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Vectors - Vector (Cross) Product of Vectors

Grade 12ICSE

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

🔑Concepts

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The Vector Product, denoted as a⃗×b⃗\vec{a} \times \vec{b}, results in a vector that is perpendicular to the plane containing both a⃗\vec{a} and b⃗\vec{b}. If you visualize a⃗\vec{a} and b⃗\vec{b} lying on a flat tabletop, the cross product vector would point straight up towards the ceiling or straight down towards the floor.

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The direction of the resulting vector 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} through the smaller angle θ\theta, your thumb points in the direction of the unit vector n^\hat{n}. This visually establishes the 'handedness' of the coordinate system.

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The magnitude of the cross product, ∣a⃗×b⃗∣|\vec{a} \times \vec{b}|, represents the area of the parallelogram formed with a⃗\vec{a} and b⃗\vec{b} as adjacent sides. Imagine stretching a flexible frame between the two vectors; the surface area within that frame is exactly the magnitude of their cross product.

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The vector product is non-commutative, meaning a⃗×b⃗=−(b⃗×a⃗)\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}). Reversing the order of vectors flips the resulting vector 180∘180^{\circ} in the opposite direction along the same axis.

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For standard unit vectors i^,j^,k^\hat{i}, \hat{j}, \hat{k}, the products follow a cyclic rule: 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}. If you visualize a circle with i,j,ki, j, k in clockwise order, following the arrow gives a positive result, while going against the arrow (e.g., j^×i^\hat{j} \times \hat{i}) gives a negative result (−k^-\hat{k}).

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If two non-zero vectors a⃗\vec{a} and b⃗\vec{b} are parallel or collinear, their cross product is the null vector 0⃗\vec{0}. This is because the angle θ\theta between them is 0∘0^{\circ} or 180∘180^{\circ}, and since sin⁡(0)=0\sin(0) = 0, the magnitude of the product becomes zero, representing a 'flattened' parallelogram with no area.

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The cross product is distributive over addition, meaning a⃗×(b⃗+c⃗)=(a⃗×b⃗)+(a⃗×c⃗)\vec{a} \times (\vec{b} + \vec{c}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}). This allows us to calculate the product of complex algebraic vector expressions component by component.

📐Formulae

a⃗×b⃗=(∣a⃗∣∣b⃗∣sin⁡θ)n^\vec{a} \times \vec{b} = (|\vec{a}| |\vec{b}| \sin \theta) \hat{n}, where 0≤θ≤π0 \leq \theta \leq \pi

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}|}

Unit vector perpendicular to the plane of a⃗\vec{a} and b⃗\vec{b}: n^=±a⃗×b⃗∣a⃗×b⃗∣\hat{n} = \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}

Area of Parallelogram with adjacent sides a⃗\vec{a} and b⃗=∣a⃗×b⃗∣\vec{b} = |\vec{a} \times \vec{b}|

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

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

Lagrange's Identity: ∣a⃗×b⃗∣2=∣a⃗∣2∣b⃗∣2−(a⃗⋅b⃗)2|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^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:

We use the determinant form: 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 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:

To find the cross product of two vectors in component form, we arrange the unit vectors in the first row and the components of the first and second vectors in the second and third rows of a 3×33 \times 3 determinant. Expanding this determinant gives the resultant vector.

Problem 2:

Find the area of a triangle having the points 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) as its vertices.

Solution:

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

Explanation:

To find the area of a triangle using vectors, we first determine the vectors representing two of its sides originating from the same vertex. The area is half the magnitude of their cross product, as the cross product's magnitude represents the area of the full parallelogram formed by those vectors.