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Concept 04 / 06Three Dimensional Geometry

Line and Plane: Angle, Intersection, Distance, Foot and Image of a Point

The angle between a line and a plane, conditions for a line to be parallel to or lie in a plane, the point where a line meets a plane, distance from a point to a plane and between parallel planes, foot of the perpendicular and image in a plane, and projection of a line on a plane.

Figure 1
Angle between a line and a plane. If the line has direction b⃗\vec{b} and the plane has normal n⃗\vec{n}, the angle ϕ\phi between the line and the plane is the complement of the angle between b⃗\vec{b} and n⃗\vec{n}: sin⁡ϕ=∣b⃗⋅n⃗∣∣b⃗∣∣n⃗∣\sin\phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}. Note sine, not cosine, which is the most common slip here. Line parallel to a plane: b⃗⋅n⃗=0\vec{b} \cdot \vec{n} = 0 (the direction is perpendicular to the normal). If in addition one point of the line lies in the plane, the whole line lies in the plane. Line perpendicular to a plane: b⃗∥n⃗\vec{b} \parallel \vec{n}. Intersection of a line and a plane. Substitute the general point (x1+aλ,y1+bλ,z1+cλ)(x_{1} + a\lambda, y_{1} + b\lambda, z_{1} + c\lambda) into the plane equation and solve for λ\lambda. If the equation has no solution the line is parallel to the plane; if it holds for all λ\lambda the line lies in the plane. Distance from a point to a plane. The perpendicular distance from (x1,y1,z1)(x_{1}, y_{1}, z_{1}) to ax+by+cz+d=0ax + by + cz + d = 0 is ∣ax1+by1+cz1+d∣a2+b2+c2\frac{|ax_{1} + by_{1} + cz_{1} + d|}{\sqrt{a^{2} + b^{2} + c^{2}}}, exactly analogous to the distance from a point to a line in the plane. In vector form: ∣p⃗⋅n⃗−q∣∣n⃗∣\frac{|\vec{p} \cdot \vec{n} - q|}{|\vec{n}|} for the plane r⃗⋅n⃗=q\vec{r} \cdot \vec{n} = q. Distance between parallel planes ax+by+cz+d1=0ax + by + cz + d_{1} = 0 and ax+by+cz+d2=0ax + by + cz + d_{2} = 0: ∣d1−d2∣a2+b2+c2\frac{|d_{1} - d_{2}|}{\sqrt{a^{2} + b^{2} + c^{2}}}, after making the normal coefficients identical. Side of a plane. Points P1,P2P_{1}, P_{2} are on the same side of ax+by+cz+d=0ax + by + cz + d = 0 iff the values of ax+by+cz+dax + by + cz + d at them have the same sign. Foot of the perpendicular and image. The perpendicular from P(x1,y1,z1)P(x_{1}, y_{1}, z_{1}) runs along the normal, so its points are (x1+at,y1+bt,z1+ct)(x_{1} + a t, y_{1} + b t, z_{1} + c t). For the foot FF and image II: x−x1a=y−y1b=z−z1c=−ax1+by1+cz1+da2+b2+c2\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c} = -\frac{ax_{1} + by_{1} + cz_{1} + d}{a^{2} + b^{2} + c^{2}} for FF, and twice that value for II. The figure shows PP, its foot on the plane and its mirror image beyond it. Projection of a line on a plane. Take two points of the line, find their feet on the plane, and join them. Alternatively, the plane containing the line and perpendicular to the given plane has normal b⃗×n⃗\vec{b} \times \vec{n}; the projection is the intersection of this plane with the given plane. Image of a line in a plane: reflect two points of the line (or the intersection point, which is its own image, and one other point). Angle bisector planes of two planes P1=0P_{1} = 0 and P2=0P_{2} = 0: a1x+b1y+c1z+d1∑a12=±a2x+b2y+c2z+d2∑a22\frac{a_{1}x + b_{1}y + c_{1}z + d_{1}}{\sqrt{\sum a_{1}^{2}}} = \pm\frac{a_{2}x + b_{2}y + c_{2}z + d_{2}}{\sqrt{\sum a_{2}^{2}}}, with the same sign rules as for the bisectors of two lines. Key Takeaways: - Line-plane angle: sin⁡ϕ=∣b⃗⋅n⃗∣∣b⃗∣∣n⃗∣\sin\phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}. - Parallel: b⃗⋅n⃗=0\vec{b} \cdot \vec{n} = 0; lies in the plane if also one point does. - Intersection: substitute the general point of the line. - Distance: ∣ax1+by1+cz1+d∣a2+b2+c2\frac{|ax_{1} + by_{1} + cz_{1} + d|}{\sqrt{a^{2} + b^{2} + c^{2}}}; parallel planes ∣d1−d2∣a2+b2+c2\frac{|d_{1} - d_{2}|}{\sqrt{a^{2} + b^{2} + c^{2}}}. - Foot: ratio =−valuea2+b2+c2= -\frac{\text{value}}{a^{2} + b^{2} + c^{2}}; image: twice. - Projection of a line: feet of two of its points. - Pitfall: using cos⁡ϕ\cos\phi for the line-plane angle. Brain-Teaser 1: Find the image of P(1,3,4)P(1, 3, 4) in the plane 2x−y+z+3=02x - y + z + 3 = 0. Solution: The value at PP is 2−3+4+3=62 - 3 + 4 + 3 = 6, and a2+b2+c2=6a^{2} + b^{2} + c^{2} = 6. For the image, x−12=y−3−1=z−41=−2×66=−2\frac{x - 1}{2} = \frac{y - 3}{-1} = \frac{z - 4}{1} = -\frac{2 \times 6}{6} = -2. So the image is (1−4,3+2,4−2)=(−3,5,2)(1 - 4, 3 + 2, 4 - 2) = (-3, 5, 2). Check: the midpoint (−1,4,3)(-1, 4, 3) satisfies −2−4+3+3=0-2 - 4 + 3 + 3 = 0. Brain-Teaser 2: Find where the line x−23=y+14=z−212\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} meets the plane x−y+z=5x - y + z = 5, and the angle between them. Solution: The general point (2+3λ,−1+4λ,2+12λ)(2 + 3\lambda, -1 + 4\lambda, 2 + 12\lambda) gives 2+3λ+1−4λ+2+12λ=52 + 3\lambda + 1 - 4\lambda + 2 + 12\lambda = 5, so 11λ=011\lambda = 0 and the point is (2,−1,2)(2, -1, 2). The angle: sin⁡ϕ=∣3−4+12∣133=11133\sin\phi = \frac{|3 - 4 + 12|}{13\sqrt{3}} = \frac{11}{13\sqrt{3}}. Note the direction vector has length 9+16+144=13\sqrt{9 + 16 + 144} = 13. Brain-Teaser 3: Find the distance between the planes 2x−2y+z+3=02x - 2y + z + 3 = 0 and 4x−4y+2z+5=04x - 4y + 2z + 5 = 0. Solution: Make the coefficients match: the second plane is 2x−2y+z+52=02x - 2y + z + \frac{5}{2} = 0. Distance =∣3−52∣4+4+1=1/23=16= \frac{|3 - \frac{5}{2}|}{\sqrt{4 + 4 + 1}} = \frac{1/2}{3} = \frac{1}{6}. Applying the formula without rescaling would give ∣3−5∣3=23\frac{|3 - 5|}{3} = \frac{2}{3}, which is wrong; the constants are only comparable when the normals are written identically.

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Line and Plane: Angle, Intersection, Distance, Foot and Image of a Point — DPP

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