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Vector Algebra

Vector (or Cross) Product of Two Vectors

Vectors can be multiplied in two ways, a scalar product where the result is a scalar and cross or vector product where is the result is a vector. In this article, we will look at the cross or vector product of two vectors.

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Explanation

We have already studied the three-dimensional right-handed rectangular coordinate system. As shown in the figure below, when the positive x-axis is rotated counter-clockwise into the positive y-axis, then a right-handed standard screw moves in the direction of the positive z-axis.

cross product

As can be seen above, in a three-dimensional right-handed rectangular coordinate system, the thumb of the right-hand points in the direction of the positive z-axis when the fingers are curled from the positive x-axis towards the positive y-axis.

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 Vector (or Cross) Product of Two Vectors

Definition

The cross or vector product of two non-zero vectors \( \vec{a} \) and \( \vec{b} \), is

\( \vec{a} \) x \( \vec{b} \) = |\( \vec{a} \)| |\( \vec{b} \)| \( \sin{\theta} \)\( \hat{n} \)

Where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \), 0 ≤ \( \theta \) ≤ \( \pi \). Also, \( \hat{n} \) 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 as shown below.

cross product

As can be seen above, when the system is rotated from \( \vec{a} \) to \( \vec{b} \), it moves in the direction of \( \hat{n} \). Also, if either \( \vec{a} \) = 0 or \( \vec{b} \) = 0, then \( \theta\) is not defined and we can say,

\( \vec{a} \) x \( \vec{b} \) = \( \vec{0} \)

Important Observations

  • \( \vec{a} \) x \( \vec{b} \) is a vector.
  • If \( \vec{a} \) and \( \vec{b} \) are two non-zero vectors, then \( \vec{a} \) x \( \vec{b} \) = 0, if and only if \( \vec{a} \) and \( \vec{b} \) are parallel (or collinear) to each other, i.e.

\( \vec{a} \) x \( \vec{b} \) = 0 ⇔ \( \vec{a} \) \( \parallel\) \( \vec{b} \)

Hence, \( \vec{a} \) x \( \vec{a} \) = 0 and \( \vec{a} \) x \( \vec{(- a)} \) = 0. This is because in the first case \( \theta\) = 0. Also, in the second case \( \theta\) = \( \pi\), giving the value of \( \sin{\theta} \) = 0.

  • If \( \theta\) = \( \frac{\pi}{2}\), then \( \vec{a} \) x \( \vec{b} \) = |\( \vec{a} \)| |\( \vec{b} \)|
  • Considering observations 2 and 3 above, for mutually perpendicular vectors \( \vec{i} \), \( \vec{j} \), and \( \vec{k} \), we have

\( \vec{i} \) x \( \vec{i} \) = \( \vec{j} \) x \( \vec{j} \) = \( \vec{k} \) x \( \vec{k} \) = 0

Also,

\( \vec{i} \) x \( \vec{j} \) = \( \vec{k} \)
\( \vec{j} \) x \( \vec{k} \) = \( \vec{i} \)
\( \vec{k} \) x \( \vec{i} \) = \( \vec{j} \)

  • The angle between the two vectors \( \vec{a} \) and \( \vec{b} \) is,

\( \sin{\theta} \) = \( \frac{|\vec{a} × \vec{b}|}{|\vec{a}||\vec{b}|} \)

  • A cross or vector product is not commutative. We know this because \( \vec{a} \) x \( \vec{b} \) = \( \vec{- b} \) x \( \vec{a} \). Now, we know that,

\( \vec{a} \) x \( \vec{b} \) = |\( \vec{a} \)| |\( \vec{b} \)| \( \sin{\theta} \)\( \hat{n} \).

Where \( \vec{a} \), \( \vec{b} \), and \( \hat{n} \) form a right-handed system. Or, \( \theta\) is traversed from \( \vec{a} \) to \( \vec{b} \). On the other hand,

\( \vec{b} \) x \( \vec{a} \) = |\( \vec{b} \)| |\( \vec{a} \)| \( \sin{\theta} \)\( \hat{n_1} \).

Where \( \vec{b} \), \( \vec{a} \), and \( \hat{n_1} \) form a right-handed system. Or, \( \theta\) is traversed from \( \vec{b} \) to \( \vec{a} \). So, if \( \vec{a} \) and \( \vec{b} \) lie on a plane of paper, then \( \hat{n} \) and \( \hat{n_1} \) are both perpendicular to the plane of the paper. However, \( \hat{n} \) is directed above the paper and \( \hat{n_1} \) is directed below it. Or, \( \hat{n} \) = – \( \hat{n_1} \). Hence,

\( \vec{a} \) x \( \vec{b} \) = |\( \vec{a} \)| |\( \vec{b} \)| \( \sin{\theta} \)\( \hat{n} \) = – |\( \vec{a} \)||\( \vec{b} \)|\( \sin{\theta} \)\( \hat{n_1} \)

= – \( \vec{b} \) x \( \vec{a} \)

  • From the observations 4 and 6 above, we have

\( \vec{j} \) x \( \vec{i} \) = – \( \vec{k} \)
\( \vec{k} \) x \( \vec{j} \) = – \( \vec{i} \)
\( \vec{i} \) x \( \vec{k} \) = – \( \vec{j} \)

  • If \( \vec{a} \) and \( \vec{b} \) represent the two sides of a triangle, then its area is \( \frac{1}{2} \)|\( \vec{a} \) x \( \vec{b} \)|. To understand this, look at the figure given below.

cross product

By the definition of the area of a triangle, we have area of ΔABC = \( \frac{1}{2} \)(AB).(CD). We know that, AB = |\( \vec{b} \)| and CD = |\( \vec{a} \)|\( \sin{\theta} \). Therefore,

Area of ΔABC = \( \frac{1}{2} \)|\( \vec{a} \)||\( \vec{b} \)|\( \sin{\theta} \) = \( \frac{1}{2} \)|\( \vec{a} \) x \( \vec{b} \)|

  • If \( \vec{a} \) and \( \vec{b} \) represent the two adjacent sides of a parallelogram, then its area is |\( \vec{a} \) x \( \vec{b} \)|. To understand this, look at the figure given below.

cross product

By the definition of the area of a parallelogram, we have area of parallelogram ABCD = (AB).(DE). We know that, AB = |\( \vec{b} \)| and DE = |\( \vec{a} \)|\( \sin{\theta} \). Therefore,

Area of parallelogram ABCD = |\( \vec{a} \)||\( \vec{b} \)|\( \sin{\theta} \) = |\( \vec{a} \) x \( \vec{b} \)|

Property: Distributivity of a cross or vector product over addition

If \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are any three vectors and \( \lambda\) is a scalar, then

  • \( \vec{a} \) x (\( \vec{b} \) + \( \vec{c} \)) = \( \vec{a} \) x \( \vec{b} \) + \( \vec{a} \) x \( \vec{c} \)
  • \( \lambda\)(\( \vec{a} \) x \( \vec{b} \)) = (\( \lambda\)\( \vec{a} \)) x \( \vec{b} \) = \( \vec{a} \) x (\( \lambda\)\( \vec{b} \))

Component Form of Vector Product

If \( \vec{a} \) and \( \vec{b} \) are two vectors given in the component form as a1\( \hat{i} \) + a2\( \hat{j} \) + a3\( \vec{k} \) and b1\( \hat{i} \) + b2\( \hat{j} \) + b3\( \vec{k} \). Then, their cross or vector product is,

\( \vec{a} \) x \( \vec{b} \) = \( \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1&b_2&b_3 \end{vmatrix} \)

Explanation

\( \vec{a} \) x \( \vec{b} \) = (a1\( \hat{i} \) + a2\( \hat{j} \) + a3\( \vec{k} \)) x (b1\( \hat{i} \) + b2\( \hat{j} \) + b3\( \vec{k} \))

= a1b1(\( \hat{i} \) x \( \hat{i} \)) + a1b2(\( \hat{i} \) x \( \hat{j} \)) + a1b3(\( \hat{i} \) x \( \hat{k} \)) + a2b1(\( \hat{j} \) x \( \hat{i} \)) + a2b2(\( \hat{j} \) x \( \hat{j} \)) + a2b3(\( \hat{j} \) x \( \hat{k} \)) + a3b1(\( \hat{k} \) x \( \hat{i} \)) + a3b2(\( \hat{k} \) x \( \hat{j} \)) + a3b3(\( \hat{k} \) x \( \hat{k} \))

We know that, \( \vec{i} \) x \( \vec{i} \) = \( \vec{j} \) x \( \vec{j} \) = \( \vec{k} \) x \( \vec{k} \) = 0. Also,  \( \vec{j} \) x \( \vec{i} \) = – \( \vec{k} \), \( \vec{k} \) x \( \vec{j} \) = – \( \vec{i} \), \( \vec{i} \) x \( \vec{k} \) = – \( \vec{j} \). Therefore,

\( \vec{a} \) x \( \vec{b} \) = a1b2(\( \hat{i} \) x \( \hat{j} \)) – a1b3(\( \hat{k} \) x \( \hat{i} \)) – a2b1(\( \hat{i} \) x \( \hat{j} \)) + a2b3(\( \hat{j} \) x \( \hat{k} \)) + a3b1(\( \hat{k} \) x \( \hat{i} \)) – a3b2(\( \hat{j} \) x \( \hat{k} \))

We also know that, \( \vec{i} \) x \( \vec{j} \) = \( \vec{k} \), \( \vec{j} \) x \( \vec{k} \) = \( \vec{i} \), \( \vec{k} \) x \( \vec{i} \) = \( \vec{j} \). Therefore, we have,

\( \vec{a} \) x \( \vec{b} \) = a1b2\( \hat{k} \) – a1b3\( \hat{j} \) – a2b1\( \hat{k} \) + a2b3\( \hat{i} \) + a3b1\( \hat{j} \) – a3b2\( \hat{i} \)
= (a2b3 – a3b2)\( \hat{i} \) – (a1b3 – a3b1)\( \hat{j} \) + (a1b2 – a2b1)\( \hat{k} \)
= \( \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1&b_2&b_3 \end{vmatrix} \).

Solved Examples for You

Question 1: Find the area of the parallelogram whose adjacent sides are determined by the following vectors,

\( \vec{a} \) = \( \hat{i} \) – \( \hat{j} \) + 3\( \vec{k} \) and

\( \vec{b} \) = 2\( \hat{i} \) – 7\( \hat{j} \) + \( \vec{k} \).

Answer : We know that if \( \vec{a} \) and \( \vec{b} \) represent the two adjacent sides of a parallelogram, then its area is |\( \vec{a} \) x \( \vec{b} \)|. Also,

\( \vec{a} \) x \( \vec{b} \) = \( \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1&b_2&b_3 \end{vmatrix} \)

Substituting the values of a1,a2,a3,b1,b2,and b3, we get

\( \vec{a} \) x \( \vec{b} \) = \( \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&3\\2&-7&1 \end{vmatrix} \)

Solving the determinant, we get

\( \vec{a} \) x \( \vec{b} \) = \( \hat{i} \){[(-1) x 1)] – [(-7) x 3]} – \( \hat{j} \){[1 x 1)] – [2 x 3]} + \( \hat{k} \){[1 x (-7))] – [2 x (-1)]}
= 20\( \hat{i} \) + 5\( \hat{j} \) – 5\( \hat{k} \).

Also, the magnitude of \( \vec{a} \) x \( \vec{b} \) is,

|\( \vec{a} \) x \( \vec{b} \)| = \( \sqrt{20^2 + 5^2 + (-5)^2} \) = \( \sqrt{450} \) = \( \sqrt{25 × 9 × 2} \) = 15\( \sqrt{2} \).

Therefore, the area of the parallelogram is 15\( \sqrt{2} \).

Question 2: What is meant by vector product of two vectors?

Answer: The vector product of two vectors refers to a vector that is perpendicular to both of them. One can obtain its magnitude by multiplying their magnitudes by the sine of the angle that exists between them.

Question 3: Explain the characteristics of vector product?

Answer: The characteristics of vector product are as follows:

  • Vector product two vectors always happen to be a vector.
  • Vector product of two vectors happens to be noncommutative.
  • Vector product is in accordance with the distributive law of multiplication.
  • If a • b = 0 and a ≠ o, b ≠ o, then the two vectors shall be parallel to each other.

Question 4: When can we say that two vectors are parallel?

Answer: Two vectors A and B will be said to be parallel if and only if they happen to be scalar multiples of one another. A = k B, where k is a constant and not equal to zero.

Question 5: What does the square of a vector mean?

Answer: Square of a vector refers to the Dot Product with itself.

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