11. THREE – DIMENSION - Aniket Mathematics

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11. THREE – DIMENSION 1. Different Forms of Straight line : (a) Equation of a straight line through some point a and having directional vector b . r = a + λ b Cartesian Equivalent : If passing point (x1 , y1, z1) i.e, a = x1 i + y1 j + z1k & d.r’s of the line (a , b, c) i.e, directional vector b = ai + bj + ck x – x 1 = y – y1 = z – z1 a b c (b) Equation of straight line through two points a and b r = a + λ ( b – a ) Or r = b + λ ( b – a ) Cartesian Equivalent : If passing points are (x1 , y1, z1) & (x2 , y2, z2) i.e, a = x1 i + y1 j + z1k & b = x2 i + y2 j + z2k x – x 1 = y – y1 = z – z1 x2 – x1 y2 – y1 z2 – z1 Or x – x 2 = y – y2 = z – z2 x2 – x1 y2 – y1 z2 – z1 2. Collinearity of Three Points : Three points are said to be collinear if the straight line through two points passes through the third point also. c – a = λ ( b – a ) ; ( if the vector joining two point set is parallel ) x3 – x1 = y3 – y1 = z3 – z1 x2 – x1 y2 – y1 z2 – z1 3. Angle between two line : Two lines intersect at the angle their directional vectors are inclined. If, r = a1 + λ b1 and r = a2 + µb2 are two lines, then angle between the lines is the angle between b1 and b2.

If ‘θ’ is the angle between the vectors b1 = a1 i + b1 j + c1 k and b2 = a2 i + b2 j + c2 k i.e, the lines are x – x 1 = y – y1 = z – z1 a1 b1 c1 and x – x 2 = y – y2 = z – z2 a2 b2 c2 Then cos θ = b1 . b2 = a1 a2 + b1 b2 + c1 c2 | b1 | | b2 | √a1

2 + b12 + c1

2 .√a22 + b2

2 + c22

(a) If lines are perpendicular, b1 . b2 = 0 a1 a2 + b1 b2 + c1 c2 = 0 (b) If lines are parallel, b1 = λ b2 a1 = b1 = c1 a2 b2 c2


4. Shortest Distance (S.D) : If r = a1 + λ b1 and r = a2 + µb2 be any two skew lines (which are neither parallel nor intersecting). Then the S.D between the lines is | ( a2 – a1 ) . ( b1 × b2 ) | | b1 × b2 | 5. Condition of Intersection of Lines : If two lines r = a1 + λ b1 and r = a2 + µb2 are intersecting then their S.D = 0. But S.D = 0 need not imply lines are intersecting. If S.D = 0 under the condition b1 × b2 = 0 lines are parallel, b1 = λ b2 = b Then S.D = | ( a2 – a1 ) × b | | b | Moreover, If, r = a1 + λ b1 and r = a2 + µb2 are two lines, then point of intersection is given by solving the three equations in ‘λ’ and ‘µ’ achieved on equating coefficients of ‘ i ’, ‘ j ’ and ‘ k ’ in a1 + λ b1 = a2 + µb2 . However, if unique ‘λ’ and ‘µ’ exists, then lines are intersecting and point of intersection is given by the value of r for that ‘λ’ and ‘µ’. If no unique ‘λ’ and ‘µ’ exists i.e, the three equations in ‘λ’ and ‘µ’ achieved on equating coefficients of ‘ i ’, ‘ j ’ and ‘ k ’ in a1 + λ b1 = a2 + µb2 does not hold together for any value of ‘λ’ and ‘µ’, the lines are non – intersecting. 6. Different forms of planes : (a) General Form : Every linear equation in x, y, and z of the form ax + by + cz + d = 0 is a plane in space, provided a, b, c not all zero at a time. i.e, 2x + 3y + 4z + 5 = 0 , 2x + 3y = 4 , 3y + z = 2, x + 3z = 4, x = 2, y = 3 , z = 3, all are planes. (b) Normal Form : If ‘ ή ’ is unit normal vector on the plane and ‘p’ is the length of perpendicular from origin. Then equation of plane is r . ή = p. Moreover, If N is normal vector on the plane and ‘p’ is the length of perpendicular from origin. Then equation of plane is r . N = p r . N = d , | N | with d = p | N | Thus, if N = ai + bj + ck (normal vector) , and r = xi + yj + zk Then, r . N = d ax + by + cz = d, thus in general equation of a plane coefficient of x, y and z gives the d.r’s of normal to the plane. Thus, if plane is 2x + 3y + 4z + 5 = 0, then normal to the plane is N = – 2i – 3j – 4k, we take the – ve sign to make the equation of the form : – 2x – 3y – 4z = 5

(c) Normal Point Form : Equation of a plane through the point a and having normal vector N is ( r – a ) . N = 0. Thus, if N = ai + bj + ck (normal vector) , and a = x1i + y1j + z1k ( passing point) and r = xi + yj + zk ( current point), Then, equation of plane is : a( x – x1) + b(y – y1) + c(z – z1) = 0 (d) Three Point Form : Equation of a plane through three points a , b and c is ( r – a ) . {( b – a ) × ( c – a )} = 0. ( r – a ) . {( b – a ) × ( c – b )} = 0. and - - - .


Thus, if a = x1i + y1j + z1k, b = x2i + y2j + z2k and c = x3i + y3j + z3k are three points. Then equation of plane is x – x1 y – y1 z – z1 x2 – x1 y2 – y1 z2 – z1 = 0 x3 – x2 y3 – y2 z3 – z2 (e) Intercept Form : If a plane makes intercepts ‘a’, ‘b’, and ‘c’ on coordinate axes then equation of the plane is given by : x + y + z = 1 a b c 7. Condition of Coplanarity of two lines : If, r = a1 + λ b1 and r = a2 + µb2 are two lines, then the lines are said to be coplanar iff ; ( a2 – a1 ) . ( b1 × b2 ) = 0. And equation of the plane containing both the lines is ( r – a1 ) . ( b1 × b2 ) = 0 or ( r – a2 ) . ( b1 × b2 ) = 0 . Moreover, If, r = a1 + λ b1 and r = a2 + µb2 are two lines, such that( a2 – a1 ) . ( b1 × b2 ) ≠ 0. Then, equation of the planes (i) ( r – a1 ) . ( b1 × b2 ) = 0 , is a plane containing the 1st line and parallel to 2nd line. and (ii) ( r – a2 ) . ( b1 × b2 ) = 0, is a plane containing the 2nd line and parallel to 1st line. Cartesian form : if the lines are x – x 1 = y – y1 = z – z1 and x – x 2 = y – y2 = z – z2 a1 b1 c1 a2 b2 c2 And, x1 – x 2 y 1 – y2 z1 – z2 a1 b1 c1 = 0 a2 b2 c2 The given lines are coplanar and the equation of plane can be give by Either, x – x 2 y – y2 z – z2 x – x 1 y – y1 z – z1 a1 b1 c1 = 0 OR a1 b1 c1 = 0 a2 b2 c2 a2 b2 c2 Moreover, if the lines are x – x 1 = y – y1 = z – z1 and x – x 2 = y – y2 = z – z2 a1 b1 c1 a2 b2 c2 And, x1 – x 2 y 1 – y2 z1 – z2 a1 b1 c1 ≠ 0. a2 b2 c2 The given lines are non coplanar and the Equation (i) x – x 2 y – y2 z – z2 and (ii) x – x 1 y – y1 z – z1 a1 b1 c1 = 0. a1 b1 c1 = 0 a2 b2 c2 a2 b2 c2 is a plane containing the 2nd line is a plane containing the 1st line and parallel to 1st line. and parallel to 2nd line.


8. Angle Between Two Planes : If r . N1 = d1 a1x + b1y + c1z = d1 and r . N2 = d2 a2x + b2y + c2z = d2 , are any two planes. Then, angle between the planes is the angle between their normal N1 = a1i + b1j + c1k and N2 = a2i + b2j + c2k Then cos θ = N1 . N2 = a1 a2 + b1 b2 + c1 c2 | N1 | | N2 | √a1

2 + b12 + c1

2 .√a22 + b2

2 + c22

9. Angle Between Line & Planes : If r . N = d a1x + b1y + c1z = d1 be any plane and r = a + λ b be any line with directional vector b = a2i + b2j + c2k . Then, angle ‘θ’ between the planes and the line is given by: Sin θ = N . b = a1 a2 + b1 b2 + c1 c2 | N | | b | √a1

2 + b12 + c1

2 .√a22 + b2

2 + c22

10. Family of Planes : If r . N1 = d1 a1x + b1y + c1z = d1 & r . N2 = d2 a2x + b2y + c2z = d2 are any two planes, then equation of family of planes containing (through) the line of intersection of the planes are r . N1 + λ r . N2 = d1 + λ d2 a1x + b1y + c1z + λ (a2x + b2y + c2z )= d1 + λ d2 The value of ‘λ’ is obtained by another provided (given) condition.

11.Distance of Point from Plane : If r . N = d ax + by + cz = d, be any plane and P(x1 , y1 , z1) a = x1i + y1j + z1k, is any outsider point. Then perpendicular distance of the point P from the plane is given by, | a . N – d | = | ax1 + by1 + cz1 – d | | N | √a2 + b2 + c2 , We take the absolute value as the distance (length) is always + ve. Moreover, if r . N = d1 ax + by + cz = d1 & r . N = d2 ax + by + cz = d2 are any two parallel planes, then distance between these planes is | d1 – d2 | = | d1 – d2 | | N | √a2 + b2 + c2

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PROBLEMS ON 3 – D (1) Find the direction cosines of the sides of the triangle whose vertices are ( 3 , 5 , –4 ) , (–1 , 1 , 2 ) and ( – 5 , – 5 , – 2 ) . Ans : –2 , –2 , 3 ; –2 , –3 , –2 ; 4 , 5 , –1 √17 √17 √17 √17 √17 √17 √42 √42 √42

(2) Show that the points ( 2 , 3 , 4 ) , (–1 , –2 , 1 ) and ( 5 , 8 , 7 ) are collinear. (3) Find the vector and Cartesian equation of the line which passes through yhe point ( 1 , 2 , 3 ) and is parallel to the vector 3i + 2j – 2k . Ans: x – 1 = y – 2 = z – 3 ; r = i + 2j + 3k + λ ( 3i + 2j – 2k) 3 2 –2 (4) Find the vector and Cartesian equation of the line joining origin to the point ( 5 , –2 , 3 ) . Ans : r = λ (5i – 2j + 3k)

*(5) Find the value of ‘p’ so that the lines: 1 – x = 7y – 14 = z – 3 ; 7 – 7x = y – 5 = 6 – z , are perpendicular. 3 2p 2 3p 1 5 Ans : p = 70 / 11 (6)Find the angle between the pair of line : x – 2 = y – 1 = z + 3 ; x + 2 = y – 4 = z – 5 . 2 5 – 3 – 1 8 4 Ans : cos–1( 26 / 9 √38) *(7) If l1 , m1 , n1 and l2 , m2 , n2 are direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are ; m1n2 – m2 n1 , n1 l2 – n2 l1 , l1 m2 – l2 m1 .

*(8) If and are the angles made by any line with the four diagonals of any cube, then prove that : cos2 cos2 cos2 cos2 4 / 3 . (9) Find the shortest distance between the lines : x + 1 = y + 1 = z + 1 and x – 3 = y – 5 = z – 7 Ans : 2√29 7 –6 1 1 –2 1 *(10) Find the shortest distance between the lines : r = ( 1 – t )i + ( t – 2 )j + ( 3 – 2t ) k ; r = ( s + 1 )i + ( 2s – 1 )j – ( 2s + 1 ) k . Ans : 8 / √29 *(11) Find the distance between the lines : r = i + 2j – 4k + λ( 2i + 3j + 6k ) ; r = 3i + 3j – 5k + µ( 2i + 3j + 6k ). Ans : √293 / 7

*(12) Find the shortest distance between the lines, r = 6i + 2j + 2k + λ ( i – 2j + 2k ) r = – 4i – k + µ( 3i – 2j – 2k ) . Ans : 9

*(13) Find the coordinates of the point where the line through ( 3, – 4, – 5) and ( 2, – 3, 1 ) crosses the plane 2x + y + z = 7 . Ans : ( 1 , – 2 , 7 ) (14) Find the coordinates of the point where the line through ( 3, 4, 1) and ( 5, 1, 6) crosses the XY plane. Ans : (13/5 , 23/5 , 0) (15) Find the vector equation of the plane which is at a distance of 6 / √29, from origin and its normal vector from the origin is 2i – 3j + 4k. Also find its Cartesian form. Ans : 2x – 3y + 4z = 6 r . ( 2i – 3j + 4k ) = 6 *(16) Find the coordinates of the foot of perpendicular drawn from origin to the plane 2x – 3y + 4z – 6 = 0 . Ans : (12 / 29, – 18 / 29 , 24 / 29 )


*(17) Find the vector & Cartesian equation of the plane passing through the points ( 2, 5, –3 ); ( –2, –3, 5 ); ( 5, 3, –3 ). Ans : 2x + 3y + 4z = 7 ; r . (2i + 3j + 4k) = 7 (18) Find the angle between the plane 10x + 2y – 11z = 3 & the line , x + 1 = y = z – 3 . 2 3 6 Ans : sin–1( 8 / 11 ) (19) If O be the origin and the coordinates of P be ( 1, 2, – 3 ), then find the equation of the plane passing through P and perpendicular to OP. Ans : x + 2y – 3z = 14 *(20) Find the equation of the plane passing trough (–1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 , 3x + 3y + z = 0 . Ans : 7x – 8y + 3z + 25 = 0 *(21) Find the distance of the point P( 6, 5, 9 ) and the plane determined by the points A( 3, –1, 2 ), B( 5, 2, 4 ) and C( – 1 , – 1 , 6 ). Ans : 6 / √34 *(22) Show that the lines : x – a + d = y – a = z – a – d & x – b + c = y – b = z – b – c are coplanar. Also find the equation of plane . Ans : x – 2y + z = 0 *(23) Find the vector equation of the plane passing through the line of intersection of the planes : r . ( 2i + 2j – 3k ) = 7, r . ( 2i + 5j + 3k ) = 9 and through the point ( 2 , 1 , 3 ) . Ans : r . ( 38i + 68j + 3k) = 153.

*(24) Find the equation of the plane passing through the line of intersection of the planes r . ( i + 2j + 3k ) = 4, and r . ( 2i + j – k ) + 5 = 0 , and which is perpendicular to the plane r . ( 5i + 3j – 6k ) + 8 = 0 . Ans : 33x + 45y + 50z = 41 *(25) Find the equation of the plane passing through the line of intersection of the planes r . ( i + j + k ) = 1 , r . ( 2i + 3j – k ) + 4 = 0 , and parallel to the x – axis . Ans : y – 3z + 6 = 0 (26) Find the vector equation of the plane passing through the line of intersection of the planes : r . ( i + j + k ) = 6 & r . ( 2i + 3j + 4k ) = – 5 , and passing through the point ( 1, 1 , 1 ) . Ans : r . ( 20i + 23j + 26k ) = 69.

*(27) Show that the lines : x + 3 = y – 1 = z – 5 ; x + 1 = y – 2 = z – 5 are coplanar. – 3 1 5 – 1 2 5 Also find the equation of plane. Ans : x – 2y + z = 0 *(28) Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5, which is perpendicular to the plane x – y + z = 0. Ans : x – z + 2 = 0 (29) Prove that if a variable plane is making intercepts a, b, and c on the coordinate axes is at a constant distance ‘p’ from origin then x – 2 + y – 2 + z – 2 = p– 2. *(30) If the point (1, 1, p) and (–3, 0, 1) are equidistant from the plane 풓⃗ . (3i + 4j – 12k) + 13 = 0. Then prove that p = 1 or 7 / 3.

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HOTS PROBLEMS (1) Find the direction cosines of the line which are connected by the relation, l – 5m + 3n = 0 and 7 l 2 + 5m 2 – 3n 2 = 0 . Ans : ( 1/√14, 2/√14, 3/√14 ) ; ( –1/√6, 1/√6, 1/√6 )

*(2) Find the point of intersection of the lines : x – 5 = y – 7 = z + 3 ; x – 8 = y – 4 = z – 5 . If the lines are intersecting. 4 4 –5 7 1 3 Ans : (1, 3, 2) *(3) Find the foot of perpendicular of the point ( 1 , 2 , 3 ) in the line , x – 6 = y – 7 = z – 7 . 3 2 –2 Ans : (3, 5, 9)

*(4) Show that the lines: x – 1 = y + 1 = z – 1 & x + 2 = y – 1 = z + 1 do not intersect . 3 2 5 4 3 –2

*(5) Show that the line r = 2i – 2j + 3k + λ( i – j + 4k ) is parallel to the plane r . ( i + 5j + k ) = 5. Also find the distance between them. Ans : 10 / √27 (6) Find the length and the coordinates of the foot of perpendicular from the point (7, 14, 5) to the plane 2x + 4y – z = 2 . Ans : 3√21 ; (1, 2, 8)

*(7) Find length of S.D and vector equation of S.D of the lines : x + 4 = y – 4 = z – 1 ; x + 3 = y + 8 = z + 3 . 1 1 –1 2 3 3 Ans :√62 ; r = i – 2j + 3k + λ (6i – 5j + k) (8) Find the equation of the plane passing through the line of intersection of the planes 2x + 3y – z + 1 = 0 ; x + y – 2z + 3 = 0 and is perpendicular to the plane 3x – y – 2z – 4 = 0. Ans : 7x + 13y + 4z = 9

*(9) Find the equation of line which intersect the lines: x – 1 = y – 2 = z – 3 ; x + 2 = y – 3 = z + 1 and passes through (1, 1, 1). 2 3 4 1 2 4 Ans : x – 1 = y – 1 = z – 1 3 10 17

(10) Find the equation of the straight line through origin, which intersect both of the lines : r = i – 3j + 5k + λ( i + 4j + 3k ) and r = 4i – 3j + 14k + ( 2i + 2j + 4k ) . Ans : r = λ(–28i + 63j – 134k ) *(11) Find the equations of the plane passing through the line of intersection of the planes : r . ( i – j ) + 6 = 0 and r . ( 3i + 3j – 4k ) = 0 , and at a unit distance from origin . Ans : 2x + y – 2z + 3 = 0 ; x + 2y – 2z – 3 = 0 (12) Find the angles of the ΔABC whose vertices are A(–1, 3, 2); B(2, 3, 5) and C(3, 5, –2). Ans : A = 90°, B = cos–1( 1 / √3) , C = cos–1(2/√6)

*(13) If A(3, 2, 0) ; B(5, 3, 2) and C(–9, 6, –3) are the vertices of ΔABC. The bisector AD of the angle BAC meets BC on D. Find the point D and the length AD. Ans : (38/16, 57/16, 17/16) ; 13√6 / 16. (14) Verify that l1 + l2 + l3 ; m1 + m2 + m3 ; n1 + n2 + n3 can be taken as the direction cosines of a line √3 √3 √3 equally inclined to three mutually perpendicular lines with direction cosines ( l1, m1, n1); ( l2 , m2 , n2 ) and ( l3 , m3, n3 ).


(15) Show that the lines r = i + j – k + λ(3i – j) and r = 4i – k + µ(2i + 3k) intersects each other. Also find the point of intersection of the lines. Ans : 4i – k (16) Show that the lines r = i + 2j + k + λ(i – j + k) and r = i + j + k + µ(i – j + 2k) do not intersect each other. (17) Show that the lines x – 1 = y – 3 = z – 3 and x – 4 = y – 1 = z , intersect each other. 2 4 4 5 2 Also find the point of intersection. Ans :( –1, –1, –1)

*(18) Find the foot and length of perpendicular drawn from the point (1, 2, 3) to the line x – 6 = y – 7 = z – 7 . Ans : (3, 5, 9) ; 7 units 3 2 –2 (19) Find the foot and length of perpendicular drawn from the point (2, 3, 4) to the line 4 – x = y = 1 – z . Ans : 170 , 78 , 10 ; 3√101 units 2 6 3 49 49 49 7

*(20) Find the image of the point (0, 2, 3) in the line: x + 3 = y – 1 = z + 4 Ans : (4, 4, –5) 5 2 3

(21) Find the image of the point (1, 6, 3) in the line: x = y – 1 = z – 2 Ans : 71 , – 40 , 73 5 2 3 19 19 19 (22) Find the image of the point (2, –1, 5) in the line: r = 11i – 2j – 8k + λ (10i – 4j – 11k) Ans : (0, 5, 1) (23) If A(0, 6, –9) ; B(–3 , –6 , 3) and C(7, 4, –1) are three given points. Find the equation of the line AB. If D is the foot of perpendicular drawn from C to the line AB, find the coordinates of D. Ans : x = y – 6 = z + 9 ; (–1, 2 , –5) 1 4 –4

*(24) Find the value of ‘λ’ and ‘µ’ so that the points (3, λ , µ) ; (2, 0, –3) and (1, –2, –2) are collinear. Ans : λ = 2, µ = –4 (25) Find the value of ‘λ’ and ‘µ’ so that the points (λ , µ , 1) ; (–1, 4, –2) and (0, 2, –1) are collinear. Ans : λ = 2, µ = –2

*(26) Prove that the lines x = ay + b , z = cy + d and x1 = a1 y + b1, z = c1 y + d1, are perpendicular if a.a1 + c.c1 + 1 = 0.

*(27) If and are the angles made by any line with the four diagonals of any cube, then prove that : sin2 sin2 sin2 sin2 8 / 3 . (28) If are the angles made by any line with coordinate axes x – axis, y – axis and z – axis respectively, then prove that : sin2 sin2 sin2 . (29) Find the shortest distance (S.D) and line of S.D between the lines, r = 6i + 3k + λ ( 2i – j + 4k ) r = – 9i + j – 10k + µ( 4i + j + 6k ) . Ans : √38 ; r = 4i + j – k + t (–5i + 2j + 3k )

*(30)Find the shortest distance S.D and line of S.D between the lines, r =(2λ + 3)i –(7λ + 15) j + (5λ + 9)k and r = (2µ – 1)i + (1 + µ)j + (9 – 3µ)k. Ans : 4√3 ; r = –i – j – k + 4t (i + j + k )


(31) Find S.D and the vector equation of S.D of the following lines : (i) x + 4 = y – 4 = z – 1 ; x + 3 = y + 8 = z + 3 Ans : x + 5 = y – 3 = z – 2 ; √62 1 1 –1 2 3 3 6 –5 1 (ii) x – 4 = y – 0 , z = –3 ; x – 4 = y = z – 3 Ans : x – 2 = y – 1 = z + 3 ; √30 2 –1 3 1 1 –1 –2 5

*(32) Find the cartesian as well as the vector equation of the plane through the intersection of the planes r . ( 2i + 6j ) + 12 = 0 and r . ( 3i – j + 4k ) = 0, and at a unit distance from origin . Ans : r . (2i + j + 2k ) + 3 = 0 : 2x + y + 2z + 3 = 0 & r . (–i + 2j – 2k ) + 3 = 0 : x – 2y + 2z = 3 (33) Find the equation of the plane passing through the line of intersection of the planes: 2x + 3y – z – 1 = 0 ; x + y – 2z + 3 = 0 and is perpendicular to the plane 3x – y – 2z – 4 = 0. Ans : 7x + 13y + 4z = 21.

*(34) Find the equation of the plane passing through the line of intersection of the planes: 2x + y – z = 3 ; 5x – 3y + 4z + 9 = 0 and is parallel to the line x – 1 = y – 3 = z – 5 –2 4 5 Ans : 21x – 17y + 22z + 51 = 0

*(35) Find the equation of the plane through the point (2, 2, 1) ; (9, 3, 6) and is perpendicular to the plane 2x + 6y + 6z = 1 Ans : 3x + 4y – 5z = 9 (36) Find the equation of the plane passing through (0, 7, –7) and containing the line : x + 1 = y – 3 = z + 2 . Ans : x + y + z = 0 –3 2 1

(37) Find the equation of the plane passing through the points (2, 3, –4) and (1, –1, 3) and parallel to the x – axis. Ans : 7x + 4z – 5 = 0

*(38) Find the length and foot of perpendicular from the point (7, 14, 5) on the plane 2x + 4y – z = 2. Ans : 3√21 ; (1, 2, 8)

(39) Find the length and foot of perpendicular from the point (1, 1, 2) on the plane r. (2i – 2j + 4k) = 5. Ans : 13√6 / 2 ; ( –1 / 12, 25/ 12, –1 / 6 )

*(40) Find the image of the point (1, 3, 4) in the plane 2x – y + z + 3= 0. Ans : (–3, 5, 2) (41) Find the image of the point (0, 0, 0) in the plane 3x + 4y – 6z + 1= 0. Ans: (–6/61, –8/61, 12/61) (42) Find the image of the point i + 2j – j in the plane r. (3i – 5j + 4k) = 5. Ans : (73/25, –6/5, 39/25)

*(43) Find the distance of the point (0, –3, 2) from the plane x + 2y – z = 1 measured parallel to the line. x + 1 = y + 1 = z . Ans : √153 2 2 3 (44) Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0 ; measured parallel to the line x + 3 = y – 2 = z . Ans : 7 units 3 6 2

(45) Show that the lines: x = y – 2 = z + 3 ; x – 2 = y – 6 = z – 3 , Ans : x – 2y + z + 7 = 0. 1 2 3 2 3 4 are coplanar. Also, find the equation of the plane containing these two lines.


(46) Show that the lines: x – 3 = y + 4 = z – 1 ; x + 1 = y – 2 = z , Ans : 8x + y – 26z + 6 = 0 3 2 1 3 2 1 are coplanar. Also, find the equation of the plane containing these two lines. (47) Show that the lines: x – 1 = y – 2 = z – 3 ; x – 2 = y – 3 = z – 4 , Ans : x – 2y + z = 0 2 3 4 3 4 5 are coplanar. Also, find the equation of the plane containing these two lines. (48) Show that the lines: 4 – x = 3 – y = 2 – z ; x – 3 = y + 2 = z , Ans : 11x – y – 3z = 35 –1 4 –5 1 –4 5 are coplanar. Also, find the equation of the plane containing these two lines. (49) Find the angle between the lines x + 2y + z = 0 = 3x + 9y + 5z ; x – 2y + z = 0 = x + 2y – 2z. Ans : cos–1( 8 / √406 ). (50) Find the angle between the lines whose direction cosines are given by the relations: (a) 2 l – m + 2n = 0 = mn + n l + l m Ans : π / 2 (b) l + m + n = 0 = 2 l m – mn + 2 n l Ans : π / 3 (c) l + m + n = 0 = l 2 + m2 – n2 Ans : π / 3 (51) A line with direction cosines proportional to (2, 7, –5) is drawn to intersect the lines x – 5 = y – 7 = z + 2 ; x + 3 = y – 3 = z – 6 . 3 –1 1 –3 2 4 Find the coordinates of the point of intersection and the length intercepted on it. Ans : (2, 8, –3) ; (0, 1, 2) ; √78

(52) Find the equation of the line equally inclined to the axes and passing through (1, –2, 3). Ans : r = i – 2j + 3k + λ ( i + j + k) (53) Show that, if the axes are rectangular, the equation of the line through the point (x1, y1, z1) and perpendicular to the two perpendicular lines : x = y = z ; x = y = z l1 m1 n1 l2 m2 n2 is

풙 풙ퟏ풎ퟏ풏ퟐ 풎ퟐ풏ퟏ

= 풚 풚ퟏ

풍ퟐ풏ퟏ 풍ퟏ풏ퟐ =

풛 풛ퟏ풍ퟏ풎ퟐ 풍ퟐ풎ퟏ

(54) Find the equation of the line through the point (2, –1, 3) and perpendicular to the lines r = i + j – k + λ( 2i – 2j + k) and r = 2i – j – 3k + µ(i + 2j + 2k) Ans : r = 2i – j + 3k + λ (2i + j – 2k) (55) Can you find a unique equation of a plane passing through the points ( –2, 3, 5) , (1, 2, 3) and (7, 0, –1) ? Justify your answer. Ans : points are collinear. (56) Show that the points – i – j – k and 2i + 3j – k lies on the same sides of the plane r . (5i + 2j – 7k) + 9 = 0.

*(57) Find the equation of the perpendicular bisector plane of the line segment joining (1, 3, –2) and (3, 1, 6). Ans : x – y + 4z = 8. (58) If foot of perpendicular from (4, –1, 2) on a plane is (–10, 5, 4). Then find equation of such plane. Ans : 7x – 3y – z + 89 = 0 (59) Find equation of the two planes through the points (0, 4, –3) and (6, –4, 3) other than the plane through origin, which cut off the intercepts on axes whose sum is zero. Ans : 6x + 3y – 2z = 18; 2x – 3y – 6z = 6


(60) Show that the lines r = 2i – j + 5k + λ(2i – j + 4k), is parallel to plane r. (i + 6j + k) = 9. (61) If from the points P(a, b, c) perpendicular PL and PM be drawn to YZ and ZX planes; find the equation of planes OLM. Ans: x + y – z = 0. a b c

(62) Show that (–1, 4, –3) is the circumcentre of the triangle formed by the points (3, 2, –5) ; ( –3, 8 – 5 ) and (–3 , 2 , 1 ). (63) Show that the plane through the points (1 , 1 , 1) ; (1 – 1 , 1) and (–7, –3 ,–5) is perpendicular to the XZ plane (y = 0).

*(64) A variable plane which is at a constant distance ‘3p’ from origin, cuts the coordinate axes at A, B and C. Then prove that the locus of the centroid of the triangle ABC is x – 2 + y – 2 + z – 2 = p – 2.

*(65) A variable plane which always remains at a constant distance ‘p’ from origin, cuts the coordinate axes at A, B and C respectively.

(a) Prove that the locus of the centroid of the triangle ABC is x – 2 + y – 2 + z – 2 = 9. p – 2 . (b) Prove that the locus of the point of intersection of the planes drawn parallel to the coordinate

planes through the points A , B and C respectively is x – 2 + y – 2 + z – 2 = p – 2 . (c) Prove that the locus of the centroid of the tetrahedron OABC is x – 2 + y – 2 + z – 2 = 16. p – 2 .

*(66) Find the value of ‘k’ so that the points (3, 2, 1) ; (4, k, 5) ; (4, 2, –2) and (6, 5, –1) are coplanar. Ans : k = 5 (67) Find the equation of plane(s) which passes through the points (4, 2, 1), (2, 1, –1) making an angle π / 4 with plane x – 4y + z = 9 Ans: x + 2y – 2z = 6, 2x + 2y – z = 3

(68) Show that the plane r .( i + 2j – k ) = 0 , contains the line r = (–i – 2j – 5k) + λ(–2i + 3j + 4k). (69) If 4x + 4y – kz = 0 is an equation of a plane containing the line, x – 1 = y + 1 = z . Find the value ‘k’. 2 3 4 Ans : k = 5 (70) Find the equation of line which intersect the lines: x – 1 = y + 3 = z – 5 ; x – 4 = y + 3 = z – 14 and passes through origin. 1 4 3 2 2 4 Ans : x = y = z 1 –3 5

(71) Find the equation of line which intersect the lines: x – 1 = y – 2 = z – 3 ; x – 4 = y = z + 3 and passes through (2, –1, 3). 2 3 4 4 5 3 Ans : x – 2 = y + 1 = z – 3 2 3 4 (72) Find the equations of the two straight lines through origin such that each line is intersecting the

line 풙 ퟑퟐ

= 풚 ퟑퟏ

= 풛 ퟏ

at an angle of 흅 ퟑ

. Ans : 풙 ퟏ

= 풚 ퟐ

= 풛 –ퟏ

; 풙 –ퟏ

= 풚 ퟏ

= 풛 –ퟐ

*(73) Find the distance of the point ( – 2, 3 , – 4 ) from the line 풙 ퟑퟑ

= ퟐ풚 ퟑퟒ

= ퟑ풛 ퟒ

ퟓ measured

parallel to the plane 4x + 12y – 3z + 1 = 0 Ans : 17 / 2 ******

11. THREE – DIMENSION - Aniket Mathematics

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