Exercise 11.2 answers: Three Dimensional Geometry

Open NCERT p. 389One way to think about it

1. Two lines with direction cosines (l₁,m₁,n₁) and (l₂,m₂,n₂) are perpendicular exactly when l₁l₂ + m₁m₂ + n₁n₂ = 0.
2. Call the three lines L₁ = 12/13, −3/13, −4/13; L₂ = 4/13, 12/13, 3/13; L₃ = 3/13, −4/13, 12/13.
3. L₁·L₂ = (12×4 + (−3)×12 + (−4)×3)/169 = (48 − 36 − 12)/169 = 0/169 = 0, so L₁ ⊥ L₂.
4. L₂·L₃ = (4×3 + 12×(−4) + 3×12)/169 = (12 − 48 + 36)/169 = 0/169 = 0, so L₂ ⊥ L₃.
5. L₃·L₁ = (3×12 + (−4)×(−3) + 12×(−4))/169 = (36 + 12 − 48)/169 = 0/169 = 0, so L₃ ⊥ L₁.
6. All three pairwise dot products are zero, so the three lines are mutually perpendicular.

In shortAll three pairwise sums l₁l₂+m₁m₂+n₁n₂ come out to 0, so the lines are mutually perpendicular.

Watch this explained “Perpendicular is the top alone”, 7:35 into The angle between two lines, from cosines or from ratios

Open NCERT p. 389One way to think about it

1. Direction ratios of the first line, through (1,−1,2) and (3,4,−2): (3−1, 4−(−1), −2−2) = (2, 5, −4).
2. Direction ratios of the second line, through (0,3,2) and (3,5,6): (3−0, 5−3, 6−2) = (3, 2, 4).
3. Two lines are perpendicular when the sum of the products of matching direction ratios is zero.
4. (2)(3) + (5)(2) + (−4)(4) = 6 + 10 − 16 = 0.
5. The sum is zero, so the two lines are perpendicular.

In shortThe direction ratios (2, 5, −4) and (3, 2, 4) give a dot product of 0, so the lines are perpendicular.

Watch this explained “Perpendicular is the top alone”, 7:35 into The angle between two lines, from cosines or from ratios

Open NCERT p. 389One way to think about it

1. Direction ratios of the first line, through (4,7,8) and (2,3,4): (2−4, 3−7, 4−8) = (−2, −4, −4).
2. Direction ratios of the second line, through (–1,–2,1) and (1,2,5): (1−(−1), 2−(−2), 5−1) = (2, 4, 4).
3. Two lines are parallel when their direction ratios are proportional — one triple is a constant multiple of the other.
4. (2, 4, 4) = (−1) × (−2, −4, −4), so the two triples are proportional with multiplier −1.
5. Since the ratios are proportional, the two lines are parallel.

In shortThe direction ratios (−2, −4, −4) and (2, 4, 4) are exact negatives of each other, so the lines are parallel.

Watch this explained “Parallel is the ratios agreeing”, 8:59 into The angle between two lines, from cosines or from ratios

Open NCERT p. 389Matches NCERT’s answer

1. Given: the vector 3î + 2ĵ − 2k̂.
2. A line is fixed by one point on it and one direction along it.
3. The point is (1, 2, 3), with position vector î + 2ĵ + 3k̂.
4. The direction is the given vector, 3î + 2ĵ − 2k̂, so its direction ratios are 3, 2, −2.
5. Vector form: r⃗ = (î + 2ĵ + 3k̂) + λ(3î + 2ĵ − 2k̂).
6. Cartesian form, one fraction per coordinate over its direction ratio: (x−1)/3 = (y−2)/2 = (z−3)/(−2).

Answerr⃗ = (î + 2ĵ + 3k̂) + λ(3î + 2ĵ − 2k̂); Cartesian: (x−1)/3 = (y−2)/2 = (z−3)/(−2).

Watch this explained “When the direction is handed to you”, 5:37 into One point plus one direction fixes a line, in vector form and in the symmetric Cartesian form the parameter eliminates to

Open NCERT p. 389Matches NCERT’s answer

1. Given: the position vector 2î − ĵ + 4k̂ and the direction î + 2ĵ − k̂.
2. The anchor point's position vector is given directly: 2î − ĵ + 4k̂, i.e. the point (2, −1, 4).
3. The direction vector is given directly: î + 2ĵ − k̂, so its direction ratios are 1, 2, −1.
4. Vector form: r⃗ = (2î − ĵ + 4k̂) + λ(î + 2ĵ − k̂).
5. Compare coefficients of î, ĵ, k̂ to get the parametric equations: x = 2 + λ, y = −1 + 2λ, z = 4 − λ.
6. Eliminate λ from each: λ = (x−2)/1 = (y+1)/2 = (z−4)/(−1). That is the Cartesian form.

Answerr⃗ = (2î − ĵ + 4k̂) + λ(î + 2ĵ − k̂); Cartesian: (x−2)/1 = (y+1)/2 = (z−4)/(−1).

Watch this explained “One vector equation, three scalar ones”, 10:45 into One point plus one direction fixes a line, in vector form and in the symmetric Cartesian form the parameter eliminates to

Open NCERT p. 389Matches NCERT’s answer

1. The new line only needs a point and a direction; the point (–2, 4, –5) is given directly.
2. Parallel lines share their direction ratios, and the given line's denominators are already its direction ratios: 3, 5, 6.
3. So the required line has direction ratios 3, 5, 6 too.
4. Write each coordinate as (coordinate − anchor value) over its direction ratio: (x−(−2))/3 = (y−4)/5 = (z−(−5))/6.
5. That simplifies to (x+2)/3 = (y−4)/5 = (z+5)/6.

Answer(x+2)/3 = (y−4)/5 = (z+5)/6

Watch this explained “Reading it backwards”, 14:25 into One point plus one direction fixes a line, in vector form and in the symmetric Cartesian form the parameter eliminates to

Open NCERT p. 389Matches NCERT’s answer

1. In (x−a)/p = (y−b)/q = (z−c)/r, the point (a, b, c) is on the line and p, q, r are its direction ratios.
2. Here the numerators are x−5, y+4, z−6.
3. Rewrite y+4 as y−(−4), so the anchor point is (5, −4, 6).
4. The denominators 3, 7, 2 are the direction ratios directly.
5. Vector form: r⃗ = (5î − 4ĵ + 6k̂) + λ(3î + 7ĵ + 2k̂).

Answerr⃗ = (5î − 4ĵ + 6k̂) + λ(3î + 7ĵ + 2k̂)

Watch this explained “Reading it backwards”, 14:25 into One point plus one direction fixes a line, in vector form and in the symmetric Cartesian form the parameter eliminates to

Open NCERT p. 389Matches NCERT’s answer

(i) λ(3î + 2ĵ + 6k̂) … μ(î + 2ĵ + 2k̂)

1. Read off the two direction vectors: (3, 2, 6) and (1, 2, 2).
2. Dot product: 3×1 + 2×2 + 6×2 = 3 + 4 + 12 = 19.
3. Lengths: √(9+4+36) = √49 = 7, and √(1+4+4) = √9 = 3.
4. cos θ = |19| / (7×3) = 19/21.
5. θ = cos⁻¹(19/21).

Answerθ = cos⁻¹(19/21)

(ii) λ(î − ĵ − 2k̂) … μ(3î − 5ĵ − 4k̂)

1. Read off the two direction vectors: (1, −1, −2) and (3, −5, −4).
2. Dot product: 1×3 + (−1)×(−5) + (−2)×(−4) = 3 + 5 + 8 = 16.
3. Lengths: √(1+1+4) = √6, and √(9+25+16) = √50.
4. cos θ = |16| / (√6×√50) = 16/√300 = 16/(10√3) = 8/(5√3) = 8√3/15.
5. θ = cos⁻¹(8√3/15).

Answerθ = cos⁻¹(8√3/15)

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Open NCERT p. 390Matches NCERT’s answer

(i) (x-2)/2 = (y-1)/5 = (z+3)/-3 and (x+2)/-1 = (y-4)/8 = (z-5)/4

1. Read the direction ratios off the denominators: (2, 5, −3) and (−1, 8, 4).
2. Dot product: 2×(−1) + 5×8 + (−3)×4 = −2 + 40 − 12 = 26.
3. Lengths: √(4+25+9) = √38, and √(1+64+16) = √81 = 9.
4. cos θ = |26| / (9√38), which simplifies to 13√38/171.
5. θ = cos⁻¹(13√38/171).

Answerθ = cos⁻¹(13√38/171)

(ii) x/2 = y/2 = z/1 and (x-5)/4 = (y-2)/1 = (z-3)/8

1. Read the direction ratios off the denominators: (2, 2, 1) and (4, 1, 8).
2. Dot product: 2×4 + 2×1 + 1×8 = 8 + 2 + 8 = 18.
3. Lengths: √(4+4+1) = √9 = 3, and √(16+1+64) = √81 = 9.
4. cos θ = |18| / (3×9) = 18/27 = 2/3.
5. θ = cos⁻¹(2/3).

Answerθ = cos⁻¹(2/3)

Watch this explained “The same work, in Cartesian form”, 10:58 into The angle between two lines, from cosines or from ratios

Open NCERT p. 390Matches NCERT’s answer

1. Both lines are written in an awkward shape — x has a negative coefficient, and y and z carry extra factors. Rearrange each into the standard (x−a)/l form first.
2. First line: (1−x)/3 = −(x−1)/3 = (x−1)/(−3); (7y−14)/(2p) = 7(y−2)/(2p) = (y−2)/(2p/7); (z−3)/2 stays as is.
3. So the first line's direction ratios are −3, 2p/7, 2.
4. Second line: (7−7x)/(3p) = −7(x−1)/(3p) = (x−1)/(−3p/7); (y−5)/1 stays; (6−z)/5 = −(z−6)/5 = (z−6)/(−5).
5. So the second line's direction ratios are −3p/7, 1, −5.
6. Perpendicular means the dot product is zero: (−3)(−3p/7) + (2p/7)(1) + (2)(−5) = 0.
7. 9p/7 + 2p/7 − 10 = 0, so 11p/7 = 10, so p = 70/11.

Answerp = 70/11

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Open NCERT p. 390One way to think about it

1. Read the direction ratios off the denominators: first line (7, −5, 1), second line (1, 2, 3).
2. Two lines are perpendicular when the sum of the products of matching direction ratios is zero.
3. (7)(1) + (−5)(2) + (1)(3) = 7 − 10 + 3 = 0.
4. The sum is zero, so the two lines are perpendicular.

In shortThe direction ratios (7, −5, 1) and (1, 2, 3) give a dot product of 0, so the lines are perpendicular.

Watch this explained “Perpendicular is the top alone”, 7:35 into The angle between two lines, from cosines or from ratios

Open NCERT p. 390Matches NCERT’s answer

1. Given: r⃗ = (î + 2ĵ + k̂) + λ(î − ĵ + k̂) and r⃗ = 2î − ĵ − k̂ + μ(2î + ĵ + 2k̂).
2. Anchors: a⃗₁ = (1, 2, 1), a⃗₂ = (2, −1, −1). Directions: b⃗₁ = (1, −1, 1), b⃗₂ = (2, 1, 2).
3. Check they are not parallel: (1,−1,1) is not a multiple of (2,1,2), so this is the skew-lines formula.
4. Anchor gap a⃗₂ − a⃗₁ = (1, −3, −2).
5. Cross the two directions: b⃗₁ × b⃗₂ = ((−1)(2)−(1)(1), (1)(2)−(1)(2), (1)(1)−(−1)(2)) = (−3, 0, 3).
6. Length of that cross product: √(9+0+9) = √18 = 3√2.
7. Numerator: (1,−3,−2)·(−3,0,3) = −3+0−6 = −9, and |−9| = 9.
8. Shortest distance = 9 / (3√2) = 3/√2 = 3√2/2.

Answer3/√2 = 3√2/2 units

Watch this explained “Two more, and a coincidence”, 23:23 into Skew lines, what shortest distance can mean when two lines never meet, and computing it in both the skew and parallel cases

Open NCERT p. 390Matches NCERT’s answer

1. Anchors read off the equations: a⃗₁ = (−1, −1, −1), a⃗₂ = (3, 5, 7). Directions: b⃗₁ = (7, −6, 1), b⃗₂ = (1, −2, 1).
2. Anchor gap a⃗₂ − a⃗₁ = (4, 6, 8).
3. Cross the two directions: b⃗₁ × b⃗₂ = ((−6)(1)−(1)(−2), (1)(1)−(7)(1), (7)(−2)−(−6)(1)) = (−4, −6, −8).
4. Notice the crossed directions are exactly −1 times the anchor gap.
5. Length of the cross product: √(16+36+64) = √116 = 2√29.
6. Numerator: (4,6,8)·(−4,−6,−8) = −16−36−64 = −116, and |−116| = 116.
7. Shortest distance = 116 / (2√29) = 58/√29 = 2√29.

Answer2√29 units

Watch this explained “Two more, and a coincidence”, 23:23 into Skew lines, what shortest distance can mean when two lines never meet, and computing it in both the skew and parallel cases

Open NCERT p. 390Matches NCERT’s answer

1. Given: r⃗ = (î + 2ĵ + 3k̂) + λ(î − 3ĵ + 2k̂) and r⃗ = 4î + 5ĵ + 6k̂ + μ(2î + 3ĵ + k̂).
2. Anchors: a⃗₁ = (1, 2, 3), a⃗₂ = (4, 5, 6). Directions: b⃗₁ = (1, −3, 2), b⃗₂ = (2, 3, 1).
3. Anchor gap a⃗₂ − a⃗₁ = (3, 3, 3).
4. Cross the two directions: b⃗₁ × b⃗₂ = ((−3)(1)−(2)(3), (2)(2)−(1)(1), (1)(3)−(−3)(2)) = (−9, 3, 9).
5. Length of the cross product: √(81+9+81) = √171 = 3√19.
6. Numerator: (3,3,3)·(−9,3,9) = −27+9+27 = 9.
7. Shortest distance = 9 / (3√19) = 3/√19 = 3√19/19.

Answer3/√19 = 3√19/19 units

Watch this explained “The awkward one, and a whole number”, 24:34 into Skew lines, what shortest distance can mean when two lines never meet, and computing it in both the skew and parallel cases

Open NCERT p. 390Matches NCERT’s answer

1. Given: r⃗ = (1 − t)î + (t − 2)ĵ + (3 − 2t)k̂ and r⃗ = (s + 1)î + (2s − 1)ĵ − (2s + 1)k̂.
2. Both lines are written with the parameter multiplied through every term, so pull out the anchor and direction first.
3. First line: x = 1 + t(−1), y = −2 + t(1), z = 3 + t(−2). Anchor a⃗₁ = (1, −2, 3), direction b⃗₁ = (−1, 1, −2).
4. Second line: x = 1 + s(1), y = −1 + s(2), z = −1 + s(−2). Anchor a⃗₂ = (1, −1, −1), direction b⃗₂ = (1, 2, −2).
5. Anchor gap a⃗₂ − a⃗₁ = (0, 1, −4).
6. Cross the two directions: b⃗₁ × b⃗₂ = ((1)(−2)−(−2)(2), (−2)(1)−(−1)(−2), (−1)(2)−(1)(1)) = (2, −4, −3).
7. Length of the cross product: √(4+16+9) = √29.
8. Numerator: (0,1,−4)·(2,−4,−3) = 0−4+12 = 8.
9. Shortest distance = 8/√29 = 8√29/29.

Answer8/√29 = 8√29/29 units

Watch this explained “The awkward one, and a whole number”, 24:34 into Skew lines, what shortest distance can mean when two lines never meet, and computing it in both the skew and parallel cases