Vector Algebra Ex 10.4 Class 12 NCERT Solutions & Concepts

Understanding the Vector (Cross) Product

The vector product (or cross product) of two non-zero vectors $\vec{a}$ and $\vec{b}$, denoted by $\vec{a} \times \vec{b}$, is defined as a vector perpendicular to both $\vec{a}$ and $\vec{b}$. Mathematically, it is written as:

$\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin\theta \hat{n}$

Where:

• $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes of the two vectors.
• $\theta$ is the angle between them ($0 \le \theta \le \pi$).
• $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ such that $\vec{a}$, $\vec{b}$, and $\hat{n}$ form a right-handed system.

Geometrically, if $\vec{a}$ and $\vec{b}$ represent the adjacent sides of a parallelogram, then the magnitude $|\vec{a} \times \vec{b}|$ represents the area of that parallelogram. Similarly, half of this magnitude represents the area of the triangle formed by these two vectors.

Key Properties and Formulas

Non-Commutative Property

Unlike the scalar product, the vector product is not commutative. $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$.

Parallel Vectors

If two non-zero vectors $\vec{a}$ and $\vec{b}$ are parallel or collinear, then $\theta = 0$ or $\pi$, which means $\vec{a} \times \vec{b} = \vec{0}$. This also implies that $\vec{a} \times \vec{a} = \vec{0}$.

Unit Vector Products

For orthogonal unit vectors: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, and $\hat{k} \times \hat{i} = \hat{j}$. Also, $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = \vec{0}$.

Area of a Triangle

The area of a triangle with adjacent sides as vectors $\vec{a}$ and $\vec{b}$ is given by $\frac{1}{2} |\vec{a} \times \vec{b}|$. If vertices $A$, $B$, $C$ are given, the area is $\frac{1}{2} |\vec{AB} \times \vec{AC}|$.

Area of a Parallelogram

The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is $|\vec{a} \times \vec{b}|$. If diagonals $\vec{d_1}$ and $\vec{d_2}$ are given, the area is $\frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.

Step-by-Step Method to Calculate Cross Product using Determinants

1. Express Vectors in Component Form — Write down the given vectors: $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.
2. Set up the 3x3 Determinant Matrix — Construct a 3x3 determinant. The first row contains the unit vectors $\hat{i}$, $\hat{j}$, $\hat{k}$. The second row contains the coefficients of $\vec{a}$ ($a_1, a_2, a_3$), and the third row contains the coefficients of $\vec{b}$ ($b_1, b_2, b_3$).
3. Expand the Determinant — Expand along the first row: $\vec{a} \times \vec{b} = \hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1)$.

Common Mistakes and Exam Tips

• The sign of the $\hat{j}$ term: When expanding the 3x3 determinant, a very common error is to write plus instead of minus for the $\hat{j}$ component. Always remember the sign pattern: $+ - +$.
• Area of Parallelogram vs. Triangle: Students often mix up when to include the $\frac{1}{2}$ multiplier. Remember: Triangle area always has $\frac{1}{2}$; Parallelogram with adjacent sides does NOT have $\frac{1}{2}$, but a parallelogram given by its diagonals does have $\frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.
• Order of operation: Because $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$, the order in which you set up the determinant rows matters when calculating specific direction vectors (though the magnitude will remain the same).

Practice Questions with Solutions

• Q: Find $|\vec{a} \times \vec{b}|$ if $\vec{a} = \hat{i} - 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}$. A: Step 1: Write down the 3x3 determinant: $\vec{a} \times \vec{b} = \det \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{bmatrix}$ Step 2: Expand the determinant along the first row: $\vec{a} \times \vec{b} = \hat{i}[(-7)(2) - (7)(-2)] - \hat{j}[(1)(2) - (7)(3)] + \hat{k}[(1)(-2) - (-7)(3)]$ $\vec{a} \times \vec{b} = \hat{i}[-14 + 14] - \hat{j}[2 - 21] + \hat{k}[-2 + 21]$ $\vec{a} \times \vec{b} = 0\hat{i} + 19\hat{j} + 19\hat{k}$ Step 3: Calculate the magnitude: $|\vec{a} \times \vec{b}| = \sqrt{0^2 + 19^2 + 19^2} = \sqrt{2 \times 19^2} = 19\sqrt{2}$ Final answer: $19\sqrt{2}$
• Q: Find a unit vector perpendicular to both $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ where $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$. A: Step 1: Compute $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$: $\vec{a} + \vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j} + 0\hat{k}$ $\vec{a} - \vec{b} = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} = 2\hat{i} + 0\hat{j} + 4\hat{k}$ Step 2: Calculate the cross product of these two new vectors: $\vec{u} = (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \det \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{bmatrix}$ $\vec{u} = \hat{i}(16 - 0) - \hat{j}(16 - 0) + \hat{k}(0 - 8) = 16\hat{i} - 16\hat{j} - 8\hat{k}$ Step 3: Compute the magnitude of $\vec{u}$: $|\vec{u}| = \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24$ Step 4: Find the perpendicular unit vector: $\hat{\eta} = \pm \frac{\vec{u}}{|\vec{u}|} = \pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24} = \pm \left(\frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k}\right)$ Final answer: $\pm \left(\frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k}\right)$
• Q: Find the area of a triangle with vertices $A(1, 1, 2)$, $B(2, 3, 5)$, and $C(1, 5, 5)$. A: Step 1: Find the vectors representing two adjacent sides, $\vec{AB}$ and $\vec{AC}$: $\vec{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$ $\vec{AC} = (1-1)\hat{i} + (5-1)\hat{j} + (5-2)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k}$ Step 2: Calculate the cross product $\vec{AB} \times \vec{AC}$: $\vec{AB} \times \vec{AC} = \det \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{bmatrix}$ $\vec{AB} \times \vec{AC} = \hat{i}(6 - 12) - \hat{j}(3 - 0) + \hat{k}(4 - 0) = -6\hat{i} - 3\hat{j} + 4\hat{k}$ Step 3: Find the magnitude of the cross product: $|\vec{AB} \times \vec{AC}| = \sqrt{(-6)^2 + (-3)^2 + 4^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$ Step 4: Compute the area of the triangle: $\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{\sqrt{61}}{2}\text{ square units}$ Final answer: $\frac{\sqrt{61}}{2}$ square units
• Q: Find the area of a parallelogram whose adjacent sides are determined by the vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$. A: Step 1: Calculate the cross product $\vec{a} \times \vec{b}$: $\vec{a} \times \vec{b} = \det \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{bmatrix}$ $\vec{a} \times \vec{b} = \hat{i}(-1 - (-21)) - \hat{j}(1 - 6) + \hat{k}(-7 - (-2))$ $\vec{a} \times \vec{b} = 20\hat{i} + 5\hat{j} - 5\hat{k}$ Step 2: Calculate the magnitude representing the area of the parallelogram: $\text{Area} = |\vec{a} \times \vec{b}| = \sqrt{20^2 + 5^2 + (-5)^2}$ $\text{Area} = \sqrt{400 + 25 + 25} = \sqrt{450} = \sqrt{225 \times 2} = 15\sqrt{2}\text{ square units}$ Final answer: $15\sqrt{2}$ square units

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