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Vector multiplication is one of the numerous techniques in mathematics for multiplying two (or more) vectors with itself. There are two relevant concepts of vector multiplication: one in which the product is a scalar and the other in which the product is a vector. There is no vector division operation. Scalar multiplication is one of the primary operations used to define a vector space in linear algebra (or more generally, a module in abstract algebra). Scalar multiplication is the multiplication of a vector by a scalar (with a vector as the product), as opposed to the inner product of two vectors (where the product is a scalar).
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Key Terms: Algebra, Multiplication of vector and scalar quantity, Scalar, Vector, Vector Space, Component of a vector
Multiplication of a Vector by a Scalar
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When a vector is multiplied by a scalar quantity, the magnitude of the vector changes in proportion to the scalar magnitude, but the direction of the vector stays the same.
Assume we have a vector; \(\vec{a}\) if this vector is multiplied by a scalar variable k, we get a new vector with a magnitude as \(|\vec{ka}|\). If k is positive, the direction of k remains the same as the vector; if k is negative, the direction of k becomes precisely opposite the vector's direction.
Let's visualize the vector's scalar multiplication.
Assume that the values of 'k' are = 2, 3, -3, \(-\frac{1}{2}\) , etc.
We can observe from the above-given vectors that the vector's \(\vec{a}\) direction remains constant while the scalar's value is positive, and the vector's direction turns opposite when the scalar's value is negative, and the magnitude changes depending on the scalar multiple's values.
We can observe from the following discussions that
\(|\vec{ka}|=k|\vec{a}|\)
Assume the scalar multiple k is -1, and we know from scalar multiplication that the resulting vector is \(-\vec{a}\), then \(\vec{a}+(-\vec{a})=0\). The vector \(-\vec{a}\) indicates the negative or additive inverse of the vector \(\vec{a}\).
Now suppose the value of \(k=\frac{1}{|a|}\) given that the value of \(a\neq0\) then by the property of scalar multiple of vectors we have \(\vec{ka}=|k|\vec{a}=\frac{1}{|a|} \times |-\vec{a}|\)
Also, as previously discussed, if k = 0, the vector becomes zero.
Example: A vector quantity is the physical quantity force. The work done is proportional to the magnitude and direction of the force applied to the item. According to Newton's second rule of linear motion, this force is a product of a vector and a scalar number. \(F=m \times a\) is the formula for force. In the above equation, 'a’ signifies the acceleration, which is a vector quantity, and ‘m’ specifies the object's mass, which is a scalar number.
Consider the situation when a vector, say vector \(\vec{a}\) is multiplied by a scalar with a magnitude of 0.50. In this situation, the product vector is a vector that represents a vector with the same direction as vector \(\vec{a}\) and a magnitude equal to \(\frac{1}{2}\) times that of vector \(\vec{a}\) (since 0.50 represents \(\frac{1}{2}\)).
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The video below explains this:
Multiplication of Vectors Video Explanation:
Properties of Scalar Multiplication
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The following rules regulate scalar multiplication:
Let \(v\) and \(w\) be vectors.
- Scalar additivity: \((c + d)v = cv + dv; \)
- Vector additivity: \(c(v + w) = cv + cw;\)
- Scalar product compatibility with scalar multiplication: \((cd)v = c(dv);\)
- A vector is not changed by multiplying it by one: \(1v = v; \)
- Multiplying by 0 gives the zero vector: \(0v = 0;\)
- The additive inverse is obtained by multiplying by -1: \((-1 )v = -v\)
The video below explains this:
Multiplication of Vector by Scalar Detailed Video Explanation:
Magnitude of Vector
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The scalar has only magnitude, in contrast, vectors have both magnitude and direction. To determine the magnitude of a vector, we must first find the length of the vector.
The magnitude of a vector formula represented as |v|, is used to compute the length of a given vector (say v). So, in essence, this variable is the distance between the vector's initial point to endpoint.
The magnitude of a Vector Formula
Assume AB is a vector quantity and has both magnitude and direction. To determine the magnitude of the vector \(\vec{AB}\), we have to calculate the distance between the beginning of point A and endpoint B. Let A have coordinates \((x_0, y_0)\) in the XY – plane and B have coordinates \((x_1, y_1)\). As a result by using the distance formula, the magnitude of a vector \(\vec{AB}\) may be expressed as;
| \(|\vec{AB}|= \sqrt{(x_1-x_0)^2+(y_1-y_0)^2}\) |
If the beginning point is (x, y) and the endpoint is the origin, the magnitude of a vector formula is;
| \(|\vec{AB}|= \sqrt{x^2+y^2}\) |
Direction of a vector
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The direction of a vector is just the measurement of the angle formed by the horizontal line. Methodology for determining the direction of a vector \(\vec{AB}\) is;
\(tan\ \alpha = \frac{y}{x};\) endpoint at 0.
Where x represents the change in a horizontal line and y represents the change in a vertical line.
Or \(tan\ \alpha = \frac{y_1-y_0}{x_1-x_0};\) where \(?(x_0, y_0)\) is initial point and \(?(x_1, y_1)\) is the endpoint.
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Practical Applications of Multiplication of Vectors with Scalars
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The multiplication of vectors with scalars has several applications in physics. Many SI units of vector values are the vector and scalar products. The SI unit of velocity, for example, is the meter per second. Velocity is measured as a vector quantity. This is calculated by multiplying two scalar values, length and time, by a unit vector in a given direction. There are many other additional applications of vector multiplication with a scalar in Mathematics and Physics.
Fun Facts about Vector Multiplication Rules
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A scalar quantity cannot be multiplied by a vector. But a vector can be multiplied by a scalar quantity.
When a vector is multiplied by a scalar, the resulting vector has the same direction but a greater magnitude.
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Difference Between Scalar and Vector
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| Scalar | Vector |
|---|---|
| Scalar quantity has only magnitude, not direction. | Vector quantity has both direction and magnitude. |
| Scalar is one-dimensional. | Vector is multidimensional. |
| This quantity varies in proportion to the change in magnitude. | Vector changes with magnitude and direction. |
| A scalar quantity can divide another scalar quantity. | A vector cannot be divided by another vector. |
| Algebraic principles apply in this case of scalar. | Vector algebra refers to a distinct set of rules. |
| In the case of speed, time, and so on, the distance between the locations is a scalar quantity instead of the direction. | Velocity is an example as it is a measurement of the rate at which an object's location changes. |
Things to Remember
- The length of a given vector may be determined, which assists in calculating the vector's magnitude. The length of the vector is the distance between the vector's beginning and ending points.
- The scalar product, often known as the dot product, is an algebraic operation that takes two equal-length numerical sequences and outputs a single number.
- A vector product also referred to as a cross-product, is a binary operation on two vectors in three dimensions.
- Vector quantities are physical quantities that have both magnitude and direction.
- Scalar quantities are physical quantities that have only magnitude and have no directions.
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Sample Questions
Ques. If \(A\cdot B=A \times B\), then the angle between A and B is. [1 marks]
Ans. Given, \(A\cdot B=A \times B\)
Taking mod on both sides, \(|A\cdot B|=|A \times B|\)
OR \(|A||B|cos\ \theta=|A| |B|sin\ \theta\)
OR \(sin\ \theta=cos\ \theta\)
OR \(tan\ \theta = 1\)
\(\Rightarrow\ \theta= 45^{\circ}\)
Ques. If the angle between the vectors \(\vec{A}\) and \(\vec{B}\) is \(\theta\), then the value of the product \((\vec{B} \times \vec{A})\cdot \vec{A}\) equals. [1 marks]
Ans. \((\vec{B} \times \vec{A})\cdot \vec{A} = BA\ sin\ \theta\ \hat{n}\ \cdot \vec{A} = 0\)
Vector A and \(\hat{n}\) are perpendicular to each other as the direction of \(\hat{n}\) can be given by right hand screw rule.
The dot product of two vectors is zero when they are perpendicular to each other.
Ques. Find the magnitude of a 3d vector 2i + 3j + 4k. [3 marks]
Ans. As we know, the magnitude of a 3d vector xi + yj + zk = \(\sqrt{x^2 + y^2 + z^2}\)
Therefore, the magnitude of a 3d vector 2i + 3j + 4k = \(\sqrt{2^2 + 3^2 + 4^2}\)
\(= \sqrt{4+9+16}\)
\(= \sqrt{29}\)
Hence, the magnitude of a 3d vector 2i + 3j + 4k ≈ 5.38
Ques. A vector is represented in the orthogonal system as \(\vec{a} = 3\hat{i} + \hat{j} + \hat{k}\). What would be the resultant vector if \(\vec{a}\) is multiplied by 5? [3 marks]
Ans. As the vector is to be multiplied by a scalar the resultant would be,
\(5\vec{a} = 5(3\hat{i}+\hat{j}+\hat{k})\)
\(5\vec{a} = 15\hat{i}+5\hat{j}+5\hat{k}\)
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Ques. A vector is represented in the orthogonal system as \(\vec{a} = 5\hat{i} + 4\hat{j} + 6\hat{k}\). What would be the resultant vector if \(\vec{a}\) is multiplied by 2? [3 marks]
Ans. As the vector is to be multiplied by a scalar the resultant would be,
\(2\vec{a} = 2(5\hat{i}+4\hat{j}+6\hat{k})\)
\(2\vec{a} = 10\hat{i}+8\hat{j}+12\hat{k})\)
Ques. If a unit vector is represented by \(0.5\hat{i}+0.8\hat{j}+c\hat{k}\) then a value of 'c' is [3 marks]
Ans. The unit vector is given as \(\hat{A}=0.5\hat{i}+0.8\hat{j}+c\hat{k}\)
Magnitude of a unit vector is always equal to 1 i.e. \(|\hat{A}|=1\)
\(\sqrt{(0.5)^2+(0.8)^2+c^2}=1\)
Or \(c^2 =0.11\)
\(\Rightarrow\ c=\sqrt{0.11}\)
Ques. Write the projection of the vector \(\hat{i}+\hat{j}+\hat{k}\) along the vector \(\hat{j}\). All India 2014C [3 marks]
Ans. Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\) and \(\vec{b}=\hat{j}\)
The projection of \(\vec{a}\) and \(\vec{b}\) is given by
\(\frac{1}{|\vec{b}|}(\vec{a}\cdot \vec{b})=\frac{1 \times0+1 \times1+1\times0}{\sqrt{1^2}}=1\)
Hence, the projection of the vector \(\hat{i}+\hat{j}+\hat{k}\) along the vector \(\hat{j}\) is 1.
Ques. If \(\vec{a}=8\) and \(\vec{b}=3\) and \((\vec{a} \times \vec{b})=12\) find the angle between \(\vec{a}\) and \(\vec{b}\). [3 marks]
Ans. Let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\).
We know that, \(|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}| sin\ \theta\)
\(|\vec{a}||\vec{b}| sin\ \theta=12\)
\(sin\ \theta=\frac{12}{|\vec{a}||\vec{b}|}= \frac{12}{8\times3}\)
\(sin\ \theta=\frac{1}{2}\Rightarrow \theta=\frac{\pi}{6}\)
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Ques. Write the projection of the vector \(\vec{a}=2\hat{i}-\hat{j}+\hat{k}\) on the vector \(\vec{b}=\hat{i}+2\hat{j}+2\hat{k}\). Delhi 2014C [3 marks]
Ans. The projection of \(\vec{a}\) and \(\vec{b}\)is given by,
\(\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\left[ \frac{(2\hat{i}-\hat{j}+\hat{k})(\hat{i}+2\hat{j}+2\hat{k})}{\sqrt{1^2+2^2+2^2}} \right]\)
\(=\frac{2\times1+(-1)\times(2)+1\times2}{\sqrt{1+4+4}}\Rightarrow \frac{2}{\sqrt{9}}=\frac{2}{3}\)
Ques. If \(\vec{a}\) and \(\vec{b}\) are unit vectors, then find the angle between \(\vec{a}\) and \(\vec{b}\), given that \(\sqrt{3}\vec{a}-\vec{b}\) is a unit vector. Delhi 2014C [3 marks]
Ans. Given \(\vec{a}\) and \(\vec{b}\) are unit vectors then, \(|\vec{a}|=|\vec{b}|=1\)
Also \((\sqrt{3}\vec{a}-\vec{b})\) is a unit vector.
\(|\sqrt{3}\vec{a}-\vec{b}|=1 (|\sqrt{3}\vec{a}-\vec{b}|)= 1^2\)
Ques. Write the value of \(\lambda\), so that the vectors \(\vec{a}=2\hat{i}+\lambda \hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}-2\hat{j}+3\hat{k}\) are perpendicular to each other. Delhi 2013C 2008 [3 marks]
Ans. Given, \(\vec{a}=2\hat{i}+\lambda \hat{j}+\hat{k}\)
and \(\vec{b}=\hat{i}-2\hat{j}+3\hat{k}\)
Since, vectors are perpendicular.
\(\vec{a} \cdot \vec{b} = 0\)
\((2\hat{i}+\lambda \hat{j}+\hat{k})\cdot(\hat{i}-2\hat{j}+3\hat{k})=0\)
\(2-2\lambda+3=0\)
\(\lambda= \frac{5}{2}\)
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