Notes on Equation of Plane
- 1. More Notes on the Equation of a Plane Course Unit:Vectors Course organiser:Mujungu Herbert Subject: MATHEMATICS FOR YEAR ONE 2019/2020 Institution: NATIONAL TEACHERS’ COLLEGE KABALE July 30, 2020 Objectives These notes will make the students be able to; i). Determine the perpendicular distance of a point from a plane. ii). Calculate an angle between a line and a plane. iii). Calculate an angle between two planes. Contents 1 EQUATION OF A PLANE 1 2 PERPENDICULAR DISTANCE FROM A POINT TO THE PLANE 2 2.1 Vector Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2.2 Cartesian Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2.3 Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3 ANGLE BETWEEN A PLANE AND A LINE 4 3.1 Derivation of the formular used . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3.2 Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 4 ANGLE BETWEEN TWO PLANES 5 4.1 Derivation of the formular used . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 4.2 Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 5 REVISION QUESTIONS 6
- 2. 1 EQUATION OF A PLANE Figure 1: This figure is extracted from the slides at https://bit.ly/padlet-DESI for deriving the equation of a plane 1
- 3. herbertm ujungu@ gm ail.com 2 PERPENDICULAR DISTANCE FROM A POINT TO THE PLANE The shortest distance of a point from a plane is said to be along the line perpendicular to the plane or in other words, is the perpendicular distance of the point from the plane. Thus, if we take the normal vector say to the given plane, a line parallel to this vector that meets the point P gives the shortest distance of that point from the plane. If we denote the point of intersection (say R) of the line touching P, and the plane upon which it falls normally, then the point R is the point on the plane that is the closest to the point P. Here, the distance between the point P and R gives the distance of the point P to the plane. This shortest distance of a point from a plane can be obtained using the Vector method and the Cartesian Method. 2.1 Vector Method Let us consider a point A whose position vector is given by ã and a plane P, given by the equation, −→r · −→ N = d Here, N is normal to the plane P under consideration. Now, let O be the origin of the coordinate system being followed and P’ another plane parallel to the first plane, which is taken such that it passes through the point A. Here, N’ is normal to the second plane. The equation of the second plane P’ is given by, (−→r − −→a ) · ˆN = 0 or −→r · ˆN = −→a · ˆN We see that, the ON gives the distance of the plane P from the origin and ON’ gives the distance of the plane P’ from the origin. Thus, the distance between the two planes is given as, ON − ON = d = |d − −→a · ˆN| So, for a plane whose equation is given by −→r · −→ N = D, the distance of a point A whose position vector is given by −→a to the plane is given by d = |−→a · −→ N − D| | −→ N | In order to calculate the length of the plane from the origin, we substitute the position vector by 0, and thus it comes out to be d = |D| | −→ N | 2.2 Cartesian Form Consider a plane ax+by+cz+d = 0 and a point A(xo, yo, zo), the normal vector to the plane is given by n = a b c and a vector from the plane to the point is given by; 2
- 4. herbertm ujungu@ gm ail.com z x y (x, y, z) (xo, yo, zo) w n A w = − x − xo y − yo z − zo . Projecting vecw onto n gives the distance d from the point to the plane as D = |projnw| = |n · w| |n| D = |a(x − xo) + b(y − yo) + c(z − zo)| √ a2 + b2 + c2 D = | − d − axo − byo − czo| √ a2 + b2 + c2 D = axo + byo + czo + d √ a2 + b2 + c2 (1) The distance d is positive if A is on the same side of the plane as the normal vector n and negative if it is on the opposite side. If A is the origin, then; D = d n (2) 2.3 Worked Examples 1. What is the distance between the point P(1,2,3) and the plane Π : 2x + 2y − 3z + 3 = 0? Solution Using D = axo + byo + czo + d √ a2 + b2 + c2 and taking (xo, yo, zo) = (1, 2, 3) and < a, b, c >=< 2, 2, −3 >. D = |2(1) + 2(2) − 3(3) + 3| 22 + 22 + (−3)2 = | − 2| √ 17 D = 2 √ 17 Therefore the distance between the point P(1,2,3) and the plane Π : 2x + 2y − 3z + 3 = 0 is 2√ 17 . 2. What is the distance between the point P(-2,-7,-12) and the plane Π : 4x−8y−3z+5 = 0? Solution Using D = axo + byo + czo + d √ a2 + b2 + c2 and taking P(xo, yo, zo) = P(−2, −7, −12) and < a, b, c >=< 4, −8, −3 >. D = |4(−2) − 2(−7) − 3(−12) + 5| 42 + (−8)2 + (−3)2 = |89| √ 89 D = 89 √ 89 units Therefore the distance between the point P(-2,-7,-12) and the plane Π : 4x−8y−3z+5 = 0 is 89√ 89 . 3
- 5. herbertm ujungu@ gm ail.com 3 ANGLE BETWEEN A PLANE AND A LINE 3.1 Derivation of the formular used The angle between a line,r, and a plane, Π, is the angle between r and its projection onto Π, r . Plane :Π Line : r n rαβ u the angle between a line and a plane is equal to the complemenatary acute angle that forms between the direction vector (u) of the line and the normal vector ((n)) of the plane. sin α = cos β = |n · u| |n||u| α = arcsin |n · u| |n||u| (3) Note: If the line, r, and the plane, Π, are perpendicular, the direction vector of the line and the normal vector of the plane have the same direction and therefore its components are pro- portional. 3.2 Worked Examples 1. Determine the angle between the line r = x − 1 2 = y + 1 1 = z 2 and the plane Π ≡ x + y − 1 = 0. Solution Letting u =< 2, 1, 2 > and n =< 1, 1, 0 > and substituting them in α = arcsin | < 2, 1, 2 > · < 1, 1, 0 √ 22 + 12 + 22 · √ 12 + 1 α = arcsin √ 2 2 = 45o ∴ 45o is the angle between the line r = x − 1 2 = y + 1 1 = z 2 and the plane Π ≡ x+y−1 = 0. 2. Determine the angle between the line r ≡ x + 3y − z + 3 = 0 2x − y − z − 1 = 0 and the plane Π ≡ 2x − y + 3z + 1 = 0. Solution r is the line of intersection of the planes x + 3y − z + 3 = 0 and 2x − y − z − 1 = 0, which is perpendicular to both the normals of the planes. This can be got by crossing the two normals of the planes i.e < 1, 3, −1 > and < 2, −1, −1 > i.e i j k 1 3 −1 2 −1 −1 = −4i − j − 7k Let u =< −4, −1, −7 > and n =< 2, −1, 3 >. α = arcsin |n · u| ||n|u| ≈ 67.09o (4) ∴ 67.09o is the line r ≡ x + 3y − z + 3 = 0 2x − y − z − 1 = 0 and the plane Π ≡ 2x − y + 3z + 1 = 0. 4
- 6. herbertm ujungu@ gm ail.com 3. Determine the angle between the line and the plane given by:- r ≡ y = 2 3x − √ 3z = 0 and Π ≡ x = 1 respectively. Solution r ≡ x = √ 3µ y = 2 z = 3λ , u =< √ 3, 0, 3 >, n =< 1, 0.0 > (5) sin α = | √ 3 · 1 + 0 · 0 + 3 · 0| ( √ 3)2 + 02 + 32 · √ 12 + 02 + 02 = √ 3 2 √ 3 (6) α = 30o (7) ∴ 30o is between r ≡ y = 2 3x − √ 3z = 0 and Π ≡ x = 1 4 ANGLE BETWEEN TWO PLANES 4.1 Derivation of the formular used Given two planes with normal vectors n1 and n2. The dot product of these two vectors n1 and n2 is defined by; n1 · n2 = |n1||n2| cos α , where α is the angle between n1 and n2. The angle between two planes is equal to the acute angle determined by the normal vectors of the planes. α(Π1, Π2) = α(n1, n2) = arccos n1 · n2 |n1||n2| (8) Note that two planes are perpendicular if their normal vectors are orthogonal. i.e, Π ⊥ Π n1 · n2 = 0 4.2 Worked Examples 1. Determine the angle between the following planes:- Π1 ≡ 2x − y + z − 1 = 0, Π2 ≡ x + z + 3 = 0 Solution Let n1 =< 2, −1, 1 >, n2 =< 1, 0, 1 > (9) cos α = |n1 · n2| |n1||n2| = 3 √ 6 · √ 2 (10) α = 30o (11) ∴ 30o is the angle between the following planes:- Π1 ≡ 2x−y+z−1 = 0, Π2 ≡ x+z+3 = 0 5
- 7. herbertm ujungu@ gm ail.com 5 REVISION QUESTIONS 1. Find the lengths of the perpendiculars from the following points to the givenm planes: (a) (6,5,4), 7x-4y-4z+3=0 (b) (4,0,-1), 3x+6y-2z+7=0 2. Determine the angle between the line r = i+2j −k+t(i−j +k) and the plane 2x-y+z=4. 3. Find the angle between the planes. (a) 9x-2y+6z=1 and 12x+y-12z=8 (b) 4x+3y+2z=5 and 2x-4y+3z=-6 (c) 3x+6y-2z=3 and 8x-4y+z=1 4. Find the equation of the plane through (1,2,3) parallel to −1 0 4 and 2 3 0 . 5. L, M, N are points (-4,0,1), (-3,1,2) and (-2,-1,1) respectively. Write down the vectors LM and LN and obtain a vector perpendicular to both. 6. The parametric equations of two planes are x1 = 1 + p, y1 = −2p + 3q, z = 1 + p + 2q and x2 = 2 − 2n, y2 = 1 + 4n, z2 = −1 + 5m + 3n. (a) Find the cosine of the acute angle below the planes. (b) The line of intersection is l. Find in the form r = a + λb, the equation of l. (c) show that the length of the perpendicular from the point (1,5,1) to the line is √ 2. 7. (a) Given that the vectors i − λj + k and − 3i + 2j + 4k are perpendicular, determine the value of λ. (b) Find the equation of the plane containing the points A(−1, 0, 1), B(3, 2, −1) and C(−1, 1, 1). END 6