Chapter 11 – Three Dimensional Geometry

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 b c l= ,m= , n= a 2 + b2 + c2 a 2 + b2 + c2 a 2 + b2 + c2 Here a, b, c are 1, 1, 2, respectively.

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 x2 − x1 y2 − y1 z2 − z1 , , . PQ PQ PQ Here PQ = ( x2 − x1 ) 2 + ( y2 − y1 ) 2 + ( z2 − z1 ) 2 = (−1 − 2) 2 + (2 − 3) 2 + (4 − 5) 2 = 9 +1+1 = 11 Hence D.C.’s are THREE DIMENSIONAL GEOMETRY 225  −3 −1 −1   3 1 1   , ±  11 11 ,  or ±  , , .  11   11 11 11 

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  3 1  Therefore, D.C.’s of the line are cos30°, cos60°, cos90° i.e., ±  2 , 2 , 0   

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 = 4 ⇒λ = 2 λ +1 −2 λ +1 Hence, the z-coordinate of P is = –1 . λ +1

Find the distance of the point whose position vector is (2iˆ + ˆj – kˆ) from the plane r . ( iˆ – 2 ĵ + 4 k̂ ) = 9 Solution Here a = 2iˆ + ˆj – kˆ , n = iˆ – 2 ˆj + 4kˆ and d = 9 ( 2iˆ + ˆj – kˆ ).( iˆ − 2 ˆj + 4kˆ ) − 9 So, the required distance is 1 + 4 +16 226 MATHEMATICS 2− 2− 4−9 13 = = . 21 21 x +3 y − 4 z +8

Find the distance of the point (– 2, 4, – 5) from the line = = 3 5 6 Solution Here P (–2, 4, – 5) is the given point. Any point Q on the line is given by (3–3, 5+ 4(6–8 ), PQ = (3 –1) iˆ + 5λ ˆj + (6λ − 3) kˆ . Since ( ) PQ 3iˆ + 5 ˆj + 6k , we have ˆ

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 – (2iˆ + 2 ˆj + kˆ)  . (iˆ – 2 ˆj ) × (iˆ – ˆj – kˆ )  = 0     i.e. r .(2iˆ + ˆj + kˆ) = 7 or 2x + y + z – 7 = 0 ... (1) Equation of line through (3, – 4, – 5) and (2, – 3, 1) is x −3 y + 4 z +5 = = ... (2) −1 1 6 THREE DIMENSIONAL GEOMETRY 227 Any point on line (2) is (– + 3, – 4, 6 – 5). This point lies on plane (1). Therefore, 2 (– + 3) + ( – 4) + (6 – 5) – 7 = 0, i.e., z Hence the required point is (1, – 2, 7).

Find the distance of the point (–1, –5, – 10) from the point of intersection of the line r = 2 iˆ − ˆj + 2kˆ + λ (3 iˆ + 4 ˆj + 2kˆ) and the plane r . ( iˆ − ˆj + kˆ ) = 5 . Solution We have r = 2 iˆ − ˆj + 2kˆ + λ (3 iˆ + 4 ˆj + 2kˆ) and r . ( iˆ − ˆj + kˆ ) = 5 Solving these two equations, we get [(2 iˆ − ˆj + 2kˆ) + λ (3 iˆ + 4 ˆj + 2kˆ)].(iˆ – ˆj + kˆ) = 5 which gives 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 [ 2 − ( − 1)]2 + [−1+ 5]2 +[2 − (−10)]2 , i.e. 13

A plane meets the co-ordinates axis in A, B, C such that the centroid of the ∆ ABC is the point (α, β, γ). Show that the equation of the plane is x y z + + =3 α β γ Solution Let the equation of the plane be x y z + + =1 a b c Then the co-ordinate of A, B, C are (a, 0, 0), (0,b,0) and (0, 0, c) respectively. Centroid of the ∆ ABC is  x1 + x2 + x3 y1 + y2 + y3 z1 + z2 + z3   a b c  , ,  i.e.  , ,  3 3 3 3 3 3 But co-ordinates of the centroid of the ∆ ABC are (α, β, γ) (given). 228 MATHEMATICS a b c Therefore, α= , β = , γ = , i.e. a = 3α, b = 3β, c = 3γ 3 3 3 Thus, the equation of plane is + + =3 α β γ

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 a = i + 2 j – k and b = –2 i + j + k , respectively.. If θ is the angle between the lines, then a .b cos θ = a b (i + 2 j – k ) ⋅ ( –2i + j + k ) 1 = = – 1 + 2 + (–1) 2 2 2 (–2) + 1 + 1 2 2 2  1 Hence θ = cos–1  –  . THREE DIMENSIONAL GEOMETRY 229

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 formula r = a + λ ( b – a ), the equation of the line BC is ( ) ( r = – j + 3k + λ 2i – 2 j – 4k ) ⇒ xi + yi + zk = 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= The required point is obtained by substituting the value of λ, in (1), which is 230 MATHEMATICS  –5 2 19   , , .  3 3 3 x y –1 z – 2

Find the image of the point (1, 6, 3) in the line = = . 1 2 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. 1 2 3 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 x1 + 1 y +6 z +3 of PQ. Therefore, = 1, 1 =3 1 =5 2 2 2 ⇒ x1 = 1, y1 = 0, z1 = 7 Hence, the image of (1, 6, 3) in the given line is (1, 0, 7). THREE DIMENSIONAL GEOMETRY 231

Find the image of the point having position vector i + 3 j + 4k in the ( ) plane r ⋅ 2i – j + k + 3 = 0 . ( ) Solution Let the given point be P i + 3j + 4k and Q be the image of P in the plane ( ) r . 2i – j + k + 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 ( ) ( r = i + 3j + 4k + 2i – j + k ) Since Q lies on the line PQ, the position vector of Q can be expressed as (i + 3j + 4k ) + ( 2i – j + k ) , i.e., (1+ 2 ) i + (3 – ) j + ( 4 + ) k Since R is the mid point of PQ, the position vector of R is (1 + 2λ ) i + ( 3 – λ ) j + ( 4 + λ ) k  + i + 3 j + 4k      232 MATHEMATICS  λ  λ i.e., (λ +1) i +  3 –  j +  4 +  k  2  2 ( ) Again, since R lies on the plane r ⋅ 2i – j + k + 3 = 0 , we have   λ  λ   ( λ + 1) i +  3 –  j +  4 +  k  ⋅ (2i – j + k ) + 3 = 0   2   2   ⇒ λ = –2 ( ) ( ) Hence, the position vector of Q is i + 3 j + 4k –2 2i – j + k , i.e. –3i + 5 j + 2k .

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.

P is a point on the line segment joining the points (3, 2, –1) and (6, 2, –2). If x co-ordinate of P is 5, then its y co-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 + 3 6λ + 3 x - coordinate of the point P may be expressed as x = giving λ +1 = 5 so that λ +1 2λ + 2 λ = 2. Thus y-coordinate of P is λ +1 = 2 .

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 THREE DIMENSIONAL GEOMETRY 233 Solution (B) is the correct answer.

The distance of a point P (a, b, c) from x-axis is

• (A)a2 + c2
• (B)a 2 + b2
• (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 b2 + c2 .

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.

A line makes equal angles with co-ordinate axis. Direction cosines of this line are  1 1 1 

• (A)± (1, 1, 1)
• (B)± , ,   3 3 3 1 1 1  1 −1 −1 
• (C)± , , 
• (D)± , ,  3 3 3  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. Since cos2 α + cos2 α + cos2 α = 1. Therefore, cos α = ± Fill in the blanks in each of the Examples from 20 to 22. 3

If a line makes angles , and with x, y, z axis, respectively, then 2 4 4 its direction cosines are _______ 234 MATHEMATICS 3  1 1  Solution The direction cosines are cos , cos , cos , i.e., ±  0, – . 2 4 4  2 2

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. π

If a line makes an angle of with each of y and z axis, then the angle which it makes with x-axis is _________ Solution Let it makes angle α with x-axis. Then cos2α + cos + cos 2 2 =1 4 4 which after simplification gives α = . State whether the following statements are True or False in Examples 23 and 24.

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.

The vector equation of the line passing through the points (3,5,4) and (5,8,11) is r = 3iˆ + 5 ˆj + 4kˆ + ( 2iˆ + 3 ˆj + 7kˆ ) Solution The position vector of the points (3,5,4) and (5,8,11) are a = 3iˆ + 5 ˆj + 4kˆ,b = 5iˆ + 8 ˆj + 11kˆ , and therefore, the required equation of the line is given by r = 3iˆ + 5 ˆj + 4kˆ + ( 2iˆ + 3 ˆj + 7kˆ ) Hence, the statement is true. THREE DIMENSIONAL GEOMETRY 235