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

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**Example text**

N = afa. Then the equation r ·a = ap can be written in the form r·n = p. This equation represents a plane which is perpendicular to the unit vector n and which is at a distance p from the origin. The vector n is the unit normal to the plane. Since a = an, the equation r ·a = d is equivalent to the equation r·n = dfa. Therefore the perpendicular distance from the origin to the plane r·a = dis dfa. -------+ There is an alternative form for the vector equation of a plane. In Fig. 4, OA = a, AB = b and AC = c where a, b and c are independent vector~.

Find the velocity and acceleration of the particle at timet. Show that when t = 2/5 the velocity and acceleration are perpendicular to each other. The velocity and acceleration are resolved into components along and 52 Advanced Mathematics 2 22 23 24 perpendicular to the vector i - 3j + 2k. Find the velocity and acceleration components parallel to this vector when t = 2/5. [AEB] The equations of two planes P 1 and P 2 are 3x + 4y = 7 and 2x - y + 2z = 1 respectively. Show that the acute angle between the planes is cos- 1 (2/ 15).

The unit vectors i, j and k are in the directions of the axes Ox, Oy and Oz respectively. If Pis the point (x, y, z) its position vector r is given by r = xi + yj + zk. Let a = a 1 i + a 2 j + a 3 k and b = b 1 i be put in terms of row vectors, or in terms of column vectors + b2 j + b3 k. The equation of the line can By equating components we obtain the three equations x = a1 + y tb 1 , = a2 + tb 2 , z = a3 + tb 3 • This is the parametric form of the cartesian equations of the straight line. It follows that since each fraction is equal tot.