11.1 Overview 11.1.1 Direction cosines of a line are the cosines of the angles made by the line with positive directions of the co-ordinate axes. 11.1.2 If l, m, nare the direction cosines of a line, then l2 + m2 + n2 = 1 11.1.3 Direction cosines of a line joining two points P (x1, y1 , z1) and Q (x2, y2, z2) are x − xy − yz − z2121 21 ,PQ, PQ ,PQ 11.1.4 Direction ratios of a line are the numbers which are proportional to the direction cosines of the line. 11.1.5 If l, m, nare the direction cosines and a, b, care the direction ratios of a line, abc;m ; nthen l 222 222 222abc abc abc 11.1.6 Skew lines are lines in the space which are neither parallel nor interesecting. They lie in the different planes. 11.1.7 Angle between skew lines is the angle between two intersecting lines drawn from any point (preferably through the origin) parallel to each of the skew lines. 11.1.8 If l1, m1, n1 and l2, m2, n2 are the direction cosines of two lines and θ is the acute angle between the two lines, then cosθ = ll + mm + nn 12 1212 11.1.9 If a1, b1, c1 and a2, b2, c2 are the directions ratios of two lines and θ is the acute angle between the two lines, then aa bbc++ c12 1212cos θ= 222 222aaa ++ . bbb ++ 123 123 11.1.10 Vector equation of a line that passes through the given point whose position GGGG Gvector is aand parallel to a given vector bis ra b.=+λ 11.1.11 Equation of a line through a point (x1, y1, z1) and having directions cosines l, m, n (or, direction ratios a, band c) is −⎛xx yy zz x−x yy zz − −−−⎞111 111==== or ⎜⎟.l mn ⎝ abc⎠11.1.12 The vector equation of a line that passes through two points whose positionsGGGGG Gvectors are aand bis =+(ba).raλ− 11.1.13 Cartesian equation of a line that passes through two points (x1, y1, z1) and (x2, y2, z2) is x−x yy −− zz 1 11== − .x−xyyzz −21 2121 G GGG GG11.1.14 If θ is the acute angle between the lines raλ raλ=+band =+b, then11 22 GG GG .bb .bb12 12 or θ=cos –1θ is given by cos θ= .GG GGbb21 b1 b2 x−xyy zz x−x yy −−− − zz 111 222====11.1.15 If and are equations of twolmn lmn111 122 lines, then the acute angle θ between the two lines is given by cosθ = ll mm n++ n.12 1212 11.1.16 The shortest distance between two skew lines is the length of the line segment perpendicular to both the lines. G GGG GG11.1.17 The shortest distance between the lines =+1 b1 and ra=+2 bisraλλ2 GG GG bb . a– a12 21GG .bb12 x−x yy −− zz 1 11==11.1.18 Shortest distance between the lines: andabc111 x−x yy −− zz 2 22==isabc222 x−xy−yz−z21 2121 abc111 abc222 222(bc−bc ) +(ca −ca ) +(ab−ab )1221 1221 12 21 G GGG11.1.19 Distance between parallel lines GG 1and ra2isra b =+λb GGG ba2– a1G .b 11.1.20 The vector equation of a plane which is at a distance pfrom the origin, where G nˆ is the unit vector normal to the plane, is . ˆ=rn p. 11.1.21 Equation of a plane which is at a distance pfrom the origin with direction cosines of the normal to the plane as l, m, nis lx+ my+ nz= p. G11.1.22 The equation of a plane through a point whose position vector is aand GGG GG GGGperpendicular to the vector nis (–ran =0 rn d, where dan).or .==..11.1.23 Equation of a plane perpendicular to a given line with direction ratios a, b, c and passing through a given point (x1, y1, z1) is a(x– x1) + b(y– y1) + c(z– z1) = 0. 11.1.24 Equation of a plane passing through three non-collinear points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is xx– yy – zz – 1 11 –––xxyyzz = 021 2121 . – yyzz –xx – 31 3131 11.1.25Vector equation of a plane that contains three non-collinear points having position GGG GGG vectors a G , b G , c Gis (– ). ( – × ca –)⎤=0ra⎡ ba )( ⎣⎦ 11.1.26Equation of a plane that cuts the co-ordinates axes at (a, 0, 0), (0, b, 0) and xyz (0, 0, c) is ++=1.abc 11.1.27Vector equation of any plane that passes through the intersection of planes GG GGGG GGand rn d is − −=0 , where λ is any non-zerorn d.= .= (.rn d ) +λ (r.nd )11 221122 constant. 11.1.28Cartesian equation of any plane that passes through the intersection of two given planes Ax+ By+ Cz+ D = 0 and Ax+ By+ Cz+ D = 0 is11112222(A1x+ B1y+ C1z+ D1) + λ ( A2x+ B2y+ C2z+ D2) = 0. G G GGGGGG GG 11.1.29Two lines ra1 1 and = 2 +λ bare coplanar if aa 2 1).( bb1 × )=0b ra 2(– 2 x– x yy zz –– x– x – zz yy – 111 22211.1.30Two lines and are coplanar if abcabc111 222 –––xxyyzz 21 2121 abc = 0111 , abc222 GG11.1.31In vector form, if θ is the acute angle between the two planes, .1 = d1rn and GG .nnGG –1 12rn.2 = d2 , then θ=cos G G .nn1 2 GGGG G11.1.32The acute angle θ between the line ra b .=and plane rn dis given by GG .bn sin θ=G G . .bn 11.2 Solved Examples Short Answer (S.A.) Example 1 If the direction ratios of a line are 1, 1, 2, find the direction cosines of the line. Solution The direction cosines are given by a bcl= , m= , n= 222 222 222abc a++ca++b++ bc Here a, b, c are 1, 1, 2, respectively. 1 12 , n=Therefore, l= , m= 222 222 222112 1 ++12 1 ++2++ 1 1 12 ⎛112 ⎞±,i.e., l= , m= , n= i.e. ⎜ ⎟ are D.C’s of the line. 6 66 ⎝6,6 6 ⎠Example 2 Find the direction cosines of the line passing through the points P (2, 3, 5) and Q (–1, 2, 4). Solution The direction cosines of a line passing through the points P (x1, y1, z1) and Q (x2, y2, z2) are x−xy−yz−z21 2121 .PQ , PQ,PQ 222= (1 2) (23) (45) = 911 = 11−−+−+−++Hence D.C.’s are ⎛−3 −1 −1 ⎞⎛ 3 11 ⎞ , ,⎜ ⎟ or ±⎜ ⎟±⎜ ⎟⎝ 11, 11 11 ⎠⎝ 11, 11 11 ⎠. Example 3 If a line makes an angle of 30°, 60°, 90° with the positive direction of x, y, z-axes, respectively, then find its direction cosines. Solution The direction cosines of a line which makes an angle of α, β, γ with the axes, are cosα, cosβ, cosγ ⎛ 31⎞ ,,0Therefore, D.C.’s of the line are cos30°, cos60°, cos90° i.e., ±⎜ 22 ⎟ ⎝⎠ Example 4 The x-coordinate of a point on the line joining the points Q (2, 2, 1) and R (5, 1, –2) is 4. Find its z-coordinate. Solution Let the point P divide QR in the ratio λ : 1, then the co-ordinate of P are ⎛5λ+2 λ+2 –2λ+1⎞ ,,⎜⎟λ+1 λ+1 λ+1⎝⎠ But x– coordinate of P is 4. Therefore, 5λ+2 42=⇒λ= λ+1 21−λ+ =–1Hence, the z-coordinate of P is .λ+1 Example 5 Find the distance of the point whose position vector is (2ˆˆ+– ˆ) fromijkthe plane r G . ( iˆ – 2ˆj+ 4 kˆ) = 9 Solution Here a G = 2ˆˆ ˆ, Gˆ ˆ and d= 9ijk+ ––2ˆ 4ni jk ˆˆˆˆˆ ˆ(2 + – ).(−2 +4 )−9ijki jkSo, the required distance is 1416++ 224−−−9 13 = = 21 21 . x +3 y −4 z +8Example 6 Find the distance of the point (– 2, 4, – 5) from the line == 356 Solution Here P (–2, 4, – 5) is the given point. Any point Q on the line is given by (3λ –3, 5λ + 4, (6λ –8 ), JJJGPQ = (3λ –1) iˆ5ˆj (6 −3) kˆ+λ+ λ. JJJG ˆˆSince PQ ⊥ (35i ++ˆj6k ), we have 3 (3λ –1) + 5( 5λ) + 6 (6λ –3 ) = 0 39λ + 25λ+ 36λ = 21, i.e. λ = 10 JJJG1 15 12 ˆ− iˆ+ ˆj − kThus PQ = 1010 10 JJJG 1PQ =Hence 10 Example 7 Find the coordinates of the point where the line through (3, – 4, – 5) and (2, –3, 1) crosses the plane passing through three points (2, 2, 1), (3, 0, 1) and (4, –1, 0) Solution Equation of plane through three points (2, 2, 1), (3, 0, 1) and (4, –1, 0) is ⎡r G iˆ++2ˆjkˆ)⎤⎡(ˆ–2 ) (ˆ ×i –ˆj ˆ(–(2 . ij ˆ– k )⎤=0⎣⎦⎣ ⎦ i.e. r G.(2ˆˆj ˆ) = or 2x + y + z – 7 = 0 ... (1)i ++k 7 Equation of line through (3, – 4, – 5) and (2, – 3, 1) is x −3 y +4 z +5 == ... (2)−11 6 Any point on line (2) is (– λ + 3, λ – 4, 6λ – 5). This point lies on plane (1). Therefore, 2 (– λ + 3) + (λ – 4) + (6λ – 5) – 7 = 0, i.e., λ = z Hence the required point is (1, – 2, 7). Long Answer (L.A.) Example 8 Find the distance of the point (–1, –5, – 10) from the point of intersection Gof the line Gˆˆ ˆ ˆ ˆ ˆ and the plane .(ˆˆˆ .2 −+ +2 λ(3i jk4 rijk=5r ijk= ++2) −+) GSolution We have Gˆˆ ˆ ˆ ˆ ˆ and .(ˆˆˆ2 −+ +2 λ(3i jk4 rijk=5r ijk= ++2) −+) ˆ ˆˆSolving these two equations, we get [(2ij kˆˆ−+2) +λ(3 ˆ++4ˆ 2 )].(ˆ– ˆ +) =5i jkijk which gives λ = 0. Therefore, the point of intersection of line and the plane is (2, 1,2) and the other given point is (– 1, – 5, – 10). Hence the distance between these two points is Example 9 A plane meets the co-ordinates axis in A, B, C such that the centroid of the Δ ABC is the point (α, β, γ). Show that the equation of the plane is xyz+ β + γ = 3α Solution Let the equation of the plane be x yz + + = 1 abcThen the co-ordinate of A, B, C are (a, 0, 0), (0,b,0) and (0, 0, c) respectively. Centroid of the Δ ABC is xxxyyyzzzabc12312 3123 ,, ,,i.e.333 333 But co-ordinates of the centroid of the Δ ABC are (α, β, γ) (given). a bc Therefore, α = , β = , γ = , i.e. a = 3α, b = 3β, c = 3γ3 33 Thus, the equation of plane is xyz++ = 3αβγ Example 10 Find the angle between the lines whose direction cosines are given by the equations: 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0. Solution Eliminating m from the given two equations, we get ⇒ 2n2 + 3 ln + l2 = 0 ⇒ (n + l) (2n + l) = 0 ⇒ either n = – l or l = – 2n Now if l = – n, then m = – 2n and if l = – 2n, then m = n. Thus the direction ratios of two lines are proportional to – n, –2n, n and –2n, n, n, i.e. 1, 2, –1 and –2, 1, 1. So, vectors parallel to these lines are G aG = i + 2 j – k and b = –2i + j + k, respectively. If θ is the angle between the lines, then GG a.bGGcos θ = a b (+2– k )⋅(–2 j ) 1ij i ++ k – = = 222 2221 2 (–1) (–2) ++ 16++ 1 1 –Hence θ = cos–1 .6 Example 11 Find the co-ordinates of the foot of perpendicular drawn from the point A (1, 8, 4) to the line joining the points B (0, –1, 3) and C (2, –3, –1). Solution Let L be the foot of perpendicular drawn from the points A (1, 8, 4) to the line passing through B and C as shown in the Fig. 11.2. The equation of line BC by using G G GGformula r= a+ λ ( b– a), the equation of the line BC is G r= (– jk +3 )+λ (2 –2 –4 )i jk ⇒ xiyizk= 2 i–2 1 i 3–4 k Comparing both sides, we get x= 2λ, y= – (2λ + 1), z= 3 – 4λ (1) Thus, the co-ordinate of L are (2λ, – (2λ + 1), (3 – 4λ), so that the direction ratios of the line AL are (1 – 2λ), 8 + (2λ + 1), 4 – (3 – 4λ), i.e. 1 – 2λ, 2λ + 9, 1 + 4λ Since AL is perpendicular to BC, we have, (1 – 2λ) (2 – 0) + (2λ + 9) (–3 + 1) + (4λ + 1) (–1 –3) = 0 –5 ⇒λ = 6 The required point is obtained by substituting the value of λ, in (1), which is ⎛ –5 219 ⎞ , , .⎜⎟⎝ 333 ⎠ xy –1 z –2 Example 12 Find the image of the point (1, 6, 3) in the line .12 3 Solution Let P (1, 6, 3) be the given point and let L be the foot of perpendicular from P to the given line. The coordinates of a general point on the given line are x –0 y –1 z–2 , i.e., x = λ, y = 2λ + 1, z = 3λ + 2. 123 If the coordinates of L are (λ, 2λ + 1, 3λ + 2), then the direction ratios of PL are λ – 1, 2λ – 5, 3λ – 1. But the direction ratios of given line which is perpendicular to PL are 1, 2, 3. Therefore, (λ – 1) 1 + (2λ – 5) 2 + (3λ – 1) 3 = 0, which gives λ = 1. Hence coordinates of L are (1, 3, 5). Let Q (x1, y1, z1) be the image of P (1, 6, 3) in the given line. Then L is the mid-point x 1 y 6 z 311 1of PQ. Therefore, 1, 3 522 2 ⇒ x1 = 1, y1 = 0, z1 = 7 Hence, the image of (1, 6, 3) in the given line is (1, 0, 7). Example 13 Find the image of the point having position vector i 3j 4k in the plane r 2– jk 3.i 0 i ljkSolution Let the given point be P 3 4 and Q be the image of P in the plane r . 2i – jk 3 0 as shown in the Fig. 11.4. Then PQ is the normal to the plane. Since PQ passes through P and is normal to the given plane, so the equation of PQ is given by G r = i 3 j 4k 2i – jk Since Q lies on the line PQ, the position vector of Q can be expressed as i 3j 4k 2i – jk , i.e., (12λi + – j ( +)( 3 λ)+ + 4 λ)k Since R is the mid point of PQ, the position vector of R is ⎡(12+λ+ ) ( 3– )j (4 λ)k ⎤⎡ ++ + i 3 j 4k ⎤i λ++⎣ ⎦⎣ ⎦ 2 ⎛λ⎞⎛ λ⎞(λ+1) i+ 3– j+ 4+ ki.e., ⎜⎟⎜ ⎟⎝ 2 ⎠⎝ 2 ⎠ G ijk =0Again, since R lies on the plane r⋅(2– + )+3 , we have ⎧ ⎛ λ⎞lj⎛ λ⎞⎫ ilj⎨(λ+1)i+⎜3– ⎟ +⎜4+⎟k⎬⋅(2 – +k)+3=0 ⎩⎝ 2 ⎠⎝ 2 ⎠⎭ ⇒λ = –2 + Hence, the position vector of Q is (i jk3 +4)–2 2–ijk, i.e. –3i+5j+2k . Objective Type Questions Choose the correct answer from the given four options in each of the Examples 14 to 19. Example 14 The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by (A) (2, 0, 0) (B) (0, 5, 0) (C) (0, 0, 7) (D) (0, 5, 7) Solution (A) is the correct answer. Example 15 P is a point on the line segment joining the points (3, 2, –1) and (6, 2, –2). If xco-ordinate of P is 5, then its yco-ordinate is (A)2 (B) 1 (C) –1 (D) –2 Solution (A) is the correct answer. Let P divides the line segment in the ratio of λ : 1, 6λ+36λ+3 x- coordinate of the point P may be expressed as x= giving =5 so thatλ+1 λ+1 2λ+2 =2λ = 2. Thus y-coordinate of P is .λ+1 Example 16 If α, β, γ are the angles that a line makes with the positive direction of x, y, z axis, respectively, then the direction cosines of the line are. (A) sin α, sin β, sin γ (B) cos α, cos β, cos γ (C) tan α, tan β, tan γ (D) cos2 α, cos2 β, cos2 γ Solution (B) is the correct answer. Example 17 The distance of a point P (a, b, c) from x-axis is 22 22(A) ac (B) ab (C) b2 c2 (D) b2 + c2 Solution (C) is the correct answer. The required distance is the distance of P (a, b, c) from Q (a, o, o), which is Example 18 The equations of x-axis in space are (A) x = 0, y = 0 (B) x = 0, z = 0 (C) x = 0 (D) y = 0, z = 0 Solution (D) is the correct answer. On x-axis the y- co-ordinate and z- co-ordinates are zero. Example 19 A line makes equal angles with co-ordinate axis. Direction cosines of this line are 111 (A) ± (1, 1, 1) (B) ±⎜⎛ ⎞ ⎝ ⎛ 111 ⎞⎛ 1 −1 −1 ⎞ ,,,(C) ±⎜ ⎟ (D) ±⎜ ⎟⎝ 333 ⎠⎝ 3,3 3 ⎠ Solution (B) is the correct answer. Let the line makes angle α with each of the axis. Then, its direction cosines are cos α, cos α, cos α. 1Since cos2 α + cos2 α + cos2 α = 1. Therefore, cos α = Fill in the blanks in each of the Examples from 20 to 22. 3Example 20 If a line makes angles ,and with x, y, z axis, respectively, then 24 4its direction cosines are _______ 3 ⎛ 11 ⎞ Solution The direction cosines are cos , cos , cos , i.e., ±⎜ 0,–24 4 ⎝ Example 21 If a line makes angles α, β, γ with the positive directions of the coordinate axes, then the value of sin2 α + sin2 β + sin2 γ is _______ Solution Note that sin2 α + sin2 β + sin2 γ = (1 – cos2α) + (1 – cos2β) + (1 – cos2γ) = 3 – (cos2α + cos2β + cos2γ) = 2. Example 22 If a line makes an angle of with each of y and z axis, then the angle4which it makes with x-axis is _________ Solution Let it makes angle α with x-axis. Then cos2α + cos 24 + cos24 = 1 which after simplification gives α = 2 . State whether the following statements are True or False in Examples 23 and 24. Example 23 The points (1, 2, 3), (–2, 3, 4) and (7, 0, 1) are collinear. Solution Let A, B, C be the points (1, 2, 3), (–2, 3, 4) and (7, 0, 1), respectively. Then, the direction ratios of each of the lines AB and BC are proportional to – 3, 1, 1. Therefore, the statement is true. Example 24 The vector equation of the line passing through the points (3,5,4) and (5,8,11) is r r 3iˆ 5 ˆj 4kˆ(2iˆ 3ˆj 7kˆ) Solution The position vector of the points (3,5,4) and (5,8,11) are r3ˆ 5 ˆj 4 ˆ, r5iˆ 8 ˆj 11 ˆ ,aikb k and therefore, the required equation of the line is given by r r 3iˆ 5 ˆj 4kˆ(2iˆ 3ˆj 7kˆ) Hence, the statement is true. 11.3 EXERCISE Short Answer (S.A.) JJJG1. Find the position vector of a point A in space such that OA is inclined at 60º to JJJGOAOX and at 45° to OY and = 10 units. 2. Find the vector equation of the line which is parallel to the vector 3iˆ 2ˆj 6kˆ and which passes through the point (1,–2,3). 3. Show that the lines x 1 y 2 z3 234 x − 4 y −1and ==z intersect.5 2 Also, find their point of intersection. 4. Find the angle between the lines r r = 3ˆ − 2ˆj + 6kˆ +λ(2 ˆ + ˆj + kˆ and r r = (2 ˆj − 5 ) ˆ +μ iˆ + 3ˆj + kˆii 2) k (6 2) 5. Prove that the line through A (0, –1, –1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (– 4, 4, 4). 6. Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0. 7. Find the equation of a plane which bisects perpendicularly the line joining the points A (2, 3, 4) and B (4, 5, 8) at right angles. 8. Find the equation of a plane which is at a distance 3 3 units from origin and the normal to which is equally inclined to coordinate axis. 9. If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane. 10. Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7). 11. Find the equations of the two lines through the origin which intersect the line x − 3 y −3 z π == at angles of each.211 3 12. Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 0. 13. If a variable line in two adjacent positions has direction cosines l, m, n and l + δl, m + δm, n + δn, show that the small angle δθ between the two positions is given by δθ2= δl2 + δm2+ δn2 14. O is the origin and A is (a, b, c).Find the direction cosines of the line OA and the equation of plane through A at right angle to OA. 15. Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′, respectively, from the origin, prove that 1111 11++= + + .222 222abca′ b′ c′ Long Answer (L.A.) 16. Find the foot of perpendicular from the point (2,3,–8) to the line 4 x y 1 z . Also, find the perpendicular distance from the given point 263 to the line. 17. Find the distance of a point (2,4,–1) from the line x 5 y 3 z 6 1 4–9 ⎛ 3 ⎞18. Find the length and the foot of perpendicular from the point ⎜1, ,2 ⎟ to the⎝ 2 ⎠plane 2x – 2y + 4z + 5 = 0. 19. Find the equations of the line passing through the point (3,0,1) and parallel to the planes x + 2y = 0 and 3y – z = 0. 20. Find the equation of the plane through the points (2,1,–1) and (–1,3,4), and perpendicular to the plane x – 2y + 4z = 10. 21. Find the shortest distance between the lines given by G (8 3 iˆ (9 16 ) ˆ+r =+λ− + λj +λkˆand r G =15 iˆ+ ˆj +5kˆ +μ(3 i 8ˆj 5) kˆ.(10 7) 29 ˆ +− 22. Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0. 23. The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is ax + by ±( a2 +b2 tan α) z = 0. 24. Find the equation of the plane through the intersection of the planes G . ( iˆ + 3ˆj ) – 6 = 0 and G . (3 iˆ – ˆj – 4 kˆ ) = 0, whose perpendicularr r distance from origin is unity. 25. Show that the points −+3) k3(ijˆˆ++ˆ are equidistant from the plane(ijˆˆ ˆ and k) r G.(5 iˆ +2ˆj −7) kˆ += 9 0 and lies on opposite side of it. JJJG JJJG 26. AB =3ˆˆ +kˆ and CD =3ˆ 2ˆ 4kˆ are two vectors. The position vectorsij– −+ + ij ˆof the points A and C are ˆ ++ j 4ˆ ˆj 2 ˆ , respectively. Find the 67ik and – 9 +k position vector of a point P on the line AB and a point Q on the line CD such JJJGJJJGJJJGthat PQ is perpendicular to AB and CD both. 27. Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles. 28. If l1, m1, n1; l2, m2, n2; l3, m3, n3 are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to l1 + l2 + l3, m1 + m2 + m3, n1 + n2 + n3 makes equal angles with them. Objective Type Questions Choose the correct answer from the given four options in each of the Exercises from 29 to 36. 29. Distance of the point (α,β,γ) from y-axis is 2 γ2(A) β (B) β (C) β + γ (D) α+ 30. If the directions cosines of a line are k,k,k, then 11 (A) k>0 (B) 0