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The two planes ax + by +by + cz + d = 0 and a'x + b'y + c'z + d = 0 are perpendicular if
- a)aa' + bb' + cc' +dd' =0
- b)a/a' = b/b' + c/c' + d/d' =0
- c)a/a'=b/b' =c/c'
- d)aa' +bb' + cc' =0
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The two planes ax + by +by +cz + d = 0 and a'x + b'y + c'z...
Explanation:
To determine if two planes are perpendicular, we need to check the dot product of their normal vectors. Let's consider the two planes:
Plane 1: ax + by + cz + d = 0
Plane 2: ax + by + cz + d = 0
Step 1: Finding the normal vectors:
The coefficients of x, y, and z in the equations of the planes give us the normal vectors. In this case, the normal vectors for both planes are given by:
Normal vector 1: (a, b, c)
Normal vector 2: (a, b, c)
Step 2: Calculating the dot product:
The dot product of two vectors is given by the sum of the products of their corresponding components. In this case, the dot product of the normal vectors is:
(a, b, c) • (a, b, c) = a*a + b*b + c*c
Step 3: Checking for perpendicularity:
For two vectors to be perpendicular, their dot product must be zero. Therefore, we need to check if:
a*a + b*b + c*c = 0
Step 4: Simplifying the equation:
To simplify the equation further, we can divide both sides by a*a, b*b, and c*c (assuming they are not zero):
(a*a + b*b + c*c)/(a*a) = 0/(a*a)
1 + (b*b)/(a*a) + (c*c)/(a*a) = 0
Step 5: Simplifying further:
We can rewrite (b*b)/(a*a) as (b/a)^2 and (c*c)/(a*a) as (c/a)^2:
1 + (b/a)^2 + (c/a)^2 = 0
Step 6: Final conclusion:
For the equation to hold true, each of the terms on the left side must be zero. Therefore, we have:
a = 0
b/a = 0
c/a = 0
This implies that a, b, and c are all zero. Hence, the correct answer is option 'D': aa + bb + cc = 0.
To determine if two planes are perpendicular, we need to check the dot product of their normal vectors. Let's consider the two planes:
Plane 1: ax + by + cz + d = 0
Plane 2: ax + by + cz + d = 0
Step 1: Finding the normal vectors:
The coefficients of x, y, and z in the equations of the planes give us the normal vectors. In this case, the normal vectors for both planes are given by:
Normal vector 1: (a, b, c)
Normal vector 2: (a, b, c)
Step 2: Calculating the dot product:
The dot product of two vectors is given by the sum of the products of their corresponding components. In this case, the dot product of the normal vectors is:
(a, b, c) • (a, b, c) = a*a + b*b + c*c
Step 3: Checking for perpendicularity:
For two vectors to be perpendicular, their dot product must be zero. Therefore, we need to check if:
a*a + b*b + c*c = 0
Step 4: Simplifying the equation:
To simplify the equation further, we can divide both sides by a*a, b*b, and c*c (assuming they are not zero):
(a*a + b*b + c*c)/(a*a) = 0/(a*a)
1 + (b*b)/(a*a) + (c*c)/(a*a) = 0
Step 5: Simplifying further:
We can rewrite (b*b)/(a*a) as (b/a)^2 and (c*c)/(a*a) as (c/a)^2:
1 + (b/a)^2 + (c/a)^2 = 0
Step 6: Final conclusion:
For the equation to hold true, each of the terms on the left side must be zero. Therefore, we have:
a = 0
b/a = 0
c/a = 0
This implies that a, b, and c are all zero. Hence, the correct answer is option 'D': aa + bb + cc = 0.
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Community Answer
The two planes ax + by +by +cz + d = 0 and a'x + b'y + c'z...
Proof: The two planes are perpendicular if
θ = 90° or cos θ = 0
or aa' + bb' + cc' = 0
This is the condition of perpendicularity.
Remark : Condition of parallelism. Two planes are prallel if
θ = 90° or cos θ = 0
or aa' + bb' + cc' = 0
This is the condition of perpendicularity.
Remark : Condition of parallelism. Two planes are prallel if
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