# 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$$).