CUET UG Mathematics: Vectors 3D - Vectors, Dot Product, Cross Product, Scalar Triple Product

Must-Know

• The dot product of two vectors (\vec{a}) and (\vec{b}) is defined as (\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta), where (\theta) is the angle between them; e.g., if (\vec{a} = 2\hat{i} + 3\hat{j}), (\vec{b} = \hat{i} - \hat{j}), then (\vec{a} \cdot \vec{b} = 2(1) + 3(-1) = -1).
• The dot product is commutative: (\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}); this fails for cross product.
• If (\vec{a} \cdot \vec{b} = 0) and neither vector is zero, then (\vec{a} \perp \vec{b}); e.g., (\hat{i} \cdot \hat{j} = 0).
• The projection of (\vec{a}) on (\vec{b}) is (\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}); for (\vec{a} = 3\hat{i} + 4\hat{j}), (\vec{b} = \hat{i}), projection = 3.
• The scalar component of (\vec{a}) in the direction of (\vec{b}) is (\vec{a} \cdot \hat{b}), where (\hat{b}) is the unit vector in direction of (\vec{b}).
• The cross product (\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}), where (\hat{n}) is a unit vector perpendicular to the plane of (\vec{a}) and (\vec{b}) following right-hand rule; e.g., (\hat{i} \times \hat{j} = \hat{k}).
• (\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})): cross product is anti-commutative.
• If (\vec{a} \times \vec{b} = \vec{0}) and neither vector is zero, then (\vec{a} \parallel \vec{b}); e.g., (\hat{i} \times \hat{i} = 0).
• Magnitude of (\vec{a} \times \vec{b}) equals the area of the parallelogram formed by (\vec{a}) and (\vec{b}); if (\vec{a} = 2\hat{i}), (\vec{b} = 3\hat{j}), area = (6) sq units.
• Area of triangle with adjacent sides (\vec{a}) and (\vec{b}) is (\frac{1}{2}|\vec{a} \times \vec{b}|).
• For vectors (\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}), (\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}),
  (\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}); e.g., (\vec{a} = \hat{i} + \hat{j}), (\vec{b} = \hat{i} - \hat{j}), then (\vec{a} \times \vec{b} = -2\hat{k}).
• The scalar triple product ([\vec{a}\ \vec{b}\ \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{vmatrix}); value is zero if vectors are coplanar.
• Scalar triple product is cyclic: (\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b})).
• If ([\vec{a}\ \vec{b}\ \vec{c}] = 0), then (\vec{a}, \vec{b}, \vec{c}) are coplanar; e.g., (\vec{a} = \hat{i}, \vec{b} = \hat{j}, \vec{c} = \hat{i} + \hat{j}), then ([\vec{a}\ \vec{b}\ \vec{c}] = 0).
• Volume of parallelepiped with coterminous edges (\vec{a}, \vec{b}, \vec{c}) is (|[\vec{a}\ \vec{b}\ \vec{c}]|).
• Volume of tetrahedron with vertices at (\vec{a}, \vec{b}, \vec{c}, \vec{d}) is (\frac{1}{6} |(\vec{b}-\vec{a}) \cdot ((\vec{c}-\vec{a}) \times (\vec{d}-\vec{a}))|).
• (\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}): dot and cross can be interchanged in scalar triple product.
• If (\vec{a}, \vec{b}, \vec{c}) are mutually perpendicular unit vectors, then ([\vec{a}\ \vec{b}\ \vec{c}] = \pm 1); e.g., (\hat{i}, \hat{j}, \hat{k}) give ([\hat{i}\ \hat{j}\ \hat{k}] = 1).
• (\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}): vector triple product expansion (not in NCERT exercises but used in derivations — verify from NCERT).
• The dot product of a vector with itself is the square of its magnitude: (\vec{a} \cdot \vec{a} = |\vec{a}|^2); e.g., (\vec{a} = 3\hat{i} + 4\hat{j}), then (\vec{a} \cdot \vec{a} = 25).

Difficulty Level

Intermediate — requires understanding of vector algebra, geometric interpretations, and determinant-based computation, but problems are formula-driven and pattern-based per NCERT.

Common CUET Traps

• Trap: Assuming (\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}) implies (\vec{b} = \vec{c}).
  Avoid: Dot product cancellation is invalid; it only implies (\vec{a} \cdot (\vec{b} - \vec{c}) = 0), so (\vec{a} \perp (\vec{b} - \vec{c})).
• Trap: Taking magnitude of cross product without (\sin\theta) when angle is not 90°.
  Avoid: Always use (|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta); e.g., if (\theta = 30^\circ), (\sin\theta = 0.5).
• Trap: Confusing scalar triple product sign; assuming order doesn’t matter.
  Avoid: ([\vec{a}\ \vec{b}\ \vec{c}] = -[\vec{a}\ \vec{c}\ \vec{b}]); sign changes with odd permutations.

Practice MCQs

1. If (\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}) and (\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}), then (\vec{a} \cdot \vec{b}) is:
   A. 0
   B. –4
   C. 4
   D. –6
   Answer: B
   Explanation: (\vec{a} \cdot \vec{b} = 2(1) + 3(-2) + (-1)(2) = 2 - 6 - 2 = -6)
   Why others fail: Option B is –4, a common arithmetic error in sign handling.

2. The area of the parallelogram whose adjacent sides are (\hat{i} + \hat{j}) and (\hat{i} - \hat{j}) is:
   A. 1
   B. 2
   C. (\sqrt{2})
   D. 4
   Answer: B
   Explanation: (|\vec{a} \times \vec{b}| = |(\hat{i} + \hat{j}) \times (\hat{i} - \hat{j})| = |-2\hat{k}| = 2)
   Why others fail: Option A is tempting if student forgets magnitude and takes determinant value as area directly.

3. If (\vec{a} = \hat{i} + \hat{j}), (\vec{b} = \hat{j} + \hat{k}), (\vec{c} = \hat{k} + \hat{i}), then the scalar triple product ([\vec{a}\ \vec{b}\ \vec{c}]) is:
   A. 0
   B. 1
   C. 2
   D. –2
   Answer: C
   Explanation: ([\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 0 & 1 \end{vmatrix} = 1(1 - 0) - 1(0 - 1) + 0 = 1 + 1 = 2)
   Why others fail: Option A is tempting if student assumes symmetry implies coplanarity.

4. The value of (\lambda) for which the vectors (2\hat{i} - \hat{j} + \hat{k}), (\hat{i} + 2\hat{j} - 3\hat{k}), and (3\hat{i} + \lambda\hat{j} + 5\hat{k}) are coplanar is:
   A. –4
   B. 4
   C. –8
   D. 8
   Answer: A
   Explanation: Scalar triple product = 0? (\begin{vmatrix} 2 & -1 & 1 \ 1 & 2 & -3 \ 3 & \lambda & 5 \end{vmatrix} = 0)-solving gives (\lambda = -4)
   Why others fail: Option C (–8) arises from sign error in cofactor expansion.

5. If (\vec{a} \times \vec{b} = \vec{c}) and (\vec{b} \times \vec{c} = \vec{a}), and (\vec{a}, \vec{b}, \vec{c}) are non-zero, then:
   A. (|\vec{a}| = |\vec{b}|)
   B. (|\vec{b}| = |\vec{c}|)
   C. (|\vec{a}| = |\vec{c}|)
   D. (|\vec{a}| = |\vec{b}| = |\vec{c}|)
   Answer: D
   Explanation: From given, (|\vec{a}||\vec{b}|\sin\theta = |\vec{c}|) and (|\vec{b}||\vec{c}|\sin\phi = |\vec{a}|); assuming orthogonal and consistent magnitudes leads to equality.
   Why others fail: Option A is tempting if only first equation is considered.

Last?Minute Revision

• (\vec{a} \cdot \vec{b} = 0)-vectors perpendicular (if non-zero).
• (\vec{a} \times \vec{b} = 0)-vectors parallel (if non-zero).
• (\hat{i} \cdot \hat{i} = 1), (\hat{i} \cdot \hat{j} = 0).
• (\hat{i} \times \hat{j} = \hat{k}), (\hat{j} \times \hat{i} = -\hat{k}).
• Dot product-scalar; Cross product-vector; Scalar triple-scalar.
• Area of parallelogram = (|\vec{a} \times \vec{b}|).
• Area of triangle = (\frac{1}{2}|\vec{a} \times \vec{b}|).
• Volume of parallelepiped = (|[\vec{a}\ \vec{b}\ \vec{c}]|).
• Volume of tetrahedron = (\frac{1}{6}|[\vec{a}\ \vec{b}\ \vec{c}]|).
• Scalar triple product = determinant of components.
• Cyclic order: ([\vec{a}\ \vec{b}\ \vec{c}] = [\vec{b}\ \vec{c}\ \vec{a}] = [\vec{c}\ \vec{a}\ \vec{b}]).
• Anti-cyclic gives negative: ([\vec{a}\ \vec{c}\ \vec{b}] = -[\vec{a}\ \vec{b}\ \vec{c}]).
• If three vectors are coplanar, scalar triple product = 0.
• Projection of (\vec{a}) on (\vec{b}) = (\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}).
• Scalar component in direction of (\vec{b}) = (\vec{a} \cdot \hat{b}).
• (\vec{a} \cdot \vec{a} = |\vec{a}|^2).
• (\vec{a} \times \vec{b}) is perpendicular to both (\vec{a}) and (\vec{b}).
• Right-hand rule determines direction of cross product.
• Mnemonic: “DOT for angle, CROSS for area, TRIPLE for volume.”
• (\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}).