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THREE DIMENSIONAL GEOMETRY, , Chapter, , 463, , 11, , THREE DIMENSIONAL GEOMETRY,  The moving power of mathematical invention is not, reasoning but imagination. – A. DEMORGAN , 11.1 Introduction, In Class XI, while studying Analytical Geometry in two, dimensions, and the introduction to three dimensional, geometry, we confined to the Cartesian methods only. In, the previous chapter of this book, we have studied some, basic concepts of vectors. We will now use vector algebra, to three dimensional geometry. The purpose of this, approach to 3-dimensional geometry is that it makes the, study simple and elegant*., In this chapter, we shall study the direction cosines, and direction ratios of a line joining two points and also, discuss about the equations of lines and planes in space, under different conditions, angle between two lines, two, Leonhard Euler, planes, a line and a plane, shortest distance between two, (1707-1783), skew lines and distance of a point from a plane. Most of, the above results are obtained in vector form. Nevertheless, we shall also translate, these results in the Cartesian form which, at times, presents a more clear geometric, and analytic picture of the situation., , 11.2 Direction Cosines and Direction Ratios of a Line, From Chapter 10, recall that if a directed line L passing through the origin makes, angles α, β and γ with x, y and z-axes, respectively, called direction angles, then cosine, of these angles, namely, cos α, cos β and cos γ are called direction cosines of the, directed line L., If we reverse the direction of L, then the direction angles are replaced by their supplements,, i.e., π − α , π − β and π − γ . Thus, the signs of the direction cosines are reversed., * For various activities in three dimensional geometry, one may refer to the Book, “A Hand Book for designing Mathematics Laboratory in Schools”, NCERT, 2005, , 2019-20

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464, , MATHEMATICS, , Fig 11.1, , Note that a given line in space can be extended in two opposite directions and so it, has two sets of direction cosines. In order to have a unique set of direction cosines for, a given line in space, we must take the given line as a directed line. These unique, direction cosines are denoted by l, m and n., Remark If the given line in space does not pass through the origin, then, in order to find, its direction cosines, we draw a line through the origin and parallel to the given line., Now take one of the directed lines from the origin and find its direction cosines as two, parallel line have same set of direction cosines., Any three numbers which are proportional to the direction cosines of a line are, called the direction ratios of the line. If l, m, n are direction cosines and a, b, c are, direction ratios of a line, then a = λl, b=λm and c = λn, for any nonzero λ ∈ R., , Note, , Some authors also call direction ratios as direction numbers., , Let a, b, c be direction ratios of a line and let l, m and n be the direction cosines, (d.c’s) of the line. Then, , Therefore, But, Therefore, or, , l, m, =, =, a, b, l = ak, m =, 2, l + m2 + n2 =, 2, 2, k (a + b2 + c2) =, , n, = k (say), k being a constant., c, bk, n = ck, 1, 1, , k= ±, , 1, a 2 + b2 + c 2, , 2019-20, , ... (1)

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THREE DIMENSIONAL GEOMETRY, , 465, , Hence, from (1), the d.c.’s of the line are, , l =±, , a, 2, , 2, , a +b +c, , 2, , ,m= ±, , b, 2, , 2, , a +b +c, , 2, , ,n = ±, , c, 2, , a + b2 + c 2, , where, depending on the desired sign of k, either a positive or a negative sign is to be, taken for l, m and n., For any line, if a, b, c are direction ratios of a line, then ka, kb, kc; k ≠ 0 is also a, set of direction ratios. So, any two sets of direction ratios of a line are also proportional., Also, for any line there are infinitely many sets of direction ratios., 11.2.1 Relation between the direction cosines of a line, Consider a line RS with direction cosines l, m, n. Through, the origin draw a line parallel to the given line and take a, point P(x, y, z) on this line. From P draw a perpendicular, PA on the x-axis (Fig. 11.2)., , OA x, = . This gives x = lr., OP r, y = mr and z = nr, A, 2, 2, 2, 2, 2, 2, 2, x + y + z = r (l + m + n ), X, 2, 2, 2, 2, x +y +z =r, 2, l + m2 + n 2 = 1, , Let OP = r. Then cos α =, Similarly,, Thus, But, Hence, , Z, S, R, , P (x, y , z), O, , Y, P, , , r, , O, x, , A, , Fig 11.2, , 11.2.2 Direction cosines of a line passing through two points, Since one and only one line passes through two given points, we can determine the, direction cosines of a line passing through the given points P(x1, y1, z1) and Q(x2, y2, z2), as follows (Fig 11.3 (a))., , Fig 11.3, , 2019-20

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466, , MATHEMATICS, , Let l, m, n be the direction cosines of the line PQ and let it makes angles α, β and γ, with the x, y and z-axis, respectively., Draw perpendiculars from P and Q to XY-plane to meet at R and S. Draw a, perpendicular from P to QS to meet at N. Now, in right angle triangle PNQ, ∠PQN=, γ (Fig 11.3 (b)., cos γ =, , Therefore,, , z −z, NQ, = 2 1, PQ, PQ, , x2 − x1, y −y, and cos β = 2 1, PQ, PQ, Hence, the direction cosines of the line segment joining the points P(x1, y1, z1) and, Q(x2, y2, z2) are, , cos α =, , Similarly, , x2 − x1 y2 − y1 z2 − z1, ,, ,, PQ, PQ, PQ, , where, , PQ =, , ( x2 − x1 ) 2 + ( y2 − y1 ) 2 + ( z2 − z1 ), , 2, , Note The direction ratios of the line segment joining P(x , y , z ) and Q(x , y , z ), , may be taken as, 1, , 1, , 1, , 2, , 2, , 2, , x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y1 – y2, z1 – z2, Example 1 If a line makes angle 90°, 60° and 30° with the positive direction of x, y and, z-axis respectively, find its direction cosines., Solution Let the d . c .'s of the lines be l , m, n. Then l = cos 900 = 0, m = cos 600 =, , 1, ,, 2, , 3, ., 2, Example 2 If a line has direction ratios 2, – 1, – 2, determine its direction cosines., , n = cos 300 =, , Solution Direction cosines are, , 2, 2 2 + ( −1) 2 + (−2) 2, , ,, , −1, 2 2 + ( −1) 2 + (−2) 2, , ,, , −2, 2 2 + (− 1) + (−2) 2, 2, , 2 −1 −2, , ,, 3 3 3, Example 3 Find the direction cosines of the line passing through the two points, (– 2, 4, – 5) and (1, 2, 3)., or, , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 467, , Solution We know the direction cosines of the line passing through two points, P(x1, y1, z1) and Q(x2, y2, z2) are given by, x2 − x1 y2 − y1 z2 − z1, ,, ,, PQ, PQ, PQ, , where, , PQ =, , ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 + (z 2 − z1 ), , 2, , Here P is (– 2, 4, – 5) and Q is (1, 2, 3)., So, PQ = (1 − (−2)) 2 + (2 − 4)2 + (3 − (−5)) 2 =, Thus, the direction cosines of the line joining two points is, , 77, , 3, −2, 8, ,, ,, 77, 77, 77, Example 4 Find the direction cosines of x, y and z-axis., Solution The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis., Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i.e., 1,0,0., Similarly, direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively., Example 5 Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are, collinear., Solution Direction ratios of line joining A and B are, 1 – 2, – 2 – 3, 3 + 4 i.e., – 1, – 5, 7., The direction ratios of line joining B and C are, 3 –1, 8 + 2, – 11 – 3, i.e., 2, 10, – 14., It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel, to BC. But point B is common to both AB and BC. Therefore, A, B, C are, collinear points., , EXERCISE 11.1, 1. If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its, direction cosines., 2. Find the direction cosines of a line which makes equal angles with the coordinate, axes., 3. If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ?, 4. Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear., 5. Find the direction cosines of the sides of the triangle whose vertices are, (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2)., , 2019-20

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468, , MATHEMATICS, , 11.3 Equation of a Line in Space, We have studied equation of lines in two dimensions in Class XI, we shall now study, the vector and cartesian equations of a line in space., A line is uniquely determined if, (i) it passes through a given point and has given direction, or, (ii) it passes through two given points., , 11.3.1 Equation of a line through a given point and parallel to a given vector b, , Let a be the position vector of the given point, A with respect to the origin O of the, rectangular coordinate system. Let l be the, line which passes through the point A and is, , , parallel to a given vector b . Let r be the, position vector of an arbitrary point P on the, line (Fig 11.4)., , , Then AP is parallel to the vector b , i.e.,, , , Fig 11.4, AP = λ b , where λ is some real number., ,  , AP = OP – OA, But, ,  , i.e., λb = r − a, Conversely, for each value of the parameter λ, this equation gives the position, vector of a point P on the line. Hence, the vector equation of the line is given by, , , , r = a+ b, ... (1), , Remark If b = aiˆ + bjˆ + ckˆ , then a, b, c are direction ratios of the line and conversely,, , if a, b, c are direction ratios of a line, then b = aiˆ + bjˆ + ckˆ will be the parallel to, , the line. Here, b should not be confused with | b |., Derivation of cartesian form from vector form, Let the coordinates of the given point A be (x1, y1, z1) and the direction ratios of, the line be a, b, c. Consider the coordinates of any point P be (x, y, z). Then, , , r = xiˆ + yˆj + zkˆ ; a = x1 iˆ + y1 ˆj + z1 kˆ, , and, b = a iˆ + b ˆj + c kˆ, Substituting these values in (1) and equating the coefficients of iˆ, ˆj and k̂ , we get, ... (2), x = x1 + λ a; y = y1 + λ b; z = z1+ λ c, , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 469, , These are parametric equations of the line. Eliminating the parameter λ from (2),, we get, , x – x1, y – y1 z – z1, =, =, a, b, c, This is the Cartesian equation of the line., , Note, , ... (3), , If l, m, n are the direction cosines of the line, the equation of the line is, , x – x1, y – y1 z – z1, =, =, l, m, n, Example 6 Find the vector and the Cartesian equations of the line through the point, (5, 2, – 4) and which is parallel to the vector 3 iˆ + 2 ˆj − 8 kˆ ., Solution We have, , , , a = 5 iˆ + 2 ˆj − 4 kˆ and b = 3 iˆ + 2 ˆj − 8 kˆ, Therefore, the vector equation of the line is, , r = 5 iˆ + 2 ˆj − 4 kˆ + λ ( 3 iˆ + 2 ˆj − 8 kˆ ), , Now, r is the position vector of any point P(x, y, z) on the line., x iˆ + y ˆj + z kˆ = 5 iˆ + 2 ˆj − 4 kˆ + λ ( 3 iˆ + 2 ˆj − 8 kˆ), Therefore,, = (5 + 3λ )i + (2 + 2λ ) ˆj + (− 4 − 8λ ) k, Eliminating λ , we get, , x −5, y−2 z+4, =, =, 3, 2, −8, which is the equation of the line in Cartesian form., 11.3.2 Equation of a line passing through two given points, , , Let a and b be the position vectors of two, points A (x 1 , y 1 , z 1 ) and B (x 2 , y 2 , z 2 ),, respectively that are lying on a line (Fig 11.5)., , Let r be the position vector of an, arbitrary point P(x, y, z), then P is a point on,   , the line if and only if AP = r − a and,   , AB = b − a are collinear vectors. Therefore,, P is on the line if and only if,  ,  , Fig 11.5, r − a = λ (b − a ), , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 473, , , , Solution Here b1 = iˆ + 2 ˆj + 2 kˆ and b2 = 3 iˆ + 2 ˆj + 6 kˆ, The angle θ between the two lines is given by,  , b1 ⋅ b2, (iˆ + 2 ˆj + 2kˆ ) ⋅ (3 iˆ + 2 ˆj + 6 kˆ ), cos θ =   =, 1 + 4 + 4 9 + 4 + 36, b1 b2, 3 + 4 + 12 19, =, 3× 7, 21,  19 , Hence, θ = cos–1  ,  21 , Example 10 Find the angle between the pair of lines, x+3, y −1 z + 3, =, =, 3, 5, 4, x +1, y−4 z −5, =, and, =, 1, 1, 2, , =, , Solution The direction ratios of the first line are 3, 5, 4 and the direction ratios of the, second line are 1, 1, 2. If θ is the angle between them, then, cos θ =, , 3.1 + 5.1 + 4.2, 2, , 2, , 3 +5 +4, , 2, , 2, , 2, , 1 +1 + 2, , 2, , =, , 16, 16, 8 3, =, =, 15, 50 6 5 2 6, , 8 3 , Hence, the required angle is cos–1 ,  .,  15 , , 11.5 Shortest Distance between Two Lines, If two lines in space intersect at a point, then the shortest distance between them is, zero. Also, if two lines in space are parallel,, then the shortest distance between them, will be the perpendicular distance, i.e. the, length of the perpendicular drawn from a, point on one line onto the other line., Further, in a space, there are lines which, are neither intersecting nor parallel. In fact,, such pair of lines are non coplanar and, are called skew lines. For example, let us, consider a room of size 1, 3, 2 units along, Fig 11.7, x, y and z-axes respectively Fig 11.7., , 2019-20

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474, , MATHEMATICS, , The line GE that goes diagonally across the ceiling and the line DB passes through, one corner of the ceiling directly above A and goes diagonally down the wall. These, lines are skew because they are not parallel and also never meet., By the shortest distance between two lines we mean the join of a point in one line, with one point on the other line so that the length of the segment so obtained is the, smallest., For skew lines, the line of the shortest distance will be perpendicular to both, the lines., 11.5.1 Distance between two skew lines, We now determine the shortest distance between two skew lines in the following way:, Let l1 and l2 be two skew lines with equations (Fig. 11.8), , , , r = a1 + λ b1, ... (1), , , , r = a2 + µ b2, ... (2), and, , , Take any point S on l1 with position vector a1 and T on l2, with position vector a 2., Then the magnitude of the shortest distance vector, T, will be equal to that of the projection of ST along the, Q, l2, direction of the line of shortest distance (See 10.6.2)., , If PQ is the shortest distance vector between, , l1 and l2 , then it being perpendicular to both b1 and, l1, , , P, S, b2 , the unit vector n̂ along PQ would therefore be, , , Fig 11.8, b1 × b2, , ... (3), n̂ = , | b1 × b2 |, , Then, PQ = d n̂, where, d is the magnitude of the shortest distance vector. Let θ be the angle between, , , ST and PQ . Then, PQ = ST | cos θ |,  , PQ ⋅ ST, But, cos θ =  , | PQ | | ST |, , , d nˆ ⋅ ( a2 − a1 ), , , , =, (since ST = a2 − a1 ), d ST, , , , , (b1 × b2 ) ⋅ ( a2 − a1), , , =, [From (3)], ST b1 × b2, , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 477, , Solution The two lines are parallel (Why? ) We have, , , , a1 = iˆ + 2 ˆj − 4 kˆ , a2 = 3 iˆ + 3 ˆj − 5 kˆ and b = 2 iˆ + 3 ˆj + 6 kˆ, Therefore, the distance between the lines is given by, iˆ, ,  , , b × ( a2 − a1 ), , d=, =, |b |, , ˆj, , kˆ, , 2 3 6, 2 1 −1, 4 + 9 + 36, , or, , =, , | − 9 iˆ + 14 ˆj − 4 kˆ |, 293, 293, =, =, 7, 49, 49, , EXERCISE 11.2, 1. Show that the three lines with direction cosines, , 12 −3 −4, 4 12 3, 3 −4 12, ,, ,, ;, ,, ,, ;, ,, ,, are mutually perpendicular., 13 13 13 13 13 13 13 13 13, 2. Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the, line through the points (0, 3, 2) and (3, 5, 6)., 3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line, through the points (– 1, – 2, 1), (1, 2, 5)., 4. Find the equation of the line which passes through the point (1, 2, 3) and is, parallel to the vector 3 iˆ + 2 ˆj − 2 kˆ ., 5. Find the equation of the line in vector and in cartesian form that passes through, the point with position vector 2 iˆ − j + 4 kˆ and is in the direction iˆ + 2 ˆj − kˆ ., 6. Find the cartesian equation of the line which passes through the point (– 2, 4, – 5), and parallel to the line given by, , x +3 y −4 z +8, ., =, =, 3, 5, 6, , x −5 y +4 z−6, . Write its vector form., =, =, 3, 7, 2, 8. Find the vector and the cartesian equations of the lines that passes through the, origin and (5, – 2, 3)., 7. The cartesian equation of a line is, , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 479, , 11.6 Plane, A plane is determined uniquely if any one of the following is known:, (i) the normal to the plane and its distance from the origin is given, i.e., equation of, a plane in normal form., (ii) it passes through a point and is perpendicular to a given direction., (iii) it passes through three given non collinear points., Now we shall find vector and Cartesian equations of the planes., 11.6.1 Equation of a plane in normal form, Consider a plane whose perpendicular distance from the origin is d (d ≠ 0). Fig 11.10., , If ON is the normal from the origin to the plane, and n̂ is the unit normal vector, , , Z, along ON . Then ON = d n̂ . Let P be any, , point on the plane. Therefore, NP is, , perpendicular to ON .,  , Therefore, NP ⋅ ON = 0, ... (1), P( x,y,z ), , Let r be the position vector of the point P,, r,  ,   , d N, then NP = r − d nˆ (as ON + NP = OP ), Y, Therefore, (1) becomes, O, ∧, ∧, , (r − d n ) ⋅ d n = 0, , or, , ∧, ∧, , ( r − d n) ⋅ n = 0, , or, , ∧ ∧,  ∧, r ⋅n − d n⋅ n = 0, , X, Fig 11.10, , (d ≠ 0), ,  ∧, ∧ ∧, r ⋅n = d, (as n ⋅ n = 1), i.e.,, This is the vector form of the equation of the plane., Cartesian form, , … (2), , Equation (2) gives the vector equation of a plane, where n̂ is the unit vector normal to, the plane. Let P(x, y, z) be any point on the plane. Then, , , OP = r = x iˆ + y ˆj + z kˆ, Let l, m, n be the direction cosines of n̂ . Then, n̂ = l iˆ + m ˆj + n kˆ, , 2019-20

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480, , MATHEMATICS, , Therefore, (2) gives, ( x iˆ + y ˆj + z kˆ ) ⋅ (l iˆ + m ˆj + n kˆ) = d, i.e.,, lx + my + nz = d, ... (3), This is the cartesian equation of the plane in the normal form., , Note Equation (3) shows that if r ⋅ (a iˆ + b ˆj + c kˆ) = d is the vector equation, of a plane, then ax + by + cz = d is the Cartesian equation of the plane, where a, b, and c are the direction ratios of the normal to the plane., , , , 6, 29, from the origin and its normal vector from the origin is 2 iˆ − 3 ˆj + 4 kˆ . Also find its, cartesian form., , Solution Let n = 2 iˆ − 3 ˆj + 4 kˆ . Then, , Example 13 Find the vector equation of the plane which is at a distance of, , , n, 2 iˆ − 3 ˆj + 4 kˆ, 2 iˆ − 3 ˆj + 4 kˆ, nˆ =  =, =, | n|, 4 + 9 + 16, 29, Hence, the required equation of the plane is,   2 ˆ −3 ˆ, 4 ˆ, 6, r ⋅, i+, j+, k=, 29, 29 , 29,  29, , Example 14 Find the direction cosines of the unit vector perpendicular to the plane, , r ⋅ (6 iˆ − 3 ˆj − 2 kˆ ) + 1 = 0 passing through the origin., Solution The given equation can be written as, , r ⋅ ( − 6 iˆ + 3 ˆj + 2 kˆ ) = 1, , ... (1), , | − 6 iˆ + 3 ˆj + 2 kˆ | = 36 + 9 + 4 = 7, Now, Therefore, dividing both sides of (1) by 7, we get, , 1,   6, 3 ˆ 2 ˆ, r ⋅  − iˆ +, j+, k =, 7, 7, 7 ,  7, , which is the equation of the plane in the form r ⋅ nˆ = d ., This shows that nˆ = −, , 6 ˆ 3 ˆ 2 ˆ, i+ j+, k is a unit vector perpendicular to the, 7, 7, 7, , plane through the origin. Hence, the direction cosines of n̂ are − 6 , 3 , 2 ., 7 7 7, , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 481, , Example 15 Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin., Solution Since the direction ratios of the normal to the plane are 2, –3, 4; the direction, cosines of it are, , 2, 2, , 2, , 2 + (− 3) + 4, , 2, , −3, , ,, , 2, , 2, , 2 + ( − 3) + 4, , 2, , 4, , ,, , 2, , 2, , 2 + ( − 3) + 4, , 2, , , i.e.,, , 2, −3, 4, ,, ,, 29 29 29, , Hence, dividing the equation 2x – 3y + 4z – 6 = 0 i.e., 2x – 3y + 4z = 6 throughout by, , 29 , we get, −3, 2, 4, 6, x +, y +, z =, 29, 29, 29, 29, This is of the form lx + my + nz = d, where d is the distance of the plane from the, , origin. So, the distance of the plane from the origin is, , 6, ., 29, , Example 16 Find the coordinates of the foot of the perpendicular drawn from the, origin to the plane 2x – 3y + 4z – 6 = 0., Solution Let the coordinates of the foot of the perpendicular P from the origin to the, plane is (x1, y1, z1) (Fig 11.11)., Z, Then, the direction ratios of the line OP are, x1, y1, z1., P(x1 , y1 , z1), Writing the equation of the plane in the normal, form, we have, 2, 3, 4, 6, x−, y+, z=, 29, 29, 29, 29, , where,, , 2, , ,, , −3, , 29 29, cosines of the OP., , ,, , 4, 29, , O, , Y, , are the direction X, Fig 11.11, , Since d.c.’s and direction ratios of a line are proportional, we have, x1, y1, =, =, 2, −3, 29, 29, , i.e.,, , x1 =, , z1, =k, 4, 29, , −3k, 4k, 2k, , z1 =, , y1 =, 29, 29, 29, , 2019-20

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482, , MATHEMATICS, , Substituting these in the equation of the plane, we get k =, , 6, ., 29, ,  12 −18 , 24 , Hence, the foot of the perpendicular is  ,, .,  29 29 29 , , Note If d is the distance from the origin and l, m, n are the direction cosines of, , the normal to the plane through the origin, then the foot of the perpendicular is, (ld, md, nd)., 11.6.2 Equation of a plane perpendicular to a, given vector and passing through a given point, In the space, there can be many planes that are, perpendicular to the given vector, but through a given, point P(x1, y1, z1), only one such plane exists (see, Fig 11.12)., Let a plane pass through a point A with position, , Fig 11.12, , vector a and perpendicular to the vector N ., , Let r be the position vector of any point P(x, y, z) in the plane. (Fig 11.13)., Then the point P lies in the plane if and only if, , ,  , AP is perpendicular to N . i.e., AP . N = 0. But,   ,   , … (1), AP = r − a . Therefore, ( r − a ) ⋅ N = 0, This is the vector equation of the plane., Cartesian form, Let the given point A be (x1, y1, z1), P be (x, y, z), , and direction ratios of N are A, B and C. Then,, , Fig 11.13, , Now, , , , , a = x1 iˆ + y1 ˆj + z1 kˆ, r = xiˆ + y ˆj + z kˆ and N = A iˆ + B ˆj + C kˆ,   , (r – a ) ⋅ N = 0, , So, , ˆ, ˆ, ˆ, ˆ, ˆ, ˆ, , ( x − x1 ) i + ( y − y1 ) j + ( z − z1 ) k  ⋅ (A i + B j + C k ) = 0, , i.e., , A (x – x1) + B (y – y1) + C (z – z1) = 0, , Example 17 Find the vector and cartesian equations of the plane which passes through, the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1., , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 483, , , Solution We have the position vector of point (5, 2, – 4) as a = 5 iˆ + 2 ˆj − 4kˆ and the, , , normal vector N perpendicular to the plane as N = 2 iˆ +3 ˆj − kˆ,   , Therefore, the vector equation of the plane is given by (r − a ).N = 0, , or, ... (1), [ r − (5 iˆ + 2 ˆj − 4 kˆ)] ⋅ (2 iˆ + 3 ˆj − kˆ) = 0, Transforming (1) into Cartesian form, we have, [( x – 5) iˆ + ( y − 2) ˆj + ( z + 4) kˆ] ⋅ (2 iˆ + 3 ˆj − kˆ) = 0, or, 2( x − 5) + 3( y − 2) − 1( z + 4) = 0, i.e., 2x + 3y – z = 20, which is the cartesian equation of the plane., 11.6.3 Equation of a plane passing through three non collinear points,  , Let R, S and T be three non collinear points on the plane with position vectors a , b and, , c respectively (Fig 11.14)., Z, , (RS X RT), R, P, r a, O, , S, b, , c, , T, Y, , X, , Fig 11.14, , , ,  , The vectors RS and RT are in the given plane. Therefore, the vector RS × RT, , is perpendicular to the plane containing points R, S and T. Let r be the position vector, of any point P in the plane. Therefore, the equation of the plane passing through R and,  , perpendicular to the vector RS × RT is,    , (r − a ) ⋅ (RS × RT) = 0,    ,  , ( r – a ).[( b – a )×(c – a )] = 0, or, … (1), , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 485, , 11.6.4 Intercept form of the equation of a plane, In this section, we shall deduce the equation of a plane in terms of the intercepts made, by the plane on the coordinate axes. Let the equation of the plane be, Ax + By + Cz + D = 0 (D ≠ 0), ... (1), Let the plane make intercepts a, b, c on x, y and z axes, respectively (Fig 11.16)., Hence, the plane meets x, y and z-axes at (a, 0, 0),, (0, b, 0), (0, 0, c), respectively., Therefore, , Aa + D = 0 or A =, , −D, a, , Bb + D = 0 or B =, , −D, b, , −D, c, Substituting these values in the equation (1) of the, plane and simplifying, we get, Cc + D = 0 or C =, , Fig 11.16, , x y z, + + =1, a b c, which is the required equation of the plane in the intercept form., , ... (1), , Example 19 Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and, z-axis respectively., Solution Let the equation of the plane be, , x y z, + + =1, ... (1), a b c, Here, a = 2, b = 3, c = 4., Substituting the values of a, b and c in (1), we get the required equation of the, plane as, , x y z, + + = 1 or 6x + 4y + 3z = 12., 2 3 4, , 11.6.5 Plane passing through the intersection, of two given planes, Let π 1 and π 2 be two planes with equations, , , r ⋅ nˆ1 = d1 and r ⋅ nˆ2 = d2 respectively. The position, vector of any point on the line of intersection must, satisfy both the equations (Fig 11.17)., , 2019-20, , Fig 11.17

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490, , MATHEMATICS, , Example 23 Find the angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5., Solution Comparing the given equations of the planes with the equations, A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0, We get, A1 = 3, B1 = – 6, C1 = 2, A2 = 2, B2 = 2, C2 = – 2, 3 × 2 + ( −6) (2) + (2) (−2), , cos θ =, , =, , Therefore,, , (32 + (− 6)2 + (−2)2 ) ( 22 + 22 + (−2)2 ), −10, 5, 5 3, =, =, 21, 7×2 3 7 3, , 5 3 , θ = cos-1 , ,  21 , , 11.9 Distance of a Point from a Plane, Vector form, , Consider a point P with position vector a and a plane π 1 whose equation is, , r ⋅ nˆ = d (Fig 11.19)., Z, , Z, , , , 1, , 2, Q, P, a, , a, , 1, , d, , O, , d, , N, , N’, , X, (a), , N’, O, , Y, , P, Y, , N, X, , (b), , Fig 11.19, , Consider a plane π2 through P parallel to the plane π1. The unit vector normal to, , , π2 is n̂ . Hence, its equation is ( r − a ) ⋅ nˆ = 0, , , r ⋅ nˆ = a ⋅ nˆ, i.e.,, , Thus, the distance ON′ of this plane from the origin is | a ⋅ nˆ | . Therefore, the distance, PQ from the plane π1 is (Fig. 11.21 (a)), , i.e.,, ON – ON′ = | d – a ⋅ nˆ |, , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 491, , which is the length of the perpendicular from a point to the given plane., We may establish the similar results for (Fig 11.19 (b))., ,  Note, 1., , 2., , ,  , If the equation of the plane π2 is in the form r ⋅ N = d , where N is normal,  , | a⋅N − d | ., to the plane, then the perpendicular distance is, , |N|, , |d|,  , The length of the perpendicular from origin O to the plane r ⋅ N = d is , |N|, , (since a = 0)., , Cartesian form, , , Let P(x1, y1, z1) be the given point with position vector a and, Ax + By + Cz = D, be the Cartesian equation of the given plane. Then, , a = x1 iˆ + y1 ˆj + z1 kˆ, , N = A iˆ + B ˆj + C kˆ, Hence, from Note 1, the perpendicular from P to the plane is, , ( x1 iˆ + y1 ˆj + z1 kˆ ) ⋅ ( A iˆ + B ˆj + C kˆ ) − D, A 2 + B2 + C 2, , =, , A x1 + B y1 + C z1 − D, A 2 + B2 + C2, , Example 24 Find the distance of a point (2, 5, – 3) from the plane, , r ⋅ ( 6 iˆ − 3 ˆj + 2 kˆ ) = 4, , , Solution Here, a = 2 iˆ + 5 ˆj − 3 kˆ , N = 6 iˆ − 3 ˆj + 2 kˆ and d = 4., Therefore, the distance of the point (2, 5, – 3) from the given plane is, , | (2 iˆ + 5 ˆj − 3 kˆ) ⋅ (6 iˆ − 3 ˆj + 2 kˆ) − 4|, | 12 − 15 − 6 − 4 | 13, =, =, ˆ, | 6 iˆ − 3 ˆj + 2 k |, 7, 36 + 9 + 4, , 2019-20

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492, , MATHEMATICS, , 11.10 Angle between a Line and a Plane, Definition 3 The angle between a line and a plane is, the complement of the angle between the line and, normal to the plane (Fig 11.20)., Vector form If the equation of the line is, ,  , r = a + λ b and the equation of the plane is,  , r ⋅ n = d . Then the angle θ between the line and the, normal to the plane is,  , b ⋅n, , cos θ = , | b |⋅| n |, , Fig 11.20, , and so the angle φ between the line and the plane is given by 90 – θ, i.e.,, sin (90 – θ) = cos θ,  , b ⋅n, –1 b ⋅ n, i.e., sin φ = ,  or φ = sin, |b | |n|, b n, Example 25 Find the angle between the line, , x +1, y, z−3, =, =, 2, 3, 6, and the plane 10 x + 2y – 11 z = 3., Solution Let θ be the angle between the line and the normal to the plane. Converting the, given equations into vector form, we have, , r = ( – iˆ + 3 kˆ ) + λ ( 2 iˆ + 3 ˆj + 6 kˆ ), , r ⋅ ( 10 iˆ + 2 ˆj − 11 kˆ ) = 3, and, , , Here, b = 2 iˆ + 3 ˆj + 6 kˆ and n = 10 iˆ + 2 ˆj − 11 kˆ, sin φ =, , =, , (2 iˆ + 3 ˆj + 6 kˆ) ⋅ (10 iˆ + 2 ˆj − 11 kˆ ), 22 + 32 + 62, , 102 + 22 + 112, , −8, 8, − 40,  8 , =, =, or φ = sin −1  , 21, 7 × 15, 21,  21 , , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 493, , EXERCISE 11.3, 1. In each of the following cases, determine the direction cosines of the normal to, the plane and the distance from the origin., (a) z = 2, (b) x + y + z = 1, (c) 2x + 3y – z = 5, (d) 5y + 8 = 0, 2. Find the vector equation of a plane which is at a distance of 7 units from the, origin and normal to the vector 3 iˆ + 5 ˆj − 6 kˆ., 3. Find the Cartesian equation of the following planes:, , , (b) r ⋅ (2 iˆ + 3 ˆj − 4 kˆ) = 1, (a) r ⋅ (iˆ + ˆj − kˆ ) = 2, , (c) r ⋅ [( s − 2t ) iˆ + (3 − t ) ˆj + (2 s + t ) kˆ ] = 15, 4. In the following cases, find the coordinates of the foot of the perpendicular, drawn from the origin., (a) 2x + 3y + 4z – 12 = 0, (b) 3y + 4z – 6 = 0, (c) x + y + z = 1, (d) 5y + 8 = 0, 5. Find the vector and cartesian equations of the planes, (a) that passes through the point (1, 0, – 2) and the normal to the plane is, iˆ + ˆj − kˆ ., (b) that passes through the point (1,4, 6) and the normal vector to the plane is, iˆ − 2 ˆj + kˆ ., 6. Find the equations of the planes that passes through three points., (a) (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3), (b) (1, 1, 0), (1, 2, 1), (– 2, 2, – 1), 7. Find the intercepts cut off by the plane 2x + y – z = 5., 8. Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX, plane., 9. Find the equation of the plane through the intersection of the planes, 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1)., 10. Find the vector equation of the plane passing through the intersection of the, , , planes r .( 2 iˆ + 2 ˆj − 3 kˆ ) = 7 , r .( 2 iˆ + 5 ˆj + 3 kˆ ) = 9 and through the point, (2, 1, 3)., 11. 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., , 2019-20

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494, , MATHEMATICS, , 12. Find the angle between the planes whose vector equations are, , , r ⋅ (2 iˆ + 2 ˆj − 3 kˆ ) = 5 and r ⋅ (3 iˆ − 3 ˆj + 5 kˆ) = 3., 13. In the following cases, determine whether the given planes are parallel or, perpendicular, and in case they are neither, find the angles between them., (a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0, (b) 2x + y + 3z – 2 = 0, , and x – 2y + 5 = 0, , (c) 2x – 2y + 4z + 5 = 0, , and 3x – 3y + 6z – 1 = 0, , (d) 2x – y + 3z – 1 = 0, , and 2x – y + 3z + 3 = 0, , (e) 4x + 8y + z – 8 = 0, , and y + z – 4 = 0, , 14. In the following cases, find the distance of each of the given points from the, corresponding given plane., Plane, , Point, (a) (0, 0, 0), , 3x – 4y + 12 z = 3, , (b) (3, – 2, 1), , 2x – y + 2z + 3 = 0, , (c) (2, 3, – 5), , x + 2y – 2z = 9, , (d) (– 6, 0, 0), , 2x – 3y + 6z – 2 = 0, , Miscellaneous Examples, Example 26 A line makes angles α, β, γ and δ with the diagonals of a cube, prove that, cos2 α + cos2 β + cos2 γ + cos2 δ =, , 4, 3, , Solution A cube is a rectangular parallelopiped having equal length, breadth and height., Let OADBFEGC be the cube with each side of length a units. (Fig 11.21), Z, The four diagonals are OE, AF, BG and CD., The direction cosines of the diagonal OE which, C(0, 0, a), is the line joining two points O and E are, F(0, a, a), a −0, a2 + a2 + a2, , i.e.,, , 1, 1, ,, ,, 3, 3, , ,, , a−0, a2 + a2 + a2, , ,, , a−0, , (a, 0, a) G, , a 2 + a 2 + a2, , 1, 3, , E(a,a,a), , Y, B(0, a, 0), A(a, 0, 0) D(a, a, 0), O, , X, , Fig 11.21, , 2019-20

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THREE DIMENSIONAL GEOMETRY, , Similarly, the direction cosines of AF, BG and CD are, –1, , 1, , 1, , and, , cosα =, , ,, , 1, , ,, , –1, 1, 1, 1, ,, ,, ;, ,, 3, 3, 3, 3, , –1, , , respectively., 3, 3, 3, 3, Let l, m, n be the direction cosines of the given line which makes angles α, β, γ, δ, with OE, AF, BG, CD, respectively. Then, 3, , ,, , 495, , 1, 1, (l + m+ n); cos β =, (– l + m + n);, 3, 3, , 1, 1, (l – m + n); cos δ =, (l + m – n), 3, 3, Squaring and adding, we get, cos2 α + cos2 β + cos2 γ + cos2 δ, , cosγ =, , (Why?), , =, , 1, [ (l + m + n )2 + (–l + m + n)2 ] + (l – m + n)2 + (l + m –n)2], 3, , =, , 1, 4, [ 4 (l2 + m2 + n2 ) ] =, 3, 3, , (as l2 + m2 + n2 = 1), , Example 27 Find the equation of the plane that contains the point (1, – 1, 2) and is, perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8., Solution The equation of the plane containing the given point is, A (x – 1) + B(y + 1) + C (z – 2) = 0, ... (1), Applying the condition of perpendicularly to the plane given in (1) with the planes, 2x + 3y – 2z = 5 and x + 2y – 3z = 8, we have, 2A + 3B – 2C = 0 and A + 2B – 3C = 0, Solving these equations, we find A = – 5C and B = 4C. Hence, the required, equation is, – 5C (x – 1) + 4 C (y + 1) + C(z – 2) = 0, i.e., 5x – 4y – z = 7, Example 28 Find the distance between 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)., Solution Let A, B, C be the three points in the plane. D is the foot of the perpendicular, drawn from a point P to the plane. PD is the required distance to be determined, which, , , , is the projection of AP on AB × AC ., , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 497, , Adding third column to the first column, we get, b−a, 2 α, β, , b − a b +c − a − d, α, α +δ, =0, β, β+γ, , Since the first and second columns are identical. Hence, the given two lines are, coplanar., Example 30 Find the coordinates of the point where the line through the points, A (3, 4, 1) and B (5, 1, 6) crosses the XY-plane., Solution The vector equation of the line through the points A and B is, , r = 3 iˆ + 4 ˆj + kˆ + λ [ (5 − 3) iˆ + (1 − 4) ˆj + (6 − 1) kˆ ], , ... (1), r = 3 iˆ + 4 ˆj + kˆ + λ ( 2 iˆ − 3 ˆj + 5 kˆ ), Let P be the point where the line AB crosses the XY-plane. Then the position, vector of the point P is of the form x iˆ + y ˆj ., i.e., , This point must satisfy the equation (1). (Why ?), i.e., x iˆ + y ˆj = (3 + 2 λ ) iˆ + ( 4 − 3 λ) ˆj + ( 1 + 5 λ ) kˆ, Equating the like coefficients of iˆ, ˆj and kˆ , we have, x= 3+2λ, y=4–3λ, 0= 1+5λ, Solving the above equations, we get, x=, , 13, 23, and y =, 5, 5, ,  13 23 , , 0 ., Hence, the coordinates of the required point are  ,, 5, 5, , , Miscellaneous Exercise on Chapter 11, 1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the, line determined by the points (3, 5, – 1), (4, 3, – 1)., 2. If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular, lines, show that the direction cosines of the line perpendicular to both of these, are m1 n2 − m2 n1 , n1 l2 − n2 l1 , l1 m2 − l2 m1, , 2019-20

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498, , MATHEMATICS, , 3. Find the angle between the lines whose direction ratios are a, b, c and, b – c, c – a, a – b., 4. Find the equation of a line parallel to x-axis and passing through the origin., 5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and, (2, 9, 2) respectively, then find the angle between the lines AB and CD., , x −1, y − 2 z −3, x −1 y − 1 z − 6, are perpendicular,, =, =, and, =, =, −3, 2k, 2, 3k, 1, −5, find the value of k., 7. Find the vector equation of the line passing through (1, 2, 3) and perpendicular to, , the plane r . ( iˆ + 2 ˆj − 5 kˆ ) + 9 = 0 ., 8. Find the equation of the plane passing through (a, b, c) and parallel to the plane, , r ⋅ (iˆ + ˆj + kˆ) = 2., 6. If the lines, , , 9. Find the shortest distance between lines r = 6 iˆ + 2 ˆj + 2 kˆ + λ (iˆ − 2 ˆj + 2 kˆ), , and r = − 4 iˆ − kˆ + µ (3 iˆ − 2 ˆj − 2 kˆ) ., 10. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1), crosses the YZ-plane., 11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1), crosses the ZX-plane., 12. Find the coordinates of the point where the line through (3, – 4, – 5) and, (2, – 3, 1) crosses the plane 2x + y + z = 7., 13. Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular, to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0., 14. If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane, , r ⋅ (3 iˆ + 4 ˆj − 12 kˆ ) + 13 = 0, then find the value of p., 15. Find the equation of the plane passing through the line of intersection of the, , , planes r ⋅ (iˆ + ˆj + kˆ) = 1 and r ⋅ (2 iˆ + 3 ˆj − kˆ ) + 4 = 0 and parallel to x-axis., 16. 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., 17. Find the equation of the plane which contains the line of intersection of the planes, , , r ⋅ (iˆ + 2 ˆj + 3 kˆ ) − 4 = 0 , r ⋅ (2 iˆ + ˆj − kˆ ) + 5 = 0 and which is perpendicular to the, , plane r ⋅ (5 iˆ + 3 ˆj − 6 kˆ ) + 8 = 0 ., , 2019-20

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THREE DIMENSIONAL GEOMETRY, , 499, , 18. Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the, , , line r = 2 iˆ − ˆj + 2 kˆ + λ (3 iˆ + 4 ˆj + 2 kˆ) and the plane r ⋅ (iˆ − ˆj + kˆ) = 5 ., 19. Find the vector equation of the line passing through (1, 2, 3) and parallel to the, , , planes r ⋅ (iˆ − ˆj + 2 kˆ) = 5 and r ⋅ (3 iˆ + ˆj + kˆ) = 6 ., 20. Find the vector equation of the line passing through the point (1, 2, – 4) and, perpendicular to the two lines:, , x − 8 y + 19 z −10, x − 15, y − 29 z − 5 ., =, =, =, and, =, 3, − 16, 7, 3, 8, −5, 21. Prove that if a plane has the intercepts a, b, c and is at a distance of p units from, 1, 1, 1, 1, + 2 + 2 = 2 ., 2, a, b, c, p, Choose the correct answer in Exercises 22 and 23., 22. Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is, the origin, then, , (A) 2 units, , (B) 4 units, , (C) 8 units, , (D), , 2, units, 29, , 23. The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are, (A) Perpendicular, (B) Parallel, (C) intersect y-axis, , 5, , (D) passes through  0,0, , , 4, , Summary, , , , , , Direction cosines of a line are the cosines of the angles made by the line, with the positive directions of the coordinate axes., If l, m, n are the direction cosines of a line, then l2 + m2 + n2 = 1., Direction cosines of a line joining two points P(x1, y1, z1) and Q(x2, y2, z2) are, x2 − x1 y2 − y1 z2 − z1, ,, ,, PQ, PQ, PQ, , where PQ =, , , , , ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 + (z 2 − z1 ), , 2, , Direction ratios of a line are the numbers which are proportional to the, direction cosines of a line., If l, m, n are the direction cosines and a, b, c are the direction ratios of a line, , 2019-20