Three Dimensional Geometry Class 12 Maths NCERT Solutions

Step-by-step solution

1. 1The direction cosines are l = cosα, m = cosβ, n = cosγ where α, β, γ are the angles with the three positive axes.
2. 2l = cos90° = 0, m = cos135° = −1/√2, n = cos45° = 1/√2.
3. 3Check: l² + m² + n² = 0 + 1/2 + 1/2 = 1, as required of any set of direction cosines.

Step-by-step solution

1. 1Let each equal angle be α; then l = m = n = cosα.
2. 2l² + m² + n² = 1 gives 3cos²α = 1, so cosα = ±1/√3.

Step-by-step solution

1. 1cosα = cos90° = 0 and cosβ = cos60° = 1/2.
2. 2l² + m² + n² = 1 gives 0 + 1/4 + cos²γ = 1, so cos²γ = 3/4 and cosγ = ±√3/2.
3. 3Hence γ = 30° or γ = 150°.

Step-by-step solution

1. 1Direction ratios of the segment from (2,3,4) to (−1,−2,1): (−3,−5,−3).
2. 2Direction ratios of the segment from (−1,−2,1) to (5,8,7): (6,10,6).
3. 3Since (6,10,6) = −2(−3,−5,−3), the two segments are parallel; they share the point (−1,−2,1), so the three points are collinear.

Step-by-step solution

1. 1Direction ratios of AB = B − A = (−4,−4,6).
2. 2|AB| = √(16+16+36) = √68 = 2√17.
3. 3Direction cosines = (−4/(2√17), −4/(2√17), 6/(2√17)) = (−2/√17, −2/√17, 3/√17).

Step-by-step solution

1. 1Direction ratios of BC = C − B = (−4,−6,−4).
2. 2|BC| = √(16+36+16) = √68 = 2√17.
3. 3Direction cosines = (−2/√17, −3/√17, −2/√17).

Step-by-step solution

1. 1Direction ratios of CA = A − C = (8,10,−2).
2. 2|CA| = √(64+100+4) = √168 = 2√42.
3. 3Direction cosines = (8/(2√42), 10/(2√42), −2/(2√42)) = (4/√42, 5/√42, −1/√42).

Step-by-step solution

1. 1Two lines are perpendicular when the dot product of their direction ratios is zero.
2. 2(12,−3,−4)·(4,12,3) = 48 − 36 − 12 = 0.
3. 3(4,12,3)·(3,−4,12) = 12 − 48 + 36 = 0.
4. 4(12,−3,−4)·(3,−4,12) = 36 + 12 − 48 = 0.

Step-by-step solution

1. 1Direction ratios of the first line: (3−1, 4+1, −2−2) = (2,5,−4).
2. 2Direction ratios of the second line: (3−0, 5−3, 6−2) = (3,2,4).
3. 3Dot product = 2·3 + 5·2 + (−4)·4 = 6 + 10 − 16 = 0, so the lines are perpendicular.

Step-by-step solution

1. 1Direction ratios of the first line: (2−4, 3−7, 4−8) = (−2,−4,−4).
2. 2Direction ratios of the second line: (1+1, 2+2, 5−1) = (2,4,4).
3. 3Since (2,4,4) = −1·(−2,−4,−4), the direction ratios are proportional and the lines are parallel.

Step-by-step solution

1. 1A line through the point with position vector a in the direction of b is r = a + λb.
2. 2Here a = i + 2j + 3k and b = 3i + 2j − 2k.
3. 3Hence the vector equation is r = (i + 2j + 3k) + λ(3i + 2j − 2k).

Step-by-step solution

1. 1From r = (i + 2j + 3k) + λ(3i + 2j − 2k): x = 1 + 3λ, y = 2 + 2λ, z = 3 − 2λ.
2. 2Eliminate λ: (x − 1)/3 = (y − 2)/2 = (z − 3)/(−2).

Step-by-step solution

1. 1Vector form: r = a + λb with a = 2i − j + 4k and b = i + 2j − k.
2. 2So r = (2i − j + 4k) + λ(i + 2j − k).
3. 3Cartesian form: (x − 2)/1 = (y + 1)/2 = (z − 4)/(−1).

Step-by-step solution

1. 1The given line has direction ratios (3,5,6); a parallel line has the same direction ratios.
2. 2Through (−2,4,−5): (x + 2)/3 = (y − 4)/5 = (z + 5)/6.

Step-by-step solution

1. 1Read from the cartesian equation: a point on the line is (5,−4,6) and direction ratios are (3,7,2).
2. 2Vector form: r = (5i − 4j + 6k) + λ(3i + 7j + 2k).

Step-by-step solution

1. 1Take a = 2i − j − 3k (the first point).
2. 2Direction b = second point − first point = (−2−2)i + (2+1)j + (5+3)k = −4i + 3j + 8k.
3. 3Vector equation: r = (2i − j − 3k) + λ(−4i + 3j + 8k).

Step-by-step solution

1. 1The direction ratios are (−4,3,8) and the line passes through (2,−1,−3).
2. 2Cartesian form: (x − 2)/(−4) = (y + 1)/3 = (z + 3)/8.

Step-by-step solution

1. 1Direction AB = (1−3, 2−4, −7+6) = (−2,−2,−1).
2. 2Vector: r = (3i + 4j − 6k) + λ(−2i − 2j − k).
3. 3Cartesian: (x − 3)/(−2) = (y − 4)/(−2) = (z + 6)/(−1).

Step-by-step solution

1. 1The angle between two lines equals the angle between their direction vectors.
2. 2b₁·b₂ = 3·1 + 2·2 + 6·2 = 19.
3. 3|b₁| = √(9+4+36) = 7 and |b₂| = √(1+4+4) = 3.
4. 4cosθ = 19/(7·3) = 19/21.

Step-by-step solution

1. 1Direction vectors b₁ = i + 2j + 2k and b₂ = 3i + 2j + 6k.
2. 2b₁·b₂ = 3 + 4 + 12 = 19; |b₁| = 3, |b₂| = 7.
3. 3cosθ = 19/(3·7) = 19/21.

Step-by-step solution

1. 1Rewrite line 1: (x−1)/(−3) = (y−2)/(2p/7) = (z−3)/2, so its direction ratios are (−3, 2p/7, 2).
2. 2Rewrite line 2: (x−1)/(−3p/7) = (y−5)/1 = (z−6)/(−5), so its direction ratios are (−3p/7, 1, −5).
3. 3Right angles require the dot product to vanish: (−3)(−3p/7) + (2p/7)(1) + (2)(−5) = 0.
4. 49p/7 + 2p/7 − 10 = 0 ⇒ 11p/7 = 10 ⇒ p = 70/11.

Step-by-step solution

1. 1Direction ratios of the first line: (7,−5,1); of the second: (1,2,3).
2. 2Dot product = 7·1 + (−5)·2 + 1·3 = 7 − 10 + 3 = 0.
3. 3The dot product being zero, the lines are perpendicular.

Step-by-step solution

1. 1Use d = |(a₂ − a₁)·(b₁ × b₂)| / |b₁ × b₂| with a₁ = (1,2,1), a₂ = (2,−1,−1), b₁ = (1,−1,1), b₂ = (2,1,2).
2. 2a₂ − a₁ = (1,−3,−2); b₁ × b₂ = (1,−1,1)×(2,1,2) = (−3,0,3) and |b₁ × b₂| = √18 = 3√2.
3. 3(a₂ − a₁)·(b₁ × b₂) = 1·(−3) + (−3)·0 + (−2)·3 = −9, so |...| = 9.
4. 4d = 9/(3√2) = 3/√2.

Step-by-step solution

1. 1a₁ = (−1,−1,−1), b₁ = (7,−6,1); a₂ = (3,5,7), b₂ = (1,−2,1).
2. 2a₂ − a₁ = (4,6,8); b₁ × b₂ = (7,−6,1)×(1,−2,1) = (−4,−6,−8); |b₁ × b₂| = √116 = 2√29.
3. 3(a₂ − a₁)·(b₁ × b₂) = 4(−4) + 6(−6) + 8(−8) = −116, so |...| = 116.
4. 4d = 116/(2√29) = 58/√29 = 2√29.

Step-by-step solution

1. 1a₁ = (1,2,3), b₁ = (1,−3,2); a₂ = (4,5,6), b₂ = (2,3,1).
2. 2a₂ − a₁ = (3,3,3); b₁ × b₂ = (1,−3,2)×(2,3,1) = (−9,3,9); |b₁ × b₂| = √171 = 3√19.
3. 3(a₂ − a₁)·(b₁ × b₂) = 3(−9) + 3(3) + 3(9) = 9.
4. 4d = 9/(3√19) = 3/√19.

Step-by-step solution

1. 1Write as a₁ = i − 2j + 3k, b₁ = −i + j − 2k; a₂ = i − j − k, b₂ = i + 2j − 2k.
2. 2a₂ − a₁ = (0,1,−4); b₁ × b₂ = (−1,1,−2)×(1,2,−2) = (2,−4,−3); |b₁ × b₂| = √29.
3. 3(a₂ − a₁)·(b₁ × b₂) = 0·2 + 1·(−4) + (−4)(−3) = 8.
4. 4d = 8/√29.

Step-by-step solution

1. 1Write the plane in the form ax + by + cz = d: here 0·x + 0·y + 1·z = 2.
2. 2Normal vector (0,0,1) is already a unit vector, so its direction cosines are (0,0,1).
3. 3Distance of the plane from the origin = d/|n| = 2/1 = 2.

Step-by-step solution

1. 1n = (1,1,1) gives |n| = √3.
2. 2Direction cosines = (1/√3, 1/√3, 1/√3).
3. 3Distance = d/|n| = 1/√3.

Step-by-step solution

1. 1n = (2,3,−1), |n| = √(4+9+1) = √14.
2. 2Direction cosines = (2/√14, 3/√14, −1/√14).
3. 3Distance = 5/√14.

Step-by-step solution

1. 1The plane is 5y = −8, i.e. y = −8/5.
2. 2Normal (0,1,0) is a unit vector, so the direction cosines are (0,1,0).
3. 3Distance from the origin = |−8/5| = 8/5.

Step-by-step solution

1. 1The plane is r·n̂ = p where p = 7 is the distance from the origin.
2. 2|n| = √(9+25+36) = √70, so r·(3i + 5j − 6k) = 7·√70.

Step-by-step solution

1. 1For r = xi + yj + zk, r·(i + j − k) = x + y − z.
2. 2Hence the cartesian equation is x + y − z = 2.

Step-by-step solution

1. 1r·(2i + 3j − 4k) = 2x + 3y − 4z.
2. 2Hence 2x + 3y − 4z = 1.

Step-by-step solution

1. 1Take the dot product with r = xi + yj + zk.
2. 2(s − 2t)x + (3 − t)y + (2s + t)z = 15.

Step-by-step solution

1. 1The foot lies on the normal (2,3,4), so it is λ(2,3,4).
2. 2Substitute into the plane: 2(2λ) + 3(3λ) + 4(4λ) = 12 ⇒ 29λ = 12 ⇒ λ = 12/29.
3. 3Foot = (24/29, 36/29, 48/29).

Step-by-step solution

1. 1The foot is λ(0,3,4) on the normal.
2. 2Substitute: 3(3λ) + 4(4λ) = 6 ⇒ 25λ = 6 ⇒ λ = 6/25.
3. 3Foot = (0, 18/25, 24/25).

Step-by-step solution

1. 1The foot is λ(1,1,1).
2. 2Substitute: λ + λ + λ = 1 ⇒ λ = 1/3.
3. 3Foot = (1/3, 1/3, 1/3).

Step-by-step solution

1. 1The plane is 5y = −8; the foot is λ(0,5,0).
2. 2Substitute: 5(5λ) = −8 ⇒ λ = −8/25.
3. 3Foot = (0, −8/5, 0).

Step-by-step solution

1. 1For a plane through a with normal n: r·n = a·n.
2. 2a·n = (i − 2k)·(i + j − k) = 1 + 0 + 2 = 3.
3. 3Vector form: r·(i + j − k) = 3. Cartesian form: x + y − z = 3.

Step-by-step solution

1. 1Direction ratios of the segment from (1,1,−1) to (6,4,−5): (5,3,−4).
2. 2Direction ratios of the segment from (6,4,−5) to (−4,−2,3): (−10,−6,8).
3. 3Since (−10,−6,8) = −2(5,3,−4), the three points are collinear, so no unique plane contains them — infinitely many planes do.

Step-by-step solution

1. 1u = P₂ − P₁ = (0,1,1), v = P₃ − P₁ = (−3,1,−1).
2. 2n = u × v = (0,1,1)×(−3,1,−1) = (−2,−3,3).
3. 3Plane: −2(x − 1) − 3(y − 1) + 3z = 0 ⇒ 2x + 3y − 3z = 5.

Step-by-step solution

1. 1Divide through by 5: x/(5/2) + y/5 + z/(−5) = 1.
2. 2Read off the intercepts: x-intercept 5/2, y-intercept 5, z-intercept −5.

Step-by-step solution

1. 1A plane parallel to ZOX (the plane y = 0) has equation y = c, a constant.
2. 2It cuts the y-axis at 3, so c = 3 and the plane is y = 3.

Step-by-step solution

1. 1Family of planes through the line of intersection: (3x − y + 2z − 4) + λ(x + y + z − 2) = 0.
2. 2Put (2,2,1): (6 − 2 + 2 − 4) + λ(2 + 2 + 1 − 2) = 0 ⇒ 2 + 3λ = 0 ⇒ λ = −2/3.
3. 3(3x − y + 2z − 4) − (2/3)(x + y + z − 2) = 0; multiplying by 3 gives 7x − 5y + 4z − 8 = 0.

Step-by-step solution

1. 1Family: [r·(2i + 2j − 3k) − 7] + λ[r·(2i + 5j + 3k) − 9] = 0.
2. 2In coordinates: (2x + 2y − 3z − 7) + λ(2x + 5y + 3z − 9) = 0.
3. 3At (2,1,3): (−10) + λ(9) = 0 ⇒ λ = 10/9.
4. 49(2x + 2y − 3z − 7) + 10(2x + 5y + 3z − 9) = 0 ⇒ 38x + 68y + 3z − 153 = 0.

Step-by-step solution

1. 1The angle between planes is the angle between their normals: cosθ = |n₁·n₂|/(|n₁||n₂|).
2. 2n₁·n₂ = 6 − 6 − 15 = −15; |n₁| = √17, |n₂| = √43.
3. 3cosθ = 15/(√17√43) = 15/√731.

Step-by-step solution

1. 1n₁·n₂ = 2(−1) + (−3)(1) + 4(0) = −5.
2. 2|n₁| = √29, |n₂| = √2.
3. 3cosθ = 5/√58.

Step-by-step solution

1. 1n₁·n₂ = 3 + 4 − 18 = −11.
2. 2|n₁| = √14, |n₂| = √(9+4+36) = 7.
3. 3cosθ = 11/(7√14).

Step-by-step solution

1. 1n₁ = (1,1,1), n₂ = (1,0,−1).
2. 2n₁·n₂ = 1 + 0 − 1 = 0.
3. 3The normals are perpendicular, so θ = 90°.

Step-by-step solution

1. 1n₁·n₂ = 21 − 5 − 60 = −44 ≠ 0 and the normals are not proportional, so the planes are neither parallel nor perpendicular.
2. 2|n₁| = √110, |n₂| = √110.
3. 3cosθ = 44/110 = 2/5.

Step-by-step solution

1. 1n₁ = (2,1,3), n₂ = (1,−2,0).
2. 2n₁·n₂ = 2 − 2 + 0 = 0.
3. 3The normals are perpendicular, so the planes are perpendicular.

Step-by-step solution

1. 1The normals are (2,−2,4) and (3,−3,6).
2. 2(3,−3,6) = (3/2)(2,−2,4), so the normals are proportional and the planes are parallel.
3. 3The angle between parallel planes is 0°.

Step-by-step solution

1. 1Both planes have normal (2,−1,3), so they are parallel.
2. 2They are distinct planes (constant terms differ), angle between them = 0°.

Step-by-step solution

1. 1Distance = |ax₁ + by₁ + cz₁ − d|/√(a²+b²+c²) with the plane written ax + by + cz = d.
2. 2|3·0 − 4·0 + 12·0 − 3| / √(9+16+144) = 3/13.

Step-by-step solution

1. 1Write the plane as 2x − y + 2z = −3; the signed numerator is 2·3 + (−1)(−2) + 2·1 + 3.
2. 2Numerator = 6 + 2 + 2 + 3 = 13; denominator √(4+1+4) = 3.
3. 3Distance = 13/3.

Step-by-step solution

1. 1Numerator = |2 + 6 + 10 − 9| = 9.
2. 2Denominator = √(1+4+4) = 3.
3. 3Distance = 9/3 = 3.

Step-by-step solution

1. 1Write the plane as 2x − 3y + 6z = 2; numerator = |2(−6) + 0 + 0 − 2| = 14.
2. 2Denominator = √(4+9+36) = 7.
3. 3Distance = 14/7 = 2.

Step-by-step solution

1. 1Line coordinates: x = 2 + 3λ, y = −1 + 4λ, z = 2 + 12λ.
2. 2Plane equation: x − y + z = 5. Substitute: (2+3λ) − (−1+4λ) + (2+12λ) = 5 ⇒ 5 + 11λ = 5 ⇒ λ = 0.
3. 3Intersection point: (2,−1,2).
4. 4Required distance = √((2+1)² + (−1+5)² + (2+10)²) = √(9+16+144) = √169 = 13.

Step-by-step solution

1. 1Family: (x + 2y + 3z − 4) + λ(2x + y − z + 5) = 0; its normal is (1+2λ, 2+λ, 3−λ).
2. 2Perpendicular to 5x + 3y + 6z + 8 = 0 means the normals are perpendicular: 5(1+2λ) + 3(2+λ) + 6(3−λ) = 0.
3. 329 + 7λ = 0 ⇒ λ = −29/7.
4. 4(x + 2y + 3z − 4) − (29/7)(2x + y − z + 5) = 0, i.e. after multiplying by 7: 7x+14y+21z−28−58x−29y+29z+145 = 0.
5. 5Hence 51x + 15y − 50z + 173 = 0.