By C. W. Celia, A. T. F. Nice, K. F. Elliott

ISBN-10: 0333231937

ISBN-13: 9780333231937

ISBN-10: 1349035661

ISBN-13: 9781349035663

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Sample text

Obtain the equation of this plane in vector form. Let the unit vector li + mj + nk be perpendicular to the plane ABC. Then this . ~ -----+ vector will be at nght angles to the vectors AB and AC. -----+ AB = b - a = 2i + 4j + 3k. This vector and the vector li (2i ~ AC = c - a + 4j + mj + nk are at right angles. + 3k) • {li + mj + nk) = 0 21 + 4m + 3n = 0 = 3i + j + 2k. + mj + nk are at right angles. + j + 2k) · (li + mj + nk) = 0 31 + m + 2n = 0 This vector and the vector li (3i 44 Advanced Mathematics 2 From these two equations in l, m and n, 4(31 + m + (21 2n) - + 4m + 3n) 101 + 5n n + 31 Therefore li + + mj = 0 = 0 -21 = m - 41 = 0 m = l nk = i(i +j - 2k).

Geometry in three dimensions 49 Example 6 Find the cartesian equations of the line of intersection of the planes x + 2y + z = 5 and 3x - y + 2z = 1. Let the direction ratios of the line be a:b:c. Since the line will be at right angles to the normal to each plane, a+ 2b + c = 0, 3a- b + 2c = 0. These equations give a:b:c = 5:1:-7. Since the line is not parallel to the x-y plane, there will be a point on the line at which z = 0. When z = 0 the equations of the planes become X+ 2y = 5, 3x - y = 1.

Hence the required distance is 2. Example 2 Find the acute angle between the planes with vector equations r · (i + 2j - k) = 5 and r · (j - k) = 7. The angle between the planes equals the angle between the normals to the planes. The unit normal to the planer· (i + 2j - k) = 5 is (i + 2j - k)/j6. The unit normal to the plane r · (j - k) = 7 is (j - k)/j2. Let (} be the angle between the two unit normals. Then their scalar product equals cos e. cos(}= (i + 2j- k)·(j- k)/j12 = 3/)12 = J3j2. Therefore the acute angle between the planes is 30°.