the distance between the two planes is 1 unit.
Hence, the correct answer is option 'D' 11/2.

Answer

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