CUET UG Mathematics Vectors 3D 3D Geometry Direction Cosines Equations of Line and Plane

Must‑Know (15–20 detailed bullets)

• Direction cosines of a line are the cosines of the angles made by the line with the positive directions of the coordinate axes; if α, β, γ are the angles, then l = cos α, m = cos β, n = cos γ. Example: If a line makes 45°, 60°, 120° with x, y, z-axes, then l = cos 45° = 1/√2, m = cos 60° = 1/2, n = cos 120° = –1/2.

• For any line, the sum of squares of direction cosines is 1: l² + m² + n² = 1. Example: (1/2)² + (√3/2)² + 0² = 1/4 + 3/4 = 1.

• Direction ratios are any three numbers proportional to the direction cosines. If a, b, c are direction ratios, then direction cosines are a/√(a²+b²+c²), b/√(a²+b²+c²), c/√(a²+b²+c²). Example: Direction ratios 2, –3, 6 → direction cosines 2/7, –3/7, 6/7.

• The direction ratios of the line joining points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) are x₂–x₁, y₂–y₁, z₂–z₁. Example: Points (1, 2, 3) and (4, 6, 9) → direction ratios 3, 4, 6.

• The vector equation of a line passing through a point with position vector a and parallel to vector b is r = a + λb, λ ∈ ℝ. Example: Line through (1, 2, 3) parallel to 2i + 3j – k → r = (i + 2j + 3k) + λ(2i + 3j – k).

• The Cartesian equation of a line passing through (x₁, y₁, z₁) and having direction ratios a, b, c is (x–x₁)/a = (y–y₁)/b = (z–z₁)/c. Example: Through (2, –1, 4), drs 3, –2, 5 → (x–2)/3 = (y+1)/(–2) = (z–4)/5.

• The vector equation of a line passing through two points with position vectors a and b is r = a + λ(b – a). Example: Points (1, 0, 2), (3, 1, 1) → r = (i + 2k) + λ(2i + j – k).

• The angle θ between two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ is given by cos θ = |(a₁a₂ + b₁b₂ + c₁c₂)| / √(a₁²+b₁²+c₁²) √(a₂²+b₂²+c₂²). Example: Lines with drs 1, 2, 2 and 3, –2, 6 → cos θ = |3 – 4 + 12| / (3)(7) = 11/21.

• Two lines are perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 0. Example: drs 2, 3, 4 and 1, –2, 1 → 2 – 6 + 4 = 0 → lines are perpendicular.

• Two lines are parallel if a₁/a₂ = b₁/b₂ = c₁/c₂. Example: drs 2, 4, 6 and 1, 2, 3 → ratios 2/1 = 4/2 = 6/3 → parallel.

• The shortest distance between two skew lines r = a₁ + λb₁ and r = a₂ + μb₂ is |(b₁ × b₂) · (a₂ – a₁)| / |b₁ × b₂|. Example: Lines r = i + j + λ(2i – j + k) and r = 2i + j – k + μ(3i – 5j + 2k) → compute cross product and scalar triple product.

• The shortest distance between two parallel lines r = a₁ + λb and r = a₂ + μb is |(a₂ – a₁) × b| / |b|. Example: Lines r = i + j + λ(2i – j) and r = 3i – j + μ(2i – j) → vector between points is 2i – 2j, cross with b = (2i – j) → magnitude divided by |b|.

• The vector equation of a plane normal to vector n and at distance d from origin is r · n̂ = d. Example: Plane with normal i + 2j + 2k, distance 3 → r · (1/3i + 2/3j + 2/3k) = 3 → r · (i + 2j + 2k) = 9.

• The Cartesian equation of a plane with normal vector having direction ratios A, B, C and at distance d from origin is Ax + By + Cz = D, where D = ± p√(A²+B²+C²), p = distance. Example: Plane with normal 2x + 3y + 6z = 7 → distance from origin is |7|/√(4+9+36) = 7/7 = 1.

• The equation of a plane passing through point (x₁, y₁, z₁) and normal to vector with drs A, B, C is A(x–x₁) + B(y–y₁) + C(z–z₁) = 0. Example: Through (1, 2, –1), normal 2, –1, 3 → 2(x–1) –1(y–2) + 3(z+1) = 0 → 2x – y + 3z + 3 = 0.

• The vector equation of a plane passing through point a and normal to n is (r – a) · n = 0. Example: Point i + j, normal 2i – j + k → (r – i – j) · (2i – j + k) = 0.

• The equation of a plane passing through three non-collinear points (x₁,y₁,z₁), (x₂,y₂,z₂), (x₃,y₃,z₃) is given by determinant:
  | x–x₁ y–y₁ z–z₁ |
  | x₂–x₁ y₂–y₁ z₂–z₁ | = 0
  | x₃–x₁ y₃–y₁ z₃–z₁ |
  Example: Points (1,0,–1), (3,2,2), (2,1,3) → form vectors and use scalar triple product.

• The angle between two planes is the angle between their normals. If normals have drs A₁,B₁,C₁ and A₂,B₂,C₂, then cos θ = |A₁A₂ + B₁B₂ + C₁C₂| / √(ΣA₁²)√(ΣA₂²). Example: Planes x + y + z = 1 and 2x – y + z = 3 → cos θ = |2 –1 +1| / (√3)(√6) = 2/√18 = √2/3.

• Two planes are perpendicular if A₁A₂ + B₁B₂ + C₁C₂ = 0. Example: Planes 2x + 3y + 4z = 5 and x – 2y + z = 7 → 2 – 6 + 4 = 0 → perpendicular.

• The distance from a point (x₁,y₁,z₁) to the plane Ax + By + Cz + D = 0 is |Ax₁ + By₁ + Cz₁ + D| / √(A² + B² + C²). Example: Point (1,2,3), plane 2x – y + 2z – 4 = 0 → |2 –2 +6 –4| / √(4+1+4) = |2|/3 = 2/3.

Difficulty Level

Intermediate — requires visualization of 3D space and consistent application of vector algebra, but formulas are direct and NCERT-based.

Common CUET Traps

• Trap: Assuming direction cosines can be found directly from direction ratios without normalization.
  Avoid: Always divide direction ratios by √(a² + b² + c²) to get direction cosines.

• Trap: Using the acute angle formula for line-plane angle but forgetting it's sin θ = |b·n| / (|b||n|), not cos.
  Avoid: Angle between line and plane is sin θ = |direction vector · normal vector| / (magnitudes), not cosine.

• Trap: Confusing the condition for parallel lines (ratios equal) with perpendicular (dot product zero).
  Avoid: For lines: parallel → a₁/a₂ = b₁/b₂ = c₁/c₂; perpendicular → a₁a₂ + b₁b₂ + c₁c₂ = 0.

Practice MCQs

Q1. If a line makes angles 90°, 135°, and 45° with the x, y, and z-axes respectively, what are its direction cosines?
A. (0, 1/√2, 1/√2)
B. (0, –1/√2, 1/√2)
C. (1, –1/√2, 1/√2)
D. (0, 1/√2, –1/√2)
Answer: B
Explanation: cos 90° = 0, cos 135° = –1/√2, cos 45° = 1/√2.
Why others fail: Option A ignores that cos 135° is negative.

Q2. What is the vector equation of the line passing through (2, –1, 3) and parallel to the vector 3i – j + 2k?
A. r = 2i – j + 3k + λ(3i – j + 2k)
B. r = 3i – j + 2k + λ(2i – j + 3k)
C. r = (2i – j + 3k) + λ(i + j + k)
D. r = (3i – j + 2k) + λ(2i – j + 3k)
Answer: A
Explanation: Standard form r = a + λb, where a is point vector, b is direction.
Why others fail: Option B swaps the point and direction vectors.

Q3. The angle between the lines with direction ratios (2, 1, –3) and (1, –2, 1) is:
A. 30°
B. 45°
C. 60°
D. 90°
Answer: D
Explanation: a₁a₂ + b₁b₂ + c₁c₂ = 2(1) + 1(–2) + (–3)(1) = 2 – 2 – 3 = –3 ≠ 0 → not perpendicular? Wait: –3 ≠ 0 → not 90°. Recalculate: 2 – 2 – 3 = –3 → |–3| = 3; √(4+1+9)=√14, √(1+4+1)=√6 → cos θ = 3/√84 ≈ 0.327 → θ ≈ 71° — none match. But wait: check if dot product zero? 2(1) + 1(–2) + (–3)(1) = 2 – 2 – 3 = –3 ≠ 0 → not 90°. But options suggest one is correct. Recheck: perhaps typo? Let's assume drs (2,3,–1) and (1,–2,–4)? No. Wait — correct calculation:
Actually, (2)(1) + (1)(–2) + (–3)(1) = 2 – 2 – 3 = –3 → not zero. But let's compute cos θ = |–3| / (√(4+1+9) √(1+4+1)) = 3 / (√14 √6) = 3 / √84 ≈ 3/9.165 ≈ 0.327 → θ ≈ 71° — not in options.
Wait — perhaps the question meant (2,3,6) and (–3,2,0)? No.
Wait — maybe (2,1,–3) and (3,–2,1)? 6 –2 –3 =1 ≠0.
But if drs were (2,3,1) and (1,–2,4): 2–6+4=0 → 90°.
But as per given, not 90°.
But in NCERT Ex 11.2 Q1: similar with (3,–2,–2) and (–2,1,–1): dot = –6 –2 +2 = –6 ≠0.
Wait — perhaps the correct answer is not 90°. But let's assume the question is:
Actually, let's correct:
Suppose drs (2, –3, 4) and (–1, 2, 2): –2 –6 +8=0 → 90°.
But as per current, no.
Wait — maybe typo in question? But in standard CUET, such errors rare.
Alternatively, perhaps the answer is not D.
But let's recompute:
a₁a₂ + b₁b₂ + c₁c₂ = 2×1 + 1×(–2) + (–3)×1 = 2 – 2 – 3 = –3
|a| = √(4+1+9)=√14, |b|=√(1+4+1)=√6
cos θ = |–3| / (√14 √6) = 3 / √84 = 3/(2√21) ≈ 3/9.165 ≈ 0.327 → θ ≈ 71° → not in options.
But if the question said (2,3,6) and (3,–6,2): 6 –18 +12=0 → 90°.
But as per given, not.
Wait — perhaps the second drs is (1,2,1)? 2+2–3=1≠0.
Alternatively, maybe the question is:
Let’s change to a valid one:
Q3. The angle between lines with direction ratios (2, 3, 6) and (1, –2, 1) is:
A. 30°
B. 45°
C. 60°
D. 90°
Now: dot = 2(1) + 3(–2) + 6(1) = 2 – 6 + 6 = 2
|a| = √(4+9+36)=7, |b|=√(1+4+1)=√6
cos θ = |2| / (7√6) ≈ 2/17.15 ≈ 0.116 → θ ≈ 83° — not in options.
Better:
Q3. The lines with direction ratios (2, –1, 1) and (1, 1, –1) are:
A. parallel
B. perpendicular
C. at 60°
D. at 45°
Now: dot = 2(1) + (–1)(1) + (1)(–1) = 2 –1 –1 = 0 → perpendicular.
So revise:

Q3. The lines with direction ratios (2, –1, 1) and (1, 1, –1) are:
A. parallel
B. perpendicular
C. at 60°
D. at 45°
Answer: B
Explanation: a₁a₂ + b₁b₂ + c₁c₂ = 2(1) + (–1)(1) + (1)(–1) = 0 → lines are perpendicular.
Why others fail: Option A is tempting if one compares ratios 2/1, –1/1, 1/–1 → not equal.

Q4. The distance from the point (1, 2, –3) to the plane 2x – y + 2z + 4 = 0 is:
A. 1
B. 2
C. 3
D. 4
Answer: A
Explanation: |2(1) –1(2) + 2(–3) + 4| / √(4+1+4) = |2 –2 –6 +4| / 3 = |–2| / 3 = 2/3 → not matching.
Wait: 2x – y + 2z + 4 = 0 → A=2,B=–1,C=2,D=4
|2(1) + (–1)(2) + 2(–3) + 4| / √(4+1+4) = |2 –2 –6 +4| / 3 = |–2| / 3 = 2/3 → not in options.
But if plane is 2x – y + 2z = 4 → then D = –4?
Standard form Ax+By+Cz+D=0 → so 2x – y + 2z – 4 = 0 → D = –4
Then |2 –2 –6 –4| / 3 = |–10| / 3 = 10/3 → not.
If point (1,2,3): |2–2+6+4|/3=10/3.
Wait — let’s set:
Plane: 3x – 4y + 12z = 3 → D = –3
Point (1,1,1): |3–4+12–3|/√(9+16+144)=|8|/13 → not integer.
Better:
Plane: x + y + z = 1, point (1,1,1): |1+1+1–1|/√3 = 2/√3
But let’s use NCERT example:
From Ex 11.3: distance from (0,0,0) to x + y + z = 1 is 1/√3
But not integer.
Another: plane 2x – 3y + 6z – 8 = 0, point (1,1,1): |2–3+6–8|/√(4+9+36)=|–3|/7=3/7
Not good.
From NCERT Example 19: distance from (–1,–2,–3) to plane 2x – y + 2z = 4 →
|2(–1) – (–2) + 2(–3) – 4| / √(4+1+4) = |–2 +2 –6 –4| / 3 = |–10| / 3 = 10/3
Still not.
But in Ex 11.3 Q14: distance from (0,0,0) to 3x – 4y + 12z = 3 is |–3|/13 = 3/13
Wait — let’s create one with answer 1:
Plane: x + y + z = 3, point (1,1,1): |1+1+1–3|/√3 = 0 → on plane.
Plane: 2x + 2y + z = 4, point (1,1,1): |2+2+1–4|/√(4+4+1)=|1|/3 → no.
Plane: 3x + 4y + 12z = 5, point (1,1,1): |3+4+12–5|/13=14/13
Not.
Plane: x – 2y + 2z = 6, point (2,1,3): |2 –2 +6 –6|/√(1+4+4)=0/3=0
On plane.
Plane: 2x – y + 2z + 3 = 0, point (1,2,1): |2 –2 +2 +3|/3 = |5|/3 ≈1.66
But if point (1,1,1): |2–1+2+3|/3=6/3=2 → so:

Q4. The distance from the point (1, 1, 1) to the plane 2x – y + 2z + 3 = 0 is:
A. 1
B. 2
C. 3
D. 4
Answer: B
Explanation: |2(1) –1(1) + 2(1) + 3| / √(4+1+4) = |2–1+2+3|/3 = 6/3 = 2.
Why others fail: Option A is tempting if one forgets to take absolute value or miscalculates numerator.

Q5. The angle between the line (x–1)/2 = (y+2)/3 = (z–3)/(–1) and the plane 2x – y + 2z = 5 is θ. Then sin θ = ?
A. 1/3
B. 2/3
C. √5/3
D. 1
Answer: B
Explanation: Direction vector of line b = (2,3,–1), normal to plane n = (2,–1,2), sin θ = |b·n| / (|b||n|) = |4 –3 –2| / (√(4+9+1) √(4+1+4)) = |–1| / (√14 × 3) = 1/(3√14) → not matching.
Wait: b·n = 2×2 + 3×(–1) + (–1)×2 = 4 –3 –2 = –1 → |–1| = 1
|b| =