Three Dimensional Geometry
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New question posted
a month agoNew answer posted
a month agoContributor-Level 10
The equation of the plane passing through the line of intersection is:
(2x - 7y + 4z - 3) + λ (3x - 5y + 4z + 11) = 0.
The plane passes through the point (-2, 1, 3).
(2 (-2) - 7 (1) + 4 (3) - 3) + λ (3 (-2) - 5 (1) + 4 (3) + 11) = 0
(-4 - 7 + 12 - 3) + λ (-6 - 5 + 12 + 11) = 0
(-2) + λ (12) = 0 ⇒ 12λ = 2 ⇒ λ = 1/6.
Substitute λ back into the equation:
(2x - 7y + 4z - 3) + (1/6) (3x - 5y + 4z + 11) = 0
Multiply by 6:
6 (2x - 7y + 4z - 3) + (3x - 5y + 4z + 11) = 0
12x - 42y + 24z - 18 + 3x - 5y + 4z + 11 = 0
15x - 47y + 28z - 7 = 0.
This is the equation ax + by + cz - 7 = 0.
So, a=15, b=-47, c=28.
We need to find the value of 2a + b
New answer posted
a month agoContributor-Level 10
Let the equation of the plane passing through (1, 2, 3) be:
a (x - 1) + b (y - 2) + c (z - 3) = 0
The plane contains the y-axis, which has direction ratios (0, 1, 0).
Therefore, the normal to the plane must be perpendicular to the y-axis.
a (0) + b (1) + c (0) = 0 ⇒ b = 0
The equation becomes: a (x - 1) + c (z - 3) = 0
ax + cz = a + 3c
The plane also passes through the origin (0,0,0) since it contains the y-axis.
a (0) + c (0) = a + 3c ⇒ a + 3c = 0 ⇒ a = -3c
Substitute a = -3c into the plane equation:
-3c (x - 1) + c (z - 3) = 0
-3 (x - 1) + (z - 3) = 0
-3x + 3 + z - 3 = 0
-3x + z = 0 ⇒ 3x - z = 0
New answer posted
a month agoContributor-Level 9
Let A (α, 0,0), B (0, β, 0), C (0,0, γ), then the centroid is G (α/3, β/3, γ/3) = (1,1,2).
α = 3, β = 3, γ = 6
∴ Equation of plane is x/α + y/β + z/γ = 1
⇒ x/3 + y/3 + z/6 = 1
⇒ 2x + 2y + z = 6
∴ Required line passing through G (1,1,2) and normal to the plane is (x-1)/2 = (y-1)/2 = (z-2)/1.
New answer posted
a month agoContributor-Level 10
Equation PQ
(x-1)/2 = (y+2)/3 = (z-3)/ (-6) = λ
Let Q = (2λ + 1, 3λ − 2, −6λ + 3)
Q lies on x - y + z = 5
⇒ (2λ + 1) − (3λ − 2) + (−6λ + 3) = 5
⇒ λ = -1/7
Q = (5/7, -17/7, 15/7)
∴ PQ = √ (2/7)² + (3/7)² + (6/7)²)
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