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The two planes represented by ax2 - by2 + cz2 + 2fyz + 2gzx + 2 hxy = 0 are perpendicular if
- a)abc + 2fgh + af2 + bg- + ch2 = 0
- b)af + bg + ch - 0
- c)a + b + c = 0
- d)f2 + g2 + h2 - bc - ca - ab = 0
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The two planes represented by ax2- by2 + cz2+ 2fyz + 2gzx + 2 hxy = 0 ...
Let the two planes r epresented by
Condition of perpendicularity.
The two planes are perpendicular if
θ = 90° or tan θ = ∞
or a + b + c = 0
Remark: since the two planes pass through the origin, therefore the two planes are never parallel. The two planes can however coincide.
Condition of perpendicularity.
The two planes are perpendicular if
θ = 90° or tan θ = ∞
or a + b + c = 0
Remark: since the two planes pass through the origin, therefore the two planes are never parallel. The two planes can however coincide.
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The two planes represented by ax2- by2 + cz2+ 2fyz + 2gzx + 2 hxy = 0 ...
To determine whether the two planes represented by the given equation are perpendicular, we can use the dot product of their normal vectors.
The equation of a plane in general form is given by ax + by + cz + d = 0, where (a, b, c) is the normal vector to the plane.
First, let's rewrite the given equation in the standard form of a plane:
ax^2 - by^2 + cz^2 + 2fyz + 2gzx + 2hxy = 0
Dividing the equation by the coefficient of x^2, we get:
x^2 - (b/a)y^2 + (c/a)z^2 + (2f/a)yz + (2g/a)zx + (2h/a)xy = 0
Comparing this equation with the general form of a plane equation, we can determine the normal vector to the plane:
(a, b/a, c/a)
Similarly, for the second plane, the normal vector is given by (a, f/b, h/b).
To check if the two planes are perpendicular, we need to find the dot product of their normal vectors. If the dot product is zero, the planes are perpendicular.
Taking the dot product of the two normal vectors, we have:
(a)(a) + (b/a)(f/b) + (c/a)(h/b) = a^2 + f + h
The dot product is zero if and only if the equation a^2 + f + h = 0 is satisfied.
Therefore, the correct answer is option 'C': a, b, c = 0.
This means that the two planes represented by the given equation are perpendicular if a^2 + f + h = 0.
The equation of a plane in general form is given by ax + by + cz + d = 0, where (a, b, c) is the normal vector to the plane.
First, let's rewrite the given equation in the standard form of a plane:
ax^2 - by^2 + cz^2 + 2fyz + 2gzx + 2hxy = 0
Dividing the equation by the coefficient of x^2, we get:
x^2 - (b/a)y^2 + (c/a)z^2 + (2f/a)yz + (2g/a)zx + (2h/a)xy = 0
Comparing this equation with the general form of a plane equation, we can determine the normal vector to the plane:
(a, b/a, c/a)
Similarly, for the second plane, the normal vector is given by (a, f/b, h/b).
To check if the two planes are perpendicular, we need to find the dot product of their normal vectors. If the dot product is zero, the planes are perpendicular.
Taking the dot product of the two normal vectors, we have:
(a)(a) + (b/a)(f/b) + (c/a)(h/b) = a^2 + f + h
The dot product is zero if and only if the equation a^2 + f + h = 0 is satisfied.
Therefore, the correct answer is option 'C': a, b, c = 0.
This means that the two planes represented by the given equation are perpendicular if a^2 + f + h = 0.
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