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Eunice Flowers
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1 Vectors D a The equation of the line is (r a) b0 r ba b. This gives: r (i+j k)(i+j+k) (i+j k) r (i+j k) i+0j k b The equation of the line is (r a) b0 r ba b. This gives: r ( i+ j+ k) (i k) ( i+ j+ k) 0 r ( i+ j+ k) i j+ k c The equation of the line is (r a) b0 r ba b. This gives: r ( i j+k)(i j+k) ( i j+k) r ( i j+k) i j 0k a b Let a a and b b a b Then the Cartesian form of the equation of the line that passes through a point with position vector a x a y a z a and is parallel to the vector b is λ (from Core Pure Book, Section 9.) b b b a b x y z λ x y z+ λ c x y+ z λ a The line is in the direction The equation is 6 r 0 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

2 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. b The line is in the direction 7 The equation is r 7 0 c The line is in the direction 7 6 The equation is r 6 0 d The line is in the direction The equation is r 0 In each part of question one solution is illustrated, but there are alternatives. Either given point may be used for a in the equation (r a) b0 and any multiple of the direction vector may be used as b. The solutions for question are in the form (r a) b0, where a is the position vector of a point on the line and b is vector parallel to the line. Use the standard formula for the Cartesian form of the equation of the line (see solution to question ). As with question, there are alternative solutions. a x y z λ b x y z 7 λ c x+ y z 6 λ Or x+y z 6µ as i + j + k is also in the direction of the line.

3 d x y z λ Or x y z+ λ as (,, ) is a point on the line and i + j k is also in the direction of the line. A straight line with the equation ra+λb passes through the point with position vector a and is parallel to the vector b. The equation of the line can be written (r a) b0. a (r (i+j k)) (i k)0 b (r (i+j)) (i+j k)0 c (r (i+j k)) (i j k)0 6 i x y+ z x y+ z λ can be written as λ The direction of the line is parallel i+j+ k A point on the line has position vector i j+ k Therefore the vector equation of the line can be written as r i+ j+ k i j+ k i+ j+ k r i+ j+ k 9i j+ 7k ii ri j+ k+t i+j+ k Or ri j+ k+s(i+0j+k) 8 7 As (p, q, ) lies on the line with equation r 7 then p q q p q (q ) i (p ) j+ ( p q) k p p q So q 8 q and p 7 p Solution: p and q p 8 q 7 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

4 8 The line with equation r has direction i+j k,i.e. If the line passes through a point (a,a,a ) then a a a a a So a +a a a a a a a a a a a ( a a )i+(a +a )j+(a a )k a +a a a These equations have an infinite number of solutions so let a 0, then as a +a and a a this gives a and a, therefore (0,, ) lies on the line. So the line equation may be written as r j+k+t(i+j k) 9 a The direction vector for the line is (i+j k) ( i+j+7k)6i+j k Hence the direction cosines are given by: 6 l m 8 6 n b A Cartesian equation of the line is x l y m z+ n Substituting for l, m and n and dividing by 6,this simplifies to x y z+ 6 Alternatively use the fact that the Cartesian form of the equation of the line that passes through a x a y a z a point with position vector a and is parallel to the vector b is b b b The line passes through i+j k and is parallel to 6i+j k So an equation of the line is x 6 y z+ Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

5 0 a The direction vector for the x-axis is just i hence the direction cosines are, 0, 0 b The direction vector for the y-axis is just j hence the direction cosines are 0,, 0 c The direction vector for the z-axis is just k hence the direction cosines are 0, 0, d The direction vector for the line x y z is i+j+k So the direction cosines are,, a The direction vector of L is i+j+k so: l m n b The direction vector of L is i+j+k so: l m n c Let r and r be the direction vectors of the two lines. r.r The angle between the two lines θ satisfies cosθ r r ( i+ j+ k )(. i+ j+ k) So cosθ ll + mm + nn So l l +m m +n n cosθ d Given any pair of intersecting lines with direction vectors s and s, divide each direction vector by a constant to produce direction vectors r (x,y,z ) and r (x,y,z ) such that r and r So x l x as r, x + y + z x + y + z And m y,n z,l x,m y,n z Therefore cosθ r.r r r x x +y y +z z xx +yy +zz ll +mm +nn x +y +z x +y +z Use the formula cosθ l l +m m +n n (see question d) + + This gives cosθ θ cos. ( s.f.) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

6 The direction cosines for this line are by definition: l cosα m cosβ n cosγ As l +m +n, this gives cos α+cos β+cos γ Using the trigonometric identity cosx cos x this gives: cosα+cosβ+cosγ cos α +cos β +cos γ (cos α+cos β+cos γ) The direction cosines are: l m n So the angle made with the x-axis is θ cos 68. ( d.p.) 9 The angle made with the y-axis is θ cos 6. ( d.p.) 9 The angle made with the z-axis is θ cos.0 ( d.p.) 9 Two of the direction cosines are: l cos and n cos 60 To find the possible values of m use the fact that l +m +n This gives m m± So there are two solutions. In Cartesian equation form: x x x x y z y z and y z y z In vector equation form: r 0, which is equivalent to r 0 and r 0, which is equivalent to r 0 6 a The direction cosines satisfy lmn As l + m + n l l m n Note that since all the direction cosines are equal the choice of sign does not matter. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 6 b A Cartesian equation for the line is x y l m z+ n This gives x y z+ This simplifies to x y z+ 7 a The direction cosines are given by lcos7,mcos,n0 Using cos(a+b)cosacosb sinasinb cos7 cos( +0 )cos cos0 sin sin0 6 Using cos(a B)cosAcosB+sinAsinB cos cos( 0 )cos cos0 +sin sin So in surd form, the direction cosines are l 6,m 6+,n0 0 6 Hence a vector equation for the line is r b Expressing each line in a vector equation of the form ra+tb, for the wires to intersect requires finding λ and µ such that: 0 6 ( 6 ) 0 +λ µ ( 6+ ) 0 This gives: ( 6 )λ+ ( ( 6 ) )µ () ( 6+ )λ+ ( ( 6+ ) )µ () 6 µ () From (): µ From (): ( 6 )λ +( 6 ) λ This result satisfies (): ( 6+ ) ( ( 6+ ) ) 6+ Hence the lines intersect. ( ) c In reality the cable will not be completely horizontal but might have some slack due to gravity. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 Challenge π π a From the diagram, the point with spherical polar coordinates,, has a position vector: rsin π cosπ i+rsinπ sinπ j+rcosπ k So the direction cosines are: lsin π cosπ 6 msin π sinπ 6 ncos π b In general a point on the line will have a position vector: rsinϕcosθi+rsinϕsinθj+rcosϕk So the direction cosines are: lsinϕcosθ, msinϕsinθ, ncosϕ Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 8

Proof by induction ME 8

Proof by induction ME 8 n Let f ( n) 9, where n. f () 9 8, which is divisible by 8. f ( n) is divisible by 8 when n =. Assume that for n =, f ( ) 9 is divisible by 8 for. f ( ) 9 9.9 9(9 ) f ( ) f ( )