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Vector Algebra - Vector (cross) product of vectors

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of Vector Product: The vector product (or cross product) of two vectors a⃗\vec{a} and b⃗\vec{b}, denoted as a⃗×b⃗\vec{a} \times \vec{b}, is defined as ∣a⃗∣∣b⃗∣sin⁡θn^|\vec{a}| |\vec{b}| \sin \theta \hat{n}, where θ\theta is the angle between the vectors (0≤θ≤π0 \leq \theta \leq \pi) and n^\hat{n} is a unit vector perpendicular to both a⃗\vec{a} and b⃗\vec{b}. Visually, this resulting vector a⃗×b⃗\vec{a} \times \vec{b} stands perpendicular to the flat plane containing the original two vectors.

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Direction and Right-Hand Rule: The direction of the unit vector n^\hat{n} is determined by the right-hand thumb rule. If you curl the fingers of your right hand from a⃗\vec{a} towards b⃗\vec{b}, your outstretched thumb points in the direction of a⃗×b⃗\vec{a} \times \vec{b}. This visualizes the 'handedness' of the coordinate system, where reversing the order of vectors points the thumb in the opposite direction.

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Geometric Interpretation of Magnitude: The magnitude ∣a⃗×b⃗∣|\vec{a} \times \vec{b}| represents the area of a parallelogram where a⃗\vec{a} and b⃗\vec{b} are adjacent sides. Visually, if you draw these two vectors from a common origin, the 'spread' between them determines the size of the parallelogram, and the cross product's length captures this area exactly.

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Condition for Parallel Vectors: If two non-zero vectors a⃗\vec{a} and b⃗\vec{b} are parallel (or collinear), their cross product is the zero vector 0⃗\vec{0}. This occurs because the angle θ\theta is either 0∘0^\circ or 180∘180^\circ, making sin⁡θ=0\sin \theta = 0. Visually, since the vectors lie on the same line, they cannot form a parallelogram, resulting in zero area.

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Properties of Unit Vectors i^,j^,k^\hat{i}, \hat{j}, \hat{k}: The cross products of standard basis vectors follow a cyclic pattern. Visually, if you arrange i,j,ki, j, k in a circle clockwise: i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, and k^×i^=j^\hat{k} \times \hat{i} = \hat{j}. Moving in the opposite direction results in negative signs, such as j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k}, while the product of any unit vector with itself is 0⃗\vec{0}.

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Anti-Commutative Property: Vector product is not commutative. Specifically, a⃗×b⃗=−(b⃗×a⃗)\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}). This means that swapping the order of the vectors in the operation reverses the direction of the resulting vector by 180∘180^\circ, though its magnitude remains the same.

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Determinant Form: For vectors given in component form a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} and b⃗=b1i^+b2j^+b3k^\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}, the cross product is calculated using a 3×33 \times 3 determinant. This algebraic method organizes the unit vectors in the first row and the components of a⃗\vec{a} and b⃗\vec{b} in the second and third rows respectively.

📐Formulae

a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θn^\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \hat{n}

a⃗×b⃗=∣i^j^k^a1a2a3b1b2b3∣\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}

sin⁡θ=∣a⃗×b⃗∣∣a⃗∣∣b⃗∣\sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|}

Area of Parallelogram=∣a⃗×b⃗∣\text{Area of Parallelogram} = |\vec{a} \times \vec{b}|

Area of Triangle=12∣a⃗×b⃗∣\text{Area of Triangle} = \frac{1}{2} |\vec{a} \times \vec{b}|

Area of Parallelogram with diagonals d1⃗,d2⃗=12∣d1⃗×d2⃗∣\text{Area of Parallelogram with diagonals } \vec{d_1}, \vec{d_2} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|

💡Examples

Problem 1:

Find the cross product a⃗×b⃗\vec{a} \times \vec{b} if a⃗=2i^+j^+3k^\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k} and b⃗=3i^+5j^−2k^\vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k}.

Solution:

a⃗×b⃗=∣i^j^k^21335−2∣\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix} Expanding the determinant along the first row: =i^(1(−2)−3(5))−j^(2(−2)−3(3))+k^(2(5)−1(3))= \hat{i}(1(-2) - 3(5)) - \hat{j}(2(-2) - 3(3)) + \hat{k}(2(5) - 1(3)) =i^(−2−15)−j^(−4−9)+k^(10−3)= \hat{i}(-2 - 15) - \hat{j}(-4 - 9) + \hat{k}(10 - 3) =−17i^+13j^+7k^= -17\hat{i} + 13\hat{j} + 7\hat{k}

Explanation:

We use the determinant method where the unit vectors are placed in the first row, and the components of the two vectors are placed in the subsequent rows. Expansion of the determinant provides the component form of the resulting vector.

Problem 2:

Find the area of a triangle with vertices A(1,1,1)A(1, 1, 1), B(1,2,3)B(1, 2, 3) and C(2,3,1)C(2, 3, 1).

Solution:

First, find two vectors representing the sides: AB⃗=(1−1)i^+(2−1)j^+(3−1)k^=0i^+j^+2k^\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = 0\hat{i} + \hat{j} + 2\hat{k} AC⃗=(2−1)i^+(3−1)j^+(1−1)k^=i^+2j^+0k^\vec{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = \hat{i} + 2\hat{j} + 0\hat{k} Calculate the cross product AB⃗×AC⃗\vec{AB} \times \vec{AC}: AB⃗×AC⃗=∣i^j^k^012120∣=i^(0−4)−j^(0−2)+k^(0−1)=−4i^+2j^−k^\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0-4) - \hat{j}(0-2) + \hat{k}(0-1) = -4\hat{i} + 2\hat{j} - \hat{k} Calculate the magnitude: ∣AB⃗×AC⃗∣=(−4)2+22+(−1)2=16+4+1=21|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21} Area=12∣AB⃗×AC⃗∣=212 sq. units\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{\sqrt{21}}{2} \text{ sq. units}

Explanation:

To find the area of a triangle, we first find two vectors representing adjacent sides by subtracting vertex coordinates. Then, we find the magnitude of their cross product and divide by 2.