D GEOMETRY ) If α β γ are angle made by a line with positive directions of x y and z axes respectively show that i) sin α + sin β + sin γ ii) cos α + cos β + cos γ + 0 Solution:- i) are angle made by a line with positive directions of x y z axes. L.H.S. sin α + sin β + sin γ cos α + cos β + cos γ ( sin θ + cos θ ) (cos α + cos β + cos γ) ( cos α + cos β + cos γ ) R.H.S.
ii) L.H.S. cos α + cos β + cos γ + cos α + cos β + cos γ (cos α + cos β + cos γ ) () ( cos α + cos β + cos γ ) 0 ( cos θ cos θ ) +
) A line lies in the yz plane and makes an angle 0 with y axis find its inclination with z axis.and hences find its d.c.s Solution:- Since the lines lies in yz plane it is perpendicular to x axis. If α β γ are direction angels of the line then α 90 given β 0 to find γ. cos α 0 cos β Now cos α + cos β + cos γ 0 + + cos γ cos γ 4 4 cos γ ± γ 60 or 0 i.e. the line is inclined at 60 or 0 with z axis.
4 ) If α β and γ π find α where α β γ are angles made by the line with the 4 positive directions of X Y and Z axes respectively. Solution:- Let α β γ Respectively be the angles made the line. With the positive direction of x y and z axes Now α β γ π 4.(i) Now cos α + cos β + cos γ π cos α + cos α + cos 4 cos α + cos α + cos α
5 cos α 4 cos α ± cosα or α π cosα or α π
6 4) Find d.c.s of a line which is equally inclined to co-ordinate axes Solution:- Let α β γ be the directed angles of the line with co-ordinate axes Given α β γ But cos α + cos β + cos γ cos α cos α ± cosα cos β cos γ ± The d.c.s. of the lines or
7 5) Show that no line in space can make angles of π 6 & π 4 with x and y - axes Solution:- If α β γ are the directed angles of the line then But π α 6 β π 4 cos α + cos β + cos γ π cos + cos 6 π + cos γ 4 + + cos γ 4 + + cos γ 5 4 + cos γ cos γ cos γ < 0 No line can make angles π 6 and π 4 5 4 4 with x and y axis.
8 6) Find the direction cosines of line Op if P(4 ) and O is origin Solution:- OP 4i + j + k OP 6 + 9 + 4 9 D.r.s are 4 D.c.s. of OP are 4 9 9 9
9 7) Find angle between AB & CD if A( 5 ) B ( -55) C( --4) D( - ) Solution:- The direction ratios of AB are - 0 i.e. - 0. The direction ratios of CD are 0 i.e. 0. Let θ be the angle between AB and CD. a a + b b + c c cos θ a + b + c a + b + c () + ( )() + (0) (0) + + 0. + + 0 + 0. 0 0 θ 90 i.e. lines AB and CD are perpendicular.
0 8) Find k if ABC is right angle at B where A (564) B (564) B (44 ) C (8k) Solution-: A (564) B (44) C (8 K) ABC is right angled at B AB + BC AC. (i) AB + 4 + 9 4 BC 6 + 4 + (K ) 0 + K K + + K K AC 9 + 6 + (K 4) 5 + 6 + K 8K From (i) 4 + + K K 4 + K 8K 6K 6 K
9) Find k if the line with d.r.s k -6- and k - k 4 are perpendicular Solution:- Since lines perpendiculars. a a + b b + c c 0 K (K ) 6 (K) 8 0 K K 6 K 8 0 K 7K 8 0 (K 8) (K + ) 0 either K 8 0 or K + 0 K 8 or K
0) The direction cosines of two lines are determined by the relation l 5m + n 0 and 7 + 5 m n 0 find them Solution:- l 5m + n 0 l 5m n 7l + 5m n 0 7(5m n) + 5m n 0 75m 0mn + 6n + 5m n 0 80m 0mn + 60n 0 Dividing throughout by 0 we get 6m 7mn + n 0 6m 4mn mn + n 0
m (m n) n (m n) (m n)(m n) 0 m n or m n n i. when m n l 5 n n n Direction ratios l m n i.e. n n i.e. Direction cosines are 6 6 6
4 ii. n when m l 5m n 0n n Direction ratios l m n i.e. n n n i.e. Direction cosines are + 4 + 9 4 4 i.e. 4 4 4
5 ) Show that the angle between any two diagonals of a cube is cos - Solution:- Let a be the length of an edge of the cube with one vertex at the origin. The diagonals of the cube are OP AR BS and CQ. Consider the diagonals OP and AR The direction ratios of OP and AR are a 0 a 0 a 0 and 0 a a 0 a 0 i.e. a a a and a a a respectively. Let θ be the angle between OP and AR then cosθ a + a + a. (a ) + a + a a a + a a + a a
6 cosθ a a Z C (00a) R (0aa) cosθ S (a0a) P (aaa) θ cos O (000) B (0a0) Y A (a00) Q (aa0) X
7 ) A line make angle α β γ δ with four diagonal of acute prove that cos - α + cos β + cos γ + cos δ 4 Solution:- F E C G B D O A Let OABC FEDG be a cube of edge one unit such that vertex O is at the origin and edges OA OC OG along the co-ordinate axes.
8 O (000) O (000) G (00) A (00) B (0) C (00) D (0) F (0) Direction ratios of diagonal OE are. diagonal BG are. diagonal CD are. diagonal AF are. Let the line which makes angles α β γ and δ with the diagonals of the cube has direction ratios x y z. cos α + cos β + cos γ + cos δ
9 x() + y() + z() + x + y + z. x() + y() + z( ) x + y + z. + x() + y( ) + z() + x + y + z. x( ) + y + z x + y + z. (x + y + z) + (x + y z) + (x y + z) + ( x + y + z) (x + y + z ) (x + y + z + xy + yz + xz) + (x + y + z + xy xz yz) (x + y + z xy yz + zx) + (x + y + z xy xz yz) (x + y + z )
0 4(x + y + z ) (x + y + z ) cos α + cos β + cos γ + cos δ 4
) A line passing thought origin make angle θ θ θ with the plane XOY YOZ and ZOX respectively then prove that cos θ + cos θ + cos θ Ans:- Consider a line L passing through origin Z making θ with XOY plane θ with L YOZ plane and θ with ZOX plane. Let P (x y z) be any point on line L such that l (OP) r l (OP) x + y + z (by distance formula) O θ P (x y z) z Y Q X Draw PQ XY Plane and join OQ. QOP θ
Q (x y 0) l (OQ) x + y (by distance formula) cos θ l (OQ) l (OP) x + y x + y + z... (i) Similarly we can prove cos θ y + z x + y + z... (ii) and cos θ z + x x + y + z... (iii)
From (i) (ii) and (iii) we get x + y x + y + z + y + z x + y + z + z + x x + y + z x + y x + y + z + y + z x + y + z + z + x x + y + z (x + y + z ) x + y + z cos θ + cos θ + cos θ
4 4) The direction ratios of AB are. If A (4 5) and l(ab) 6 units find the coordinates of B. Solution:- Direction ratios of line AB are. Direction ratios are proportional to α α α α such that l(ab) 6 4α + 4α + α 6 9α 6 α ± Direction ratios are 4 4 or 4 4. Let B(x y z) A(4 5)
5 x 4 4 y 4 z 5 x 0 y 5 z 7 B (0 5 7) Or x 4 4 y 4 z 5 x 8 y z B (8 )
6 5) Find the direction cosines of the sides of the triangle whose vertices are ( 5 4). ( ) and ( 5 5 ). What type of triangle is it? Solution:- A ( 5 4) B ( ) C ( 5 5 ) A B b a 4i 4j + 6k BC c b 4i 6j 4k CA a c 8i + 0j k A B BC 6 + 6 + 6 68 It is an isosceles triangle.
7 And direction ratios of sides are ( ) ( ) (4 5 ). Direction cosines of sides are 7 7 7 7 7 7 4 7 5 7 7
GROUP (A) HOME WORK PROBLEMS Q-) A line makes 45 0 with X axis 90 0 with Y axis and 5 0 with Z axis. Find the direction cosines of the line. Ans. A line makes 45 0 with X-axis 90 0 with Y-axis and 5 0 with Z-axis. Given : α 45 0 Find the β 90 0 direchon γ 5 0 cosines of the line α be the angle with x axis β be the angle with y axis γ be the angle with z axis cos α cos 45 0 cos β cos 90 0 0 cos γ cos 5 0 cos (80 0 45 0 ) Hence the direction ratios of the line are and the direction cosines of the line are or Q-) Show that two line with d.r.s and 0 are perpendicular to each other. Ans. Given : let the two lines are AB and CD. Drs (AB) ( ) (L M n ) Drs (CD) ( 0) (L M n ) Consider l l + m m + n n () ( ) + () () + () (0) + 0 The given lines are perpendicular. DCS 0 Q-4) A line lies in the YZ plane and makes an angle 0 0 with Y axis find its inclination with Z axis Ans. Given : line lies in the YZ plane β 0 0 β is the angle made by the line with y axis. Q-) A line passes through A ( ) and B (5 ). Find d.r.s of line AB. Ans. Let A ( ) and B (5 ) be the given points. Then the direction ratios of the line AB are 5 i.e The direction cosines of the line AB are ( ) ( ) ( ) ± ± + + + + ± i.e ( ) ( ) ( ) ( ) ( ) + + ± α 90 0 (As the line dies in YZ plane. Hence it is ar to x axis) and r 60 0 or 0 0 Q-5) Find the angle between lines AB and CD where A ( ) B ( ) C (4 ) and D ( 0) Ans. Given : line AB and CD where A ( ) B ( ) C (4 ) D ( 0) } } AB CD
Mahesh Tutorials Science 9 Drs ( AB ) A B ( ) Drs ( CD ) C D ( ) cos θ cos θ θ θ a a + b b + c c a + b + c a + b + c + 6. 6 6. 6 cos π c 8k 64k + 4 0 8k + 64k 4 0 4k + k 7 0 4k + 4k k 7 0 k (k + 7) (k + 7) 0 (k + 7) (k ) 0 k 7 or Q-7) Find the vector of magnitude 9 which is equally inclined to the coordinate axes. Ans. let a be the vector which is equally inclined to the co-ordination axes. If α β γ are the direction angles of the line then α β γ cos α + cos β + cos γ gives cos α cos α Q-6) If angle between the lines with d.r.s is k and is π find k. Ans. Direction ratios of first line are a b k c direction ratios of second line are a b c The between them θ but angle between the lines is given by cos θ cos π 5 + 4k 6 8 4k π a a + b b + c c a + b + c a + b + c ( ) k ( ) ( ) + + ( ) ( ) ( ) + + + + k k + 4 + 4 k + ++ 4 4 k 5 + 4k 6 squaring both the sides we get (5 + 4k )6 (8 4k) 0 + 4k 64 64k + 6k 64 64k 6k 0 4k 0 cos α ± the unit vectors along the vector a are i + j + k and i + j + k i.e Now a 9 ± ( + + ) i j k a ± 9 ( + + ) i j k i.e a ± ( i + j + k ) Q-8) If l m n and l m n are the direction cosines of two lines then show that the direction cosines of the line perpendicular to them are proportional to m n m n n l n l l m l m Ans. Let l m n be the direction cosines of the line perpendicular to each of the given lines ll + mm + nn 0 ll + mm + nn 0 on solving (i) and (ii)...(i)...(ii)
0 Mahesh Tutorials Science m m l n n m l n l n l l n m m i.e 5 k + k 4 k + 4 k 4 k + 6 k k + k + k + Q-9) Show that the vector AB is perpendicular to CD where A ( 4 ) B ( ) C (0 ) and D ( 5 6). Ans. The direction of the line along vector AB are a b 4 5 c ( ) 4 The direction ratio of the line along the vector CD are a 0 b 5 c 6 4 a a + b b + c c ( )() + ( 5)() + 4(4) 6 0 + 6 0 the line along the vector AB is perpendicular to the line along vector CD. Hence AB is perpendicular to CD Q-0) If a line drawn from the point A () is perpendicular to the line joining P (46) and Q (5 4 4) then find the coordinates of the foot of the perpendicular. Ans. Let AM be the perpendicular from the point A( ) to the line PQ where P( 4 6) and Q(5 4 4) let M divides PQ inteunally in the ratio K : M m n l m n 5 k + 4 k + 4 4 k + 6 k + k + k + The direction ratios of AM are k + 4 k +4 4 k +6 k + k + k + n l m l n l m n m l Hence the direction cosines of the line perpendicular to the given lines are proportional to m n m n n l n l l m l m Q-) Find the direction cosines of vector which is perpendicular to vectors with direction ratios and 0. Ans. Let a b c are the direction ratios of the required vector which is perpendicular to the vector with the direction ratio and 0 a + b + c 0 and 0. a + b + c 0 a b c 0 0 a b c 4 0 0 a b c i.e 4 k k + k + 5 k + k + k + and the direction ratios of PQ are 5 4 4 4 6 i.e 4 0 Since AM is perpendicular to PQ 4k k + k +5 4 + 0 0 k + k + k + 6k 6k 0 0 0k 0 k M i.e M ( 4 5) a b c The direction ratios of the required vector are ( ) The direction cosines of the vector are + + + + ( ) ( ) i.e 6 8 0 Hence the co-ordination of the foot of perpendicular are ( 4 5) ( ) + +
Mahesh Tutorials Science Q-) If the direction ratios of two vectors are connected by the relations p + q + r 0 and p + q r 0 find angle between them. Ans. Given p + q + r 0...(i) and p + q r 0...(ii) from (i) P (q + r) Putting the ralue of P in eq. we get [ (q + r)] + q r 0 q + qr + r + q r 0 q + qr 0 q (q + r) 0 q 0 or q + r 0 q 0 or q r Now P (q + r) therefore if q 0 p (0 + r) r P r and q 0 Q-) ABC is a triangle where A ( 5) B ( ) and C (λ 5 µ). If the median through A is equally inclined to the axis then find the values of λ and µ Ans. Let AD be the medion of the ABC through A. Then D be the midpoint of BC the direction ratio of median AD are λ µ + 4 5 i.e λ 5 µ + 8 i.e λ 5 µ 8 the direction cosines of the median AD are l m λ 5 ( λ 5 ) + + ( µ 8) ( λ 5 ) + + ( µ 8) P r and q 0 the direction ratio of the first vector are a b 0 c If q r P ( r + r) 0 P 0 and q r q r P 0 and The direction ratio of the second rector are a 0 b c let θ be the angle between the vectors then n µ 8 ( λ 5 ) + + ( µ 8) Now the median is equally inclined to the axes. If α β r are the direction angles of the median then α β r cos α cos β cos r l m n λ 5 ( λ 5 ) + 4 + ( µ 8) cos θ a a + b b + c c ( ) a + b + c 0 + + ( ) ( ) ( ) ( ) 0 + 0 + ( ) ( ) + 0 +. 0 + +. µ 8 ( λ 5 ) + 4 + ( µ 8) λ 5 µ 8 λ 7 and µ 0 ( λ 5 ) + + ( µ 8) cos θ cos π θ π
Mahesh Tutorials Science Q-4) Find acute angle between lines having direction ratios 5 4 and Ans. Direction ratios of the lines are 5 4 and cos θ cos θ θ 5 ( ) + 4( ) + ( ) ( ) 5 +6 + 9 + 4 + 5 + 8 4 4 84 4 cos 4 7 4 7 Q-5) Can three numbers for some line? Justify. Ans. l m n l + m + n BASIC ASSIGNMENTS (BA) BA + + can be d.c.s of some lines be d.c.s Q-) Show that there is no line in space whose direction angles are 0 0 45 0 60 0 Ans. Let if possible the direction angles of a line in space be 0 0 45 0 60 0 with standard notahone α 0 0 β 45 0 γ 60 0 cos α cos 0 0 Q-) Find the value of λ for which the points (6 ) (8 7 λ) and (5 4) are collinear. Ans. Let A (6 ) B (8 7 λ) and C (5 4) be cos β cos 45 0 cos γ cos 60 0 cos α + cos β + cos γ the given points Then direction ratios of line AB are 8 6 7 ( ) λ i.e 6 λ and direction ratios of the line AC are 5 6 ( ) 4 i.e Since A B C are collinear. The direction ratio of AB and AC are in same proportion 6 λ 4 λ λ + + 4 4 λ and i.e α β γ dose not satisfy the identity cos α + cos β + cos γ Q-) Find angle between the lines whose direction cosines l m n satisfy equations 5l + m + n 0 and 5mn nl + 6lm 0. Ans. Given 5l + m + n 0 and 5mn nl + 6lm 0 From (i) m (5l + n)...(i)...(ii) Putting the value of m in equation (ii) we get 5 (5l + n) n nl 6l (5l + n) 0 5ln 5n nl 0l 8ln 0 0l 45ln 5n 0
Mahesh Tutorials Science l + ln + n 0 l + ln + ln + n 0 l (l + n) + n (l + n) 0 (l + n)(l + n) 0 l + n 0 or l + n 0 l n or n l Now m (5l + n) therefore if l n m ( 5n + n) n l l Q-4) If a line drawn from the point A ( ) is perpendicular to the line joining P ( 4 6) and Q (5 4 4) then find the coordinates of the foot of the perpendicular. Ans. Let M divides PQ internally in thed ratios k: M m n m n the direction ratios of the first line are a b c If n l m (5l 6l) l l m l m n n the direction ratios of the second line are a b c Let θ be the angle between the lines. Then cos θ θ a a + b b + c c a + b + c. a + b + c ( ) ( ) + ( ) + ( ) ( ) + +. + + ( ) + cos 6. 6 6 6 6 5 k + 4 k + 4 4 k + 6 k + k + k + BASIC ASSIGNMENTS (BA) BA Q-) Find the direction cosines of the sides of the triangle whose vertices are ( 5 4) ( ) and ( 5 5 ). What type of triangle is it? Ans. Let ABC be the triangle where A ( 5 4) the direction ratio of AM are 5 k + 4 k + 4 k + 6 k + k + k + i.e i.e B ( ) and C ( 5 5 ). the direction ratios of side AB are 5 ( 4) i.e. 4 4 6 i.e. the direction cosines of side AB are + + + + i.e. 5 k + k 4 k + 4 k k + k + 4 k + + 5 m k k k + k + k + and the direction ratios of PQ are 5 4 4 4 6 i.e 4 0 since AM is or to PQ 4k 4 k + to k + k + 6k 6k 0 0 ( ) ( ) 7 7 7 4 k + 6 k k + k + 5 k + 0 0k 0 k 0 M 6 8 0 i.e M ( 4 5) Hence the co-ordinates of the foot of perpendicular are ( 4 5) ( ) + + The direction ratios of side BC are
4 Mahesh Tutorials Science 5 ( ) 5 i.e. 4 6 4 i.e. the direction cosines of side BC are + + + + + + i.e. 7 7 7 The direction ratios of side AC are 5 5 5 ( 4) i.e. 8 0 i.e. 4 5 the direction cosines of side AC are 4 5 4 + 5 + 4 + 5 + ( ) ( ) ( ) 4 + 5 + Q-) Show that the four points (5 ) ( 4) (7 4 7) and ( 6 0) are the vertices of rhombus. Ans. Let A (5 ) B ( 4) C (7 4 7) and D ( 6 0) d. r. s. of AB are : 5 + 4 i.e. 6 d.r.s. of CD are : 7 6 + 4 0 7 i.e. 6 Thus d.r.s. of AB are same as that of CD AB CD d.r.s. of AC are 7 5 4 + 7 i.e. 6 d.r.s. of BD are + 6 + 0 4 i.e. 6 AC BD D C i.e. 4 5 4 4 4 Let θ be the angle between the side AB and AC. Then by using the formula cos θ cos θ l l + m m + n n we get 8 0 + + 7 4 7 4 7 4 7 4 Let α be the angle between the sides AB and BC. Then cos β 4 6 6 + + 7 7 7 4 7 A B ABDC is a parallelogram. d.c.s. of diagonal AD are 4 5 9 d.c.s of diagonal BC are 8 Sum of products of d.r.s. 8 ( 4) + ( ) ( 5) + () (9) + 5 + 7 0 AD BC Thus ABCD is a parallelogram in which diagonals are mutually perpendicular. ABCD is a rhombus. cos θ cos α θ α the triangle is an isosceles triangle. Hence the direction cosines of the sides of the triangle are ; 7 7 7 ; 7 7 7 4 5 ; 4 4 4 and it is an isosceles triangle. Q-) If Q be the foot of the perpendicular from Ans. P ( 4 ) on the line joining the points A ( 4) and B ( 4 5) dind coordinates of Q. P ( 4 ) A ( 4) Q ( x y z) B ( 4 5)
Mahesh Tutorials Science 5 Let Q (x y z) d.r.s. of AQ are : x y z z 4 d.r.s. of AB are : 4 5 4 i.e. Since AQ and AB are segments of the same line x y z 4 (say) x t + y t + z t + 4 Q (t + t + t + 4) d.r.s. of PQ are t + t + 4 t + 4 i.e. t t t + Since PQ AB (t ) + (t ) + (t + ) 0 4t + 4t 4 + t + 0 9t 5 0 t 5 9 Q (t + t + t + 4) 5 5 5 + + + 4 9 9 9 9 8 4 9 9 9 Q-5) If points A ( λ) B ( 0 ) and C ( 4 µ ) are collinear show that point D ( 9) is on the same line. Ans. The direction ratios of line AB are + 0 λ i.e. λ Direction ratios of line AC are 4 + µ λ i.e. µ λ Since A B C are collinear. AB AC i.e. µ µ λ + λ Solving we have µ 6 λ λ + λ Direction ratios of AB are or and Direction ratios of AD are + 9 i.e 96 Now 9 6 Hence AB AD point A is common. AB and AD are along the same line. The point D ( 9) is on the line containing A B and C. Q-4) A ( 6) B ( 4 5) and C ( 5 ) are the vertices of triangle ABC. Find m ABC. Ans. We know that the direction ratios of the line segment joining (x y z ) and (x y z ) are x x y y z z Direction ratios of BA are 4 6 5 i.e Direction ratios of BC are 5 4 5 i.e let m ABC θ then cos θ cos θ + + ( ) ( ) + +. + + 4 9. 9 0 θ 90 0