Dayaan M. answered • 17d
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Both parts come out of the same two tools: the cross product to get directions, and the dot product to impose "orthogonal" and to measure the angle. No calculus is needed anywhere in this, it is all vector geometry.
Setting up. The normals are
n₁ = (3, 5, 1), n₂ = (4, 7, -2), n₃ = (1, 1, 1)
L₀ is the intersection of P₁ and P₂, so it lies in both planes and is therefore perpendicular to both normals. That makes its direction the cross product:
d₀ = n₁ × n₂ = (5(-2) - 1(7), 1(4) - 3(-2), 3(7) - 5(4)) = (-17, 10, 1)
For a point on L₀, set z = 0 and solve the two plane equations together:
3x + 5y = 6
4x + 7y = 9
which gives x = -3, y = 3. So (-3, 3, 0) is on L₀, and you can check it satisfies both originals. Then
L₀: (x, y, z) = (-3, 3, 0) + t(-17, 10, 1)
Part 1. L₁ has to satisfy two conditions at once, and each one is a dot product set to zero. Lying in P₃ means its direction is perpendicular to n₃. Being orthogonal to L₀ means it is perpendicular to d₀. A vector perpendicular to two given vectors is their cross product:
d₁ = n₃ × d₀ = (1(1) - 1(10), 1(-17) - 1(1), 1(10) - 1(-17)) = (-9, -18, 27) = -9(1, 2, -3)
so take d₁ = (1, 2, -3). Check both conditions: d₁·n₃ = 1 + 2 - 3 = 0, and d₁·d₀ = -17 + 20 - 3 = 0. Both hold.
Now you need a point, and here is the part the problem is quietly pushing you toward. Part 2 refers to "the point of intersection of L₀ and L₁", which means these two lines are meant to meet. Where can that happen? L₁ lies in P₃, so any meeting point is in P₃. And L₀ is not parallel to P₃, since
d₀·n₃ = -17 + 10 + 1 = -6 ≠ 0
A line not parallel to a plane pierces it in exactly one point. So there is only one candidate, and L₁ is forced to pass through it. Find it by substituting L₀ into P₃:
(-3 - 17t) + (3 + 10t) + (t) = 60 → -6t = 60 → t = -10
Q = (-3 + 170, 3 - 100, 0 - 10) = (167, -97, -10)
Worth verifying, since the numbers are large: 3(167) + 5(-97) + (-10) = 501 - 485 - 10 = 6, and 4(167) + 7(-97) - 2(-10) = 668 - 679 + 20 = 9, and 167 - 97 - 10 = 60. All three planes check out, so Q really is on L₀ and on P₃.
L₁: (x, y, z) = (167, -97, -10) + s(1, 2, -3)
or componentwise, x = 167 + s, y = -97 + 2s, z = -10 - 3s. Note that this makes L₁ unique, not just unique up to choice of point, which is a slightly stronger statement than the problem asks for.
Part 2, and whether it is even possible. Before hunting for L₂, it is worth asking what angles are actually available, because L₀ is not in the plane. Every line through Q inside P₃ makes some angle with L₀, and the smallest such angle is the angle between L₀ and the plane itself. That angle satisfies
sin θ = |d₀·n₃| / (|d₀| |n₃|) = 6 / (√390 · √3) = √130 / 65 ≈ 0.17541
so θ ≈ 10.10°. The largest available angle is 90°, which is exactly what L₁ achieves. So the reachable range is about 10.10° up to 90°, and since 60° sits inside that range, yes, such a line exists. In fact two of them do, one on each side of L₁. This check is worth doing first; if the problem had asked for 5° the honest answer would be that no such line exists.
Finding them. Any direction lying in P₃ can be written using the two vectors we already have, since both are perpendicular to n₃:
d₂ = a(-5, 4, 1) + b(-1, -2, 3)
I am using (-5, 4, 1) because it is the projection of d₀ into the plane, reduced, and (-1, -2, 3) because that is d₁ again. They are perpendicular to each other, which keeps the algebra clean. The useful facts are
(-5, 4, 1)·d₀ = 85 + 40 + 1 = 126, (-1, -2, 3)·d₀ = 17 - 20 + 3 = 0
So d₂·d₀ = 126a, and |d₂|² = 42a² + 14b². Imposing cos 60° = 1/2:
126|a| / (√(42a² + 14b²) · √390) = 1/2
Square both sides and clear denominators:
63504a² = 390(42a² + 14b²) = 16380a² + 5460b²
47124a² = 5460b²
b²/a² = 561/65
So a = √65 and b = ±√561 gives the two directions:
d₂ = √65 (-5, 4, 1) ± √561 (-1, -2, 3)
Numerically those are about (-64.00, -15.12, 79.12) and (-16.63, 79.62, -62.99). I checked both: each has zero dot product with n₃, so each lies in P₃, and each gives cos of the angle with d₀ equal to exactly 1/2, so exactly 60°. The parametric equations are
L₂: (x, y, z) = (167, -97, -10) + r · d₂
with either choice of sign.
Two things worth carrying forward from this. First, "lies in a plane" and "is perpendicular to a line" are both dot-product-equals-zero conditions, and two such conditions on a direction in three dimensions is exactly what a cross product solves. Recognizing that shape saves you from setting up a system by hand. Second, when a problem asks "is there a line making angle α", check the feasible range before you start solving. The answer depends on the angle between the line and the plane, and setting up the equation blindly will hand you a contradiction with no explanation of why.
