3.1 MCQs-Vector Algebra

Vector Algebra

Basic Concepts and Definitions

1. A vector is a quantity that has:

1. Only magnitude

2. Only direction

3. Both magnitude and direction

4. Neither magnitude nor direction

Show me the answer

Answer: 3. Both magnitude and direction

Explanation:

• A vector is a mathematical object that has both magnitude (length) and direction.

• Examples: displacement, velocity, force, acceleration.

• In contrast, a scalar has only magnitude (e.g., mass, temperature, time).

• Vectors are represented by directed line segments: $\overrightarrow{AB}$ or bold letters: $\mathbf{a}$, $\mathbf{v}$.

• The magnitude of vector $\mathbf{a}$ is denoted by $|\mathbf{a}|$ or $\|\mathbf{a}\|$.

2. Which of the following is NOT a vector quantity?

1. Velocity

2. Force

3. Temperature

4. Acceleration

Show me the answer

Answer: 3. Temperature

Explanation:

• Vector quantities have both magnitude and direction:

  • Velocity: speed with direction

  • Force: strength with direction

  • Acceleration: rate of change of velocity with direction

• Scalar quantities have only magnitude:

  • Temperature: measured in degrees, no direction

  • Other scalars: mass, time, distance, speed, energy

• Speed is scalar (magnitude only), while velocity is vector (magnitude + direction).

3. Two vectors are equal if they have:

1. Same magnitude

2. Same direction

3. Same magnitude and direction

4. Same initial point

Show me the answer

Answer: 3. Same magnitude and direction

Explanation:

• Two vectors $\mathbf{a}$ and $\mathbf{b}$ are equal if:

  1. They have the same magnitude: $|\mathbf{a}| = |\mathbf{b}|$

  2. They have the same direction: they are parallel and point the same way

• The initial point (starting point) doesn't matter for vector equality.

• Vectors are free vectors - they can be moved parallel to themselves without changing the vector.

• Example: $\overrightarrow{AB} = \overrightarrow{CD}$ if AB and CD have same length and same direction.

4. The magnitude of a vector $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ is:

1. 3

2. 4

3. 5

4. 7

Show me the answer

Answer: 3. 5

Explanation:

• For a vector $\mathbf{a} = x\mathbf{i} + y\mathbf{j}$ in 2D, the magnitude is: $|\mathbf{a}| = \sqrt{x^2 + y^2}$

• Here, $x = 3$, $y = 4$

• $|\mathbf{a}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

• This is a 3-4-5 right triangle.

• In 3D: for $\mathbf{a} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, $|\mathbf{a}| = \sqrt{x^2 + y^2 + z^2}$

Vector Operations

5. The sum of two vectors $\mathbf{a}$ and $\mathbf{b}$ can be found using:

1. Triangle law

2. Parallelogram law

3. Both triangle and parallelogram laws

4. Neither triangle nor parallelogram laws

Show me the answer

Answer: 3. Both triangle and parallelogram laws

Explanation:

• Triangle Law: Place vectors head-to-tail. The sum is the vector from the tail of the first to the head of the second.

• Parallelogram Law: Place vectors tail-to-tail. Their sum is the diagonal of the parallelogram formed.

• Both give the same result: $\mathbf{a} + \mathbf{b}$.

• Properties of vector addition:

  • Commutative: $\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}$

  • Associative: $(\mathbf{a} + \mathbf{b}) + \mathbf{c} = \mathbf{a} + (\mathbf{b} + \mathbf{c})$

  • Additive identity: $\mathbf{a} + \mathbf{0} = \mathbf{a}$

6. If $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}$ and $\mathbf{b} = \mathbf{i} - 2\mathbf{j}$, then $\mathbf{a} + \mathbf{b}$ is:

1. $3\mathbf{i} + \mathbf{j}$

2. $3\mathbf{i} - \mathbf{j}$

3. $\mathbf{i} + 5\mathbf{j}$

4. $\mathbf{i} + \mathbf{j}$

Show me the answer

Answer: 1. $3\mathbf{i} + \mathbf{j}$

Explanation:

• To add vectors in component form, add corresponding components: $\mathbf{a} + \mathbf{b} = (2\mathbf{i} + 3\mathbf{j}) + (\mathbf{i} - 2\mathbf{j})$

• Add i-components: $2 + 1 = 3$

• Add j-components: $3 + (-2) = 1$

• Therefore: $\mathbf{a} + \mathbf{b} = 3\mathbf{i} + 1\mathbf{j} = 3\mathbf{i} + \mathbf{j}$

• Verification: $(2,3) + (1,-2) = (3,1)$

7. The negative of a vector $\mathbf{a}$ is:

1. A vector with same magnitude but opposite direction

2. A vector with different magnitude but same direction

3. A vector with zero magnitude

4. Not defined

Show me the answer

Answer: 1. A vector with same magnitude but opposite direction

Explanation:

• The negative of vector $\mathbf{a}$, denoted $-\mathbf{a}$, has:

  • Same magnitude: $|-\mathbf{a}| = |\mathbf{a}|$

  • Opposite direction

• If $\mathbf{a} = \overrightarrow{AB}$, then $-\mathbf{a} = \overrightarrow{BA}$.

• Vector subtraction: $\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b})$

• Example: If $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}$, then $-\mathbf{a} = -2\mathbf{i} - 3\mathbf{j}$.

8. If $\mathbf{a} = 4\mathbf{i} - 3\mathbf{j}$ and $\mathbf{b} = -2\mathbf{i} + 5\mathbf{j}$, then $2\mathbf{a} - 3\mathbf{b}$ is:

1. $2\mathbf{i} - 21\mathbf{j}$

2. $14\mathbf{i} - 21\mathbf{j}$

3. $14\mathbf{i} + 21\mathbf{j}$

4. $2\mathbf{i} + 21\mathbf{j}$

Show me the answer

Answer: 2. $14\mathbf{i} - 21\mathbf{j}$

Explanation:

• First compute $2\mathbf{a} = 2(4\mathbf{i} - 3\mathbf{j}) = 8\mathbf{i} - 6\mathbf{j}$

• Compute $3\mathbf{b} = 3(-2\mathbf{i} + 5\mathbf{j}) = -6\mathbf{i} + 15\mathbf{j}$

• Now: $2\mathbf{a} - 3\mathbf{b} = (8\mathbf{i} - 6\mathbf{j}) - (-6\mathbf{i} + 15\mathbf{j})$

• = $(8\mathbf{i} - 6\mathbf{j}) + (6\mathbf{i} - 15\mathbf{j})$ (subtracting = adding negative)

• = $(8+6)\mathbf{i} + (-6-15)\mathbf{j}$

• = $14\mathbf{i} - 21\mathbf{j}$

Position Vectors and Section Formula

9. The position vector of point P(3,4) is:

1. $3\mathbf{i}$

2. $4\mathbf{j}$

3. $3\mathbf{i} + 4\mathbf{j}$

4. $4\mathbf{i} + 3\mathbf{j}$

Show me the answer

Answer: 3. $3\mathbf{i} + 4\mathbf{j}$

Explanation:

• A position vector gives the location of a point relative to the origin O(0,0).

• For point P(x,y), the position vector is: $\overrightarrow{OP} = x\mathbf{i} + y\mathbf{j}$

• Here, P(3,4): x=3, y=4

• Therefore: $\overrightarrow{OP} = 3\mathbf{i} + 4\mathbf{j}$

• In 3D: for P(x,y,z), $\overrightarrow{OP} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$

10. If A and B have position vectors $\mathbf{a}$ and $\mathbf{b}$, then the position vector of the midpoint of AB is:

1. $\frac{\mathbf{a} + \mathbf{b}}{2}$

2. $\frac{\mathbf{a} - \mathbf{b}}{2}$

3. $\mathbf{a} + \mathbf{b}$

4. $\mathbf{a} - \mathbf{b}$

Show me the answer

Answer: 1. $\frac{\mathbf{a} + \mathbf{b}}{2}$

Explanation:

• Let M be the midpoint of AB.

• From triangle OAB: $\overrightarrow{OM} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB}$

• But $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$

• So: $\overrightarrow{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{\mathbf{a} + \mathbf{b}}{2}$

• Alternatively: M divides AB in ratio 1:1 internally.

• General section formula: Point dividing AB in ratio m:n has position vector $\frac{m\mathbf{b} + n\mathbf{a}}{m+n}$

11. The point dividing the line joining A(1,2) and B(3,4) in the ratio 2:1 internally has position vector:

1. $\frac{5}{3}\mathbf{i} + \frac{8}{3}\mathbf{j}$

2. $\frac{7}{3}\mathbf{i} + \frac{10}{3}\mathbf{j}$

3. $\frac{7}{3}\mathbf{i} - \frac{10}{3}\mathbf{j}$

4. $\frac{5}{3}\mathbf{i} - \frac{8}{3}\mathbf{j}$

Show me the answer

Answer: 2. $\frac{7}{3}\mathbf{i} + \frac{10}{3}\mathbf{j}$

Explanation:

• Section formula: Point dividing AB in ratio m:n internally has position vector: $\frac{m\mathbf{b} + n\mathbf{a}}{m+n}$

• Here: A(1,2), so $\mathbf{a} = \mathbf{i} + 2\mathbf{j}$

• B(3,4), so $\mathbf{b} = 3\mathbf{i} + 4\mathbf{j}$

• m:n = 2:1, so m=2, n=1

• Position vector = $\frac{2(3\mathbf{i} + 4\mathbf{j}) + 1(\mathbf{i} + 2\mathbf{j})}{2+1}$

• = $\frac{(6\mathbf{i} + 8\mathbf{j}) + (\mathbf{i} + 2\mathbf{j})}{3}$

• = $\frac{7\mathbf{i} + 10\mathbf{j}}{3} = \frac{7}{3}\mathbf{i} + \frac{10}{3}\mathbf{j}$

• Coordinates: $\left(\frac{7}{3}, \frac{10}{3}\right)$

Dot Product (Scalar Product)

12. The dot product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is defined as:

1. $|\mathbf{a}| |\mathbf{b}|$

2. $|\mathbf{a}| |\mathbf{b}| \cos\theta$

3. $|\mathbf{a}| |\mathbf{b}| \sin\theta$

4. $|\mathbf{a}| + |\mathbf{b}|$

Show me the answer

Answer: 2. $|\mathbf{a}| |\mathbf{b}| \cos\theta$

Explanation:

• The dot product (scalar product) of vectors $\mathbf{a}$ and $\mathbf{b}$ is: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos\theta$ where $\theta$ is the angle between them ($0 \leq \theta \leq \pi$).

• Properties:

  • Commutative: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$

  • Distributive: $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$

  • Scalar multiplication: $(k\mathbf{a}) \cdot \mathbf{b} = k(\mathbf{a} \cdot \mathbf{b})$

• Result is a scalar (hence "scalar product").

13. If $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}$ and $\mathbf{b} = \mathbf{i} - \mathbf{j}$, then $\mathbf{a} \cdot \mathbf{b}$ equals:

1. -1

2. 1

3. 5

4. 6

Show me the answer

Answer: 1. -1

Explanation:

• In component form: $\mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y$ (for 2D)

• For $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}$: $a_x = 2$, $a_y = 3$

• For $\mathbf{b} = \mathbf{i} - \mathbf{j}$: $b_x = 1$, $b_y = -1$

• $\mathbf{a} \cdot \mathbf{b} = (2)(1) + (3)(-1) = 2 - 3 = -1$

• In 3D: $\mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z$

14. Two vectors are perpendicular if their dot product is:

1. 0

2. 1

3. -1

4. Maximum

Show me the answer

Answer: 1. 0

Explanation:

• For perpendicular (orthogonal) vectors, the angle $\theta = 90^\circ = \pi/2$ radians.

• $\cos 90^\circ = 0$

• Therefore: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos 90^\circ = 0$

• Conversely, if $\mathbf{a} \cdot \mathbf{b} = 0$ and neither vector is zero, then they are perpendicular.

• Example: $\mathbf{i} \cdot \mathbf{j} = 0$ (unit vectors along x and y axes are perpendicular).

15. The angle between vectors $\mathbf{a} = \mathbf{i} + \mathbf{j}$ and $\mathbf{b} = \mathbf{i} - \mathbf{j}$ is:

1. $0^\circ$

2. $45^\circ$

3. $90^\circ$

4. $180^\circ$

Show me the answer

Answer: 3. $90^\circ$

Explanation:

• Compute dot product: $\mathbf{a} \cdot \mathbf{b} = (1)(1) + (1)(-1) = 1 - 1 = 0$

• Since dot product = 0, vectors are perpendicular.

• Therefore, angle = $90^\circ$.

• Alternatively, using formula: $\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}$

• $|\mathbf{a}| = \sqrt{1^2 + 1^2} = \sqrt{2}$

• $|\mathbf{b}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$

• $\cos\theta = \frac{0}{\sqrt{2} \cdot \sqrt{2}} = 0 \Rightarrow \theta = 90^\circ$

Cross Product (Vector Product)

16. The cross product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is:

1. A scalar

2. A vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$

3. A vector parallel to both $\mathbf{a}$ and $\mathbf{b}$

4. Zero if $\mathbf{a}$ and $\mathbf{b}$ are perpendicular

Show me the answer

Answer: 2. A vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$

Explanation:

• The cross product (vector product) $\mathbf{a} \times \mathbf{b}$ is:

  • A vector (hence "vector product")

  • Perpendicular to both $\mathbf{a}$ and $\mathbf{b}$

  • Magnitude: $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin\theta$

  • Direction: Given by right-hand rule

• Properties:

  • Anti-commutative: $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$

  • Distributive: $\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}$

17. The magnitude of $\mathbf{a} \times \mathbf{b}$ equals the area of:

1. A triangle with sides $\mathbf{a}$ and $\mathbf{b}$

2. A parallelogram with adjacent sides $\mathbf{a}$ and $\mathbf{b}$

3. A rectangle with sides $\mathbf{a}$ and $\mathbf{b}$

4. A circle with diameter $\mathbf{a} + \mathbf{b}$

Show me the answer

Answer: 2. A parallelogram with adjacent sides $\mathbf{a}$ and $\mathbf{b}$

Explanation:

• $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin\theta$

• This equals the area of the parallelogram with adjacent sides $\mathbf{a}$ and $\mathbf{b}$.

• Area of parallelogram = base × height = $|\mathbf{a}| \times (|\mathbf{b}| \sin\theta)$

• Area of triangle formed by $\mathbf{a}$ and $\mathbf{b}$ = $\frac{1}{2}|\mathbf{a} \times \mathbf{b}|$

• This geometric interpretation is very useful in physics and engineering.

18. If $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$ and $\mathbf{b} = \mathbf{i} - \mathbf{j} + 2\mathbf{k}$, then $\mathbf{a} \times \mathbf{b}$ is:

1. $7\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}$

2. $7\mathbf{i} + 3\mathbf{j} - 5\mathbf{k}$

3. $7\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}$

4. $-7\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}$

Show me the answer

Answer: 1. $7\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}$

Explanation:

• Cross product using determinant: $\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 1 \\ 1 & -1 & 2 \end{vmatrix}$

• Expand: $= \mathbf{i}\begin{vmatrix} 3 & 1 \\ -1 & 2 \end{vmatrix} - \mathbf{j}\begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} + \mathbf{k}\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix}$

• $= \mathbf{i}(3\cdot2 - 1\cdot(-1)) - \mathbf{j}(2\cdot2 - 1\cdot1) + \mathbf{k}(2\cdot(-1) - 3\cdot1)$

• $= \mathbf{i}(6 + 1) - \mathbf{j}(4 - 1) + \mathbf{k}(-2 - 3)$

• $= 7\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}$

19. Two vectors are parallel if their cross product is:

1. Zero

2. Maximum

3. Equal to their dot product

4. A unit vector

Show me the answer

Answer: 1. Zero

Explanation:

• For parallel vectors, the angle $\theta = 0^\circ$ or $180^\circ$

• $\sin 0^\circ = \sin 180^\circ = 0$

• Therefore: $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin\theta = 0$

• So $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ (zero vector)

• Conversely, if $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ and neither vector is zero, then they are parallel.

• Example: $\mathbf{i} \times \mathbf{i} = \mathbf{0}$ (same vector)

• $\mathbf{i} \times (-\mathbf{i}) = \mathbf{0}$ (opposite direction)

Scalar Triple Product

20. The scalar triple product $[\mathbf{a} \ \mathbf{b} \ \mathbf{c}]$ is defined as:

1. $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$

2. $\mathbf{a} \times (\mathbf{b} \cdot \mathbf{c})$

3. $(\mathbf{a} \cdot \mathbf{b}) \times \mathbf{c}$

4. $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}$

Show me the answer

Answer: 1. $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$

Explanation:

• The scalar triple product of vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is: $[\mathbf{a} \ \mathbf{b} \ \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$

• It's a scalar quantity.

• Properties:

  • Cyclic property: $[\mathbf{a} \ \mathbf{b} \ \mathbf{c}] = [\mathbf{b} \ \mathbf{c} \ \mathbf{a}] = [\mathbf{c} \ \mathbf{a} \ \mathbf{b}]$

  • Changing order changes sign: $[\mathbf{a} \ \mathbf{b} \ \mathbf{c}] = -[\mathbf{a} \ \mathbf{c} \ \mathbf{b}]$

  • If any two vectors are equal, the product is zero.

• Geometrically: $|[\mathbf{a} \ \mathbf{b} \ \mathbf{c}]|$ = volume of parallelepiped with edges $\mathbf{a}, \mathbf{b}, \mathbf{c}$

21. The volume of the parallelepiped with edges $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is given by:

1. $|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$

2. $|\mathbf{a} \times (\mathbf{b} \cdot \mathbf{c})|$

3. $|(\mathbf{a} \cdot \mathbf{b}) \times \mathbf{c}|$

4. $|\mathbf{a} \cdot \mathbf{b} \cdot \mathbf{c}|$

Show me the answer

Answer: 1. $|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$

Explanation:

• The absolute value of the scalar triple product gives the volume of the parallelepiped: Volume = $|[\mathbf{a} \ \mathbf{b} \ \mathbf{c}]| = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$

• Volume of tetrahedron with vertices at origin and ends of $\mathbf{a}, \mathbf{b}, \mathbf{c}$ = $\frac{1}{6}|[\mathbf{a} \ \mathbf{b} \ \mathbf{c}]|$

• If $[\mathbf{a} \ \mathbf{b} \ \mathbf{c}] = 0$, then the vectors are coplanar (volume = 0).

• Example: For unit vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$: $[\mathbf{i} \ \mathbf{j} \ \mathbf{k}] = \mathbf{i} \cdot (\mathbf{j} \times \mathbf{k}) = \mathbf{i} \cdot \mathbf{i} = 1$ Volume of unit cube = 1.

22. If $\mathbf{a} = \mathbf{i} + \mathbf{j}$, $\mathbf{b} = \mathbf{j} + \mathbf{k}$, $\mathbf{c} = \mathbf{k} + \mathbf{i}$, then $[\mathbf{a} \ \mathbf{b} \ \mathbf{c}]$ equals:

1. 0

2. 1

3. 2

4. 3

Show me the answer

Answer: 3. 2

Explanation:

• Compute $\mathbf{b} \times \mathbf{c}$: $\mathbf{b} = \mathbf{j} + \mathbf{k} = (0,1,1)$ $\mathbf{c} = \mathbf{k} + \mathbf{i} = (1,0,1)$ $\mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} = \mathbf{i}(1\cdot1 - 1\cdot0) - \mathbf{j}(0\cdot1 - 1\cdot1) + \mathbf{k}(0\cdot0 - 1\cdot1)$ $= \mathbf{i}(1) - \mathbf{j}(0-1) + \mathbf{k}(0-1) = \mathbf{i} + \mathbf{j} - \mathbf{k}$

• Now $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (\mathbf{i} + \mathbf{j}) \cdot (\mathbf{i} + \mathbf{j} - \mathbf{k})$

• $= (1,1,0) \cdot (1,1,-1) = 1\cdot1 + 1\cdot1 + 0\cdot(-1) = 1 + 1 + 0 = 2$

• Therefore, $[\mathbf{a} \ \mathbf{b} \ \mathbf{c}] = 2$

Vector Triple Product

23. The vector triple product $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$ equals:

1. $(\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

2. $(\mathbf{a} \cdot \mathbf{b})\mathbf{c} - (\mathbf{a} \cdot \mathbf{c})\mathbf{b}$

3. $(\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{c})\mathbf{b}$

4. $(\mathbf{a} \cdot \mathbf{b})\mathbf{c} + (\mathbf{a} \cdot \mathbf{c})\mathbf{b}$

Show me the answer

Answer: 1. $(\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

Explanation:

• The BAC-CAB rule for vector triple product: $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

• Mnemonic: "BAC minus CAB"

• Note: $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \neq (\mathbf{a} \times \mathbf{b}) \times \mathbf{c}$ in general.

• The result lies in the plane containing $\mathbf{b}$ and $\mathbf{c}$.

• Useful identities:

  • $(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a}$

  • Jacobi identity: $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) + \mathbf{b} \times (\mathbf{c} \times \mathbf{a}) + \mathbf{c} \times (\mathbf{a} \times \mathbf{b}) = \mathbf{0}$

Unit Vectors and Direction Cosines

24. A unit vector in the direction of $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ is:

1. $3\mathbf{i} + 4\mathbf{j}$

2. $\frac{3}{5}\mathbf{i} + \frac{4}{5}\mathbf{j}$

3. $\frac{4}{5}\mathbf{i} + \frac{3}{5}\mathbf{j}$

4. $\frac{1}{5}\mathbf{i} + \frac{1}{5}\mathbf{j}$

Show me the answer

Answer: 2. $\frac{3}{5}\mathbf{i} + \frac{4}{5}\mathbf{j}$

Explanation:

• A unit vector has magnitude 1.

• Unit vector in direction of $\mathbf{a}$: $\hat{a} = \frac{\mathbf{a}}{|\mathbf{a}|}$

• Here, $|\mathbf{a}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

• Therefore: $\hat{a} = \frac{3\mathbf{i} + 4\mathbf{j}}{5} = \frac{3}{5}\mathbf{i} + \frac{4}{5}\mathbf{j}$

• Check magnitude: $\sqrt{(3/5)^2 + (4/5)^2} = \sqrt{9/25 + 16/25} = \sqrt{25/25} = 1$

25. If a vector makes angles $\alpha, \beta, \gamma$ with the x, y, z axes respectively, then $\cos^2\alpha + \cos^2\beta + \cos^2\gamma$ equals:

1. 0

2. 1

3. 2

4. 3

Show me the answer

Answer: 2. 1

Explanation:

• $\cos\alpha, \cos\beta, \cos\gamma$ are called direction cosines of the vector.

• For any vector: $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

• If $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$, then: $\cos\alpha = \frac{a_1}{|\mathbf{a}|}, \ \cos\beta = \frac{a_2}{|\mathbf{a}|}, \ \cos\gamma = \frac{a_3}{|\mathbf{a}|}$

• Therefore: $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = \frac{a_1^2 + a_2^2 + a_3^2}{|\mathbf{a}|^2} = \frac{|\mathbf{a}|^2}{|\mathbf{a}|^2} = 1$

• Direction ratios: proportional to direction cosines (e.g., $a_1 : a_2 : a_3$)

Projection of Vectors

26. The projection of vector $\mathbf{a}$ on vector $\mathbf{b}$ is:

1. $\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$

2. $\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}$

3. $\frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{b}|}$

4. $\frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}|}$

Show me the answer

Answer: 1. $\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$

Explanation:

• Scalar projection of $\mathbf{a}$ on $\mathbf{b}$: $\text{comp}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$

• Vector projection of $\mathbf{a}$ on $\mathbf{b}$: $\text{proj}_{\mathbf{b}}\mathbf{a} = \left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\right)\mathbf{b}$

• Geometrically: It's the length of the "shadow" of $\mathbf{a}$ on $\mathbf{b}$.

• If $\theta$ is the angle between them: $\text{comp}_{\mathbf{b}}\mathbf{a} = |\mathbf{a}|\cos\theta$

• Example: Projection of $\mathbf{i} + \mathbf{j}$ on $\mathbf{i}$: $\frac{(1,1,0) \cdot (1,0,0)}{1} = \frac{1}{1} = 1$

Collinear and Coplanar Vectors

27. Two vectors are collinear if:

1. Their dot product is zero

2. Their cross product is zero

3. They have equal magnitude

4. They are perpendicular

Show me the answer

Answer: 2. Their cross product is zero

Explanation:

• Collinear vectors lie on the same line (parallel or anti-parallel).

• For collinear vectors: $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ (zero vector)

• Equivalently: $\mathbf{a} = k\mathbf{b}$ for some scalar k

• If $\mathbf{a} = (a_1, a_2, a_3)$ and $\mathbf{b} = (b_1, b_2, b_3)$ are collinear, then: $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$ (provided no denominator is zero)

• Example: $\mathbf{a} = 2\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}$ and $\mathbf{b} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$ are collinear ($\mathbf{a} = 2\mathbf{b}$).

28. Three vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are coplanar if:

1. $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0$

2. $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{0}$

3. $\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c}$

4. $\mathbf{b} \cdot \mathbf{c} = 0$

Show me the answer

Answer: 1. $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0$

Explanation:

• Coplanar vectors lie in the same plane.

• The scalar triple product $[\mathbf{a} \ \mathbf{b} \ \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ gives the volume of the parallelepiped.

• If the vectors are coplanar, the volume is zero.

• Therefore: $[\mathbf{a} \ \mathbf{b} \ \mathbf{c}] = 0$ for coplanar vectors.

• Equivalently: $\mathbf{a} = p\mathbf{b} + q\mathbf{c}$ for some scalars p, q (one vector is a linear combination of the other two).

• Example: $\mathbf{i}, \mathbf{j}, \mathbf{i}+\mathbf{j}$ are coplanar.

Applications and Problem Solving

29. The work done by a force $\mathbf{F}$ in moving an object through displacement $\mathbf{d}$ is:

1. $\mathbf{F} \times \mathbf{d}$

2. $\mathbf{F} \cdot \mathbf{d}$

3. $|\mathbf{F} \times \mathbf{d}|$

4. $|\mathbf{F} \cdot \mathbf{d}|$

Show me the answer

Answer: 2. $\mathbf{F} \cdot \mathbf{d}$

Explanation:

• Work done = Force component in direction of displacement × displacement

• Mathematically: $W = \mathbf{F} \cdot \mathbf{d} = |\mathbf{F}| |\mathbf{d}| \cos\theta$

• Where $\theta$ is the angle between force and displacement.

• Work is a scalar quantity (dot product gives scalar).

• If force is perpendicular to displacement ($\theta = 90^\circ$), work = 0.

• If force is in direction of displacement ($\theta = 0^\circ$), work = $|\mathbf{F}| |\mathbf{d}|$ (maximum).

30. The torque $\boldsymbol{\tau}$ about a point O due to a force $\mathbf{F}$ acting at point P with position vector $\mathbf{r}$ is:

1. $\mathbf{r} \cdot \mathbf{F}$

2. $\mathbf{r} \times \mathbf{F}$

3. $\mathbf{F} \times \mathbf{r}$

4. $\mathbf{r} \cdot \mathbf{F} \cdot \mathbf{r}$

Show me the answer

Answer: 2. $\mathbf{r} \times \mathbf{F}$

Explanation:

• Torque (moment of force) = position vector × force

• $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$

• Magnitude: $|\boldsymbol{\tau}| = |\mathbf{r}| |\mathbf{F}| \sin\theta$

• Where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{F}$.

• Torque is a vector quantity (cross product gives vector).

• Direction: Perpendicular to both $\mathbf{r}$ and $\mathbf{F}$ (right-hand rule).

• Maximum torque when $\mathbf{r}$ and $\mathbf{F}$ are perpendicular ($\theta = 90^\circ$).

• Zero torque when $\mathbf{r}$ and $\mathbf{F}$ are parallel ($\theta = 0^\circ$ or $180^\circ$).