Vector Algebra

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Vector Algebra — Quick Facts

Scalars vs Vectors:

• Scalar: quantity with only magnitude (e.g. mass, temperature, speed)
• Vector: quantity with both magnitude and direction (e.g. velocity, force, displacement)

A vector $\vec{a}$ has magnitude $|\vec{a}|$ and direction. Graphically, it is represented by a directed line segment with an arrowhead indicating direction.

Position Vector: For a point $P(x, y, z)$ with respect to the origin $O$, the position vector is: $$\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}$$ where $\hat{i}$, $\hat{j}$, $\hat{k}$ are unit vectors along the $x$, $y$, and $z$ axes respectively.

Unit Vector: A unit vector along $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$, so that $|\hat{a}| = 1$. The magnitude of $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is: $$|\vec{r}| = \sqrt{x^2 + y^2 + z^2}$$

⚡ NDA Exam Tip: Many NDA questions ask you to find the magnitude of a vector or verify that two vectors are equal. Always check both the magnitude and direction. A common mistake is forgetting the square root when computing magnitude.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding and problem-solving fluency.

Vector Addition

Triangle Law: Placing the tail of vector $\vec{b}$ at the head of vector $\vec{a}$ gives the resultant $\vec{R} = \vec{a} + \vec{b}$.

Parallelogram Law: If $\vec{a}$ and $\vec{b}$ share a common tail, completing the parallelogram, the diagonal from the common tail gives $\vec{R} = \vec{a} + \vec{b}$.

Laws of Vector Addition:

• Commutative: $\vec{a} + \vec{b} = \vec{b} + \vec{a}$
• Associative: $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$
• Identity: $\vec{a} + \vec{0} = \vec{a}$
• Inverse: $\vec{a} + (-\vec{a}) = \vec{0}$

Subtraction: $\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$.

Scalar (Dot) Product

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

In component form, if $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$: $$\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$$

Properties of Dot Product:

• $\vec{a} \cdot \vec{a} = |\vec{a}|^2$
• $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ (commutative)
• $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$
• $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$
• $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$ (orthogonal unit vectors)

NDA Worked Example: Find $\vec{a} \cdot \vec{b}$ when $\vec{a} = 3\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 4\hat{j} - 2\hat{k}$. $$\vec{a} \cdot \vec{b} = 3(1) + (-2)(4) + (1)(-2) = 3 - 8 - 2 = -7$$

Vector (Cross) Product

The cross product $\vec{a} \times \vec{b}$ produces a vector perpendicular to both $\vec{a}$ and $\vec{b}$: $$\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta \ \hat{n}$$ where $\hat{n}$ is the unit normal vector given by the right-hand rule.

In determinant form: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\hat{i} - (a_1 b_3 - a_3 b_1)\hat{j} + (a_1 b_2 - a_2 b_1)\hat{k}$$

Properties of Cross Product:

• $\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$ (anti-commutative)
• $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$
• $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$ equals the area of the parallelogram formed by $\vec{a}$ and $\vec{b}$
• $\vec{a} \times \vec{a} = \vec{0}$
• $\hat{i} \times \hat{j} = \hat{k},\ \hat{j} \times \hat{k} = \hat{i},\ \hat{k} \times \hat{i} = \hat{j}$

NDA Worked Example: Find the area of the parallelogram with adjacent sides $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$. $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -1 \ 3 & -1 & 2 \end{vmatrix} = (2(2)-(-1)(-1))\hat{i} - (1(2)-(-1)(3))\hat{j} + (1(-1)-2(3))\hat{k}$$ $$= (4-1)\hat{i} - (2+3)\hat{j} + (-1-6)\hat{k} = 3\hat{i} - 5\hat{j} - 7\hat{k}$$ $$|\vec{a} \times \vec{b}| = \sqrt{9 + 25 + 49} = \sqrt{83} \text{ square units}$$

Angle Between Two Vectors: $$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{a_1 b_1 + a_2 b_2 + a_3 b_3}{\sqrt{a_1^2 + a_2^2 + a_3^2}\sqrt{b_1^2 + b_2^2 + b_3^2}}$$

Projection of $\vec{a}$ on $\vec{b}$: $$\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory, derivations, and advanced problem types for thorough NDA preparation.

Scalar Triple Product

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

It equals the volume of the parallelepiped having $\vec{a}$, $\vec{b}$, $\vec{c}$ as adjacent edges.

In determinant form: $$\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{vmatrix}$$

Properties:

• $[\vec{a} \ \vec{b} \ \vec{c}] = [\vec{b} \ \vec{c} \ \vec{a}] = [\vec{c} \ \vec{a} \ \vec{b}]$ (cyclic permutation leaves value unchanged)
• $[\vec{a} \ \vec{b} \ \vec{c}] = -[\vec{b} \ \vec{a} \ \vec{c}]$ (swapping two vectors changes sign)
• $[\vec{a} \ \vec{b} \ \vec{c}] = 0$ if any two vectors are parallel or if the three vectors are coplanar

NDA Worked Example: Find the volume of the parallelepiped with edges $\vec{a} = \hat{i} - \hat{j}$, $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{c} = 3\hat{i} - 2\hat{k}$. $$[\vec{a} \ \vec{b} \ \vec{c}] = \begin{vmatrix} 1 & -1 & 0 \ 2 & 1 & -1 \ 3 & 0 & -2 \end{vmatrix} = 1[1(-2)-(-1)(0)] - (-1)[2(-2)-(-1)(3)] + 0$$ $$= 1[-2 - 0] + 1[-4 + 3] = -2 - 1 = -3$$ Volume $= |[\vec{a} \ \vec{b} \ \vec{c}]| = 3$ cubic units.

Vector Triple Product

The vector triple product formula: $$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$$

This result appears in physics problems involving magnetic fields and rotational dynamics.

Application: Coplanarity of Four Points

Points $A$, $B$, $C$, $D$ are coplanar if and only if $(\vec{AB} \times \vec{AC}) \cdot \vec{AD} = 0$, i.e. the scalar triple product of $\vec{AB}$, $\vec{AC}$, $\vec{AD}$ equals zero.

Straight Line and Plane Problems in NDA

Given two vectors $\vec{a}$ and $\vec{b}$ representing sides of a parallelogram:

• Diagonal 1: $\vec{a} + \vec{b}$
• Diagonal 2: $\vec{a} - \vec{b}$

The area of the parallelogram is $|\vec{a} \times \vec{b}|$. The area of the triangle with the same edges is $\frac{1}{2}|\vec{a} \times \vec{b}|$.

NDA Worked Example (2020): Given $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$, find the angle between them. $$\vec{a} \cdot \vec{b} = 2(1) + 1(-1) + (-1)(2) = 2 - 1 - 2 = -1$$ $$|\vec{a}| = \sqrt{4+1+1} = \sqrt{6}, \quad |\vec{b}| = \sqrt{1+1+4} = \sqrt{6}$$ $$\cos\theta = \frac{-1}{\sqrt{6} \cdot \sqrt{6}} = \frac{-1}{6} \Rightarrow \theta = \cos^{-1}\left(-\frac{1}{6}\right) \approx 99.59°$$

Previous Year NDA Question Patterns:

Year	Topic Asked
2022	Find magnitude of a vector; dot product of two vectors
2021	Cross product; area of parallelogram
2020	Angle between vectors using dot product; projection
2019	Scalar triple product; volume of parallelepiped
2018	Verify coplanarity of four points; unit vector perpendicular to two given vectors

Vector Algebra typically yields 4–6 questions per NDA paper, particularly from dot product, cross product, and magnitude calculations. Always represent vectors in component form first — this makes all subsequent operations straightforward. The scalar triple product determinant method is a reliable technique that avoids errors in computing cross products of cross products.