Class 12 Maths - Vectors

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Published on February 28, 2014

Author: ednexa

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Class 12 Maths - Vectors by ednexa.com

Vectors 1. ABCD is a parallelogram. P, Q are the midpoints of the sides AB and CD respectively. Show that DP and BQ trisect AC and trisected by AC. Sol: Let DP and BQ intersect AC at M and N respectively. Let a , b , c, d , p, q , m , n be the position vectors of the points A, B, C, D, P, Q, M, N respectively with respect to some origin O. □ ABCD is a parallelogram.  AB  DC     bacd ...(i) P is the midpoint of AB   a b  p 2     2p  a  b ...(ii) Q is the midpoint of CD   cd q  2     2q  c  d ...(iii)  For eliminating b from (i) and (ii), subtract (i) from (ii) b  a  2p b  a  c d     _________________ 2 a  2p  c  d  2p  d  2a  c  2p  d 2a  c  2 1 2 1 This means that M divides DP internally in the ratio 2 : 1 and M divides CA internally in the ratio 2 : 1. For eliminating d from (i) and (iii), add (i) and (iii), …(iv)

c  d  2q c d b a ____________ 2c  2q  b  a  2q  b  2c  a  2q  b 2c  a  2 1 2 1 This means that N divides BQ internally in the ratio 2: 1 and N divides AC internally in the ratio 2 : 1. From (iv) and (v), if follows that DP and BQ trisect AC and are trisected by AC. 2. If in a tetrahedron, the two pairs of opposite edges are perpendicular, then show that the edges in the third pair is also perpendicular. Sol: Let O-ABC be a tetrahedron. Then (OA, BC), (OB, CA) and (OC, AB) are the pair of opposite edges. Take O as the origin of reference and let a, b and c be the position vectors of the vertices A, B and C respectively. Then OA  a, OB  b, OC  c, AB  b  a, BC  c  b and CA  a  c. Now, suppose the pair (OA, BC) and (OB, CA) are perpendicular to each other. Then

OA. BC  0, i.e., a. (c  b)  0  a. c  a. b  0 ...(i) and OB. CA  0, i.e., b. (a  c)  0  b. a  b. c  0  a. b  b. c  0 Adding (i) and (2), we get, ...(ii) a. c  b. c  0  c. b  c. a  0 i.e., c. (b  a)  0  OC. AB  0 ∴ the third pair (OC, AB) is perpendicular. 3. Show that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half the length of that of the third side. Sol: Let ABC be a triangle and M and N be the midpoints of the AB and AC. Then we have to show MN parallel to BC and 1 (MN)  . (BC). 2 Let a, b, c, m and n be the position vectors of A, B, C, M and N respectively. Since M and N are the midpoints of AB and AC respectively,

ab a c and n  2 2  MN  n  m m ac ab     2   2  1  (a  c  a  b) 2 1  (c  b) 2 1  BC 2 Thus MN is non –zero scalar multiple of BC .  MN is parallel to BC. seg MN is parallel to seg BC. 1 Also, | MN |  | BC | 2 1  (MN)  . (BC). 2 Keep on visiting www.ednexa.com for more study material. -Team Ednexa

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