By Tabachnikov S.
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When it operates on a scalar function, it turns ∇ϕ into a vector with deﬁnite magnitude and direction. It also has a deﬁnite physical meaning. 1. Show that ∇r = r and ∇f (r) = rdf /dr, where r is a unit vector along the position vector r = xi + yj + zk and r is the magnitude of r. 1. ∂ ∂ ∂ +j +k r, ∂x ∂y ∂z ∂ ix ix ∂r 1/2 =i x2 + y 2 + z 2 = = , i 1/2 2 2 2 ∂x ∂x r (x + y + z ) xi + yj + zk r ix jy kz + + = = = r. ∇r = r r r r r ∇r = i etc. ∂f ∂f ∂f +j +k , ∂x ∂y ∂z ∂f df ∂r df x i =i =i , etc.
10i + 3j + 11k, −10i − 3j − 11k. 8. Find the area of a parallelogram having diagonals A = 3i + j − 2k and B=i− √ 3j + 4k. Ans. 5 3. 9. Evaluate (2i − 3j) · [(i + j − k) × (3j − k)]. Ans. 4. 10. Find the volume of the parallelepied whose edges are represented by A = 2i − 3j + 4k, B = i + 2j − k, and C = 3i − j + 2k. Ans. 7. 11. Find the constant a such that the vectors 2i − j + k, i + 2j − 3k and 3i + aj + 5k are coplanar. Ans. a = −4. 12. Show that (a) (b × c) × (c × a) = c (a · b × c) ; 2 (b) (a × b) · (b × c) × (c × a) = (a · b × c) .
To ﬁnd the perpendicular distance between them is a diﬃcult problem in analytical geometry. With vectors, it is relatively easy. Let the equations of two such lines be r = r1 + tv1 , r = r2 + t v2 . 3 Lines and Planes 25 Let a on line 1 and b on line 2 be the end points of the common perpendicular on these two lines. 47) with t = t2 . Accordingly ra = r1 + t1 v1 , rb = r2 + t2 v2 . 49) Since rb − ra is perpendicular to both v1 and v2 , it must be in the direction of (v1 × v2 ) . If d is the length of rb − ra , then rb − ra = v1 × v2 d.