NEET Physics Solutions For Class 11 Chapter 10 Mathematical Tools – Multiplication Of Vectors

Multiplication Of Vectors

1. The Scalar Product:

The scalar product or dot product of any two vectors \(\vec{A} \text { and } \vec{B}\), denoted as \(\vec{A} \cdot \vec{B}\)(read \(\vec{A}\).\(\vec{B}\)) is defined as the product of their magnitude with cosine of angle between them. Thus, \(\overrightarrow{\mathrm{A}}\). \(\vec{B}\) = AB cos θ {here θ is the angle between the vectors}

The Scalar Product Properties:

It is always a scalar which is positive if the angle between the vectors is acute (i.e. < 90º) and negative if the angle between them is obtuse (i.e. 90º < θ ≤ 180º)

It is commutative, i.e., \(\vec{A} \cdot \vec{B}=\vec{A} \cdot \vec{B}\)

It is distributive, i.e., \(\vec{A} \cdot(\vec{B}+\vec{C})=\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{C}\)

As by definition \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}\)= AB cos θ.

The angle between the vectors θ = \(\cos ^{-1}\left[\frac{\vec{A} \cdot \vec{B}}{A B}\right]\)

⇒ \(\vec{A}.\vec{B}\) = A (B cos θ) = B (A cos θ)

Geometrically, B cos θ is the projection of \(\vec{B} \text { onto } \vec{A}\), and A cos θ is the projection of \(\vec{A} \text { onto } \vec{B}\) shown. So \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}\) is the product of the magnitude of \(\overrightarrow{\mathrm{A}}\) and the component of \(\overrightarrow{\mathrm{B}}\) along \(\overrightarrow{\mathrm{A}}\) and vice versa.

Component of \(\vec{B} \text { along } \overrightarrow{\mathrm{A}}\) = B cosθ = \(\frac{\vec{A} \cdot \vec{B}}{A}=\vec{A} \cdot \vec{B}\)

Component of \(\vec{A} \text { along } \overrightarrow{\mathrm{B}}\) along = \(\frac{\vec{A} \cdot \vec{B}}{B}=\vec{A} \cdot \vec{B}\)

Scalar product of two vectors will be maximum when cos θ = max = 1, i.e., θ = 0º, i.e., vectors are parallel

⇒ \((\vec{A} \cdot \vec{B})_{\max }=A B\)

If the scalar product of two nonzero vectors vanishes then the vectors are perpendicular. The scalar product of a vector by itself is termed as self dot product and is given by

⇒ \((\vec{A})^2=\vec{A} \cdot \vec{A}=A A \cos \theta=A A \cos 0^{\circ}=A^2\)

⇒ \(A=\sqrt{\vec{A} \cdot \vec{A}}\)

In case of unit vector \(\hat{n}\)

⇒ \(\hat{n} . \hat{n}=1 \times 1 \times \cos 0^{\circ}=1\)

⇒ \(\hat{n} \cdot \hat{n}=\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1\)

In case of orthogonal unit vectors \(\hat{\mathrm{i}}, \hat{\mathrm{j}} \text { and } \hat{\mathrm{k}}\); \(\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}=0\)

⇒ \(\vec{A} \cdot=\left(\hat{i} A_x+\hat{j} A_y+\hat{k} A_z\right) \cdot\left(\hat{i} B_x+\hat{j} B_y+\hat{k} B_z\right)\)

⇒ \(\left[A_x B_x+A_y B_y+A_z B_z\right]\)

Question 1. If the Vectors \(\vec{P}=a \hat{i}+a \hat{j}+3 \hat{k} \text { and } \vec{Q}=a \hat{i}-2 \hat{j}-\hat{k}\) are perpendicular to each other. Find the value of a?

Answer:

If vectors \(\vec{P} \text { and } \vec{Q}\) are perpendicular

⇒ \(\vec{P} \cdot \vec{Q}=0\)

⇒ \((a \hat{i}+a \hat{j}+3 \hat{k}) \cdot(a \hat{i}-2 \hat{j}-\hat{k})=0\)

⇒ \(a^2-2 a-3=0\)

⇒ \(a^2-3 a+a-3=0\)

⇒  a(a-3)+1(a-3) = 0

⇒ (a + 1)(a – 3) = 0

⇒ a + 1 = 0  or  a – 3 = 0

⇒ a = -1 or a = 3

⇒ a = -1,3

Question 2. Find the component of \(3 \hat{i}+4 \hat{j} \text { along } \hat{i}+\hat{j}\)?

Answer:

Component of \(\overrightarrow{\mathrm{A}}\)along \(\overrightarrow{\mathrm{B}}\) is given by \(\frac{\vec{A} \cdot \vec{B}}{B}\) hence required component

⇒ \(\frac{(3 \hat{i}+4 \hat{j}) \cdot(\hat{i}+\hat{j})}{\sqrt{2}}\)

⇒ \(\frac{7}{\sqrt{2}}\)

Question 3. Find angle between \(\vec{A}=3 \hat{i}+4 \hat{j} \text { and } \vec{B}=12 \hat{i}+5 \hat{j}\)

Answer:

We have cos θ = \(\frac{\vec{A} \cdot \vec{B}}{A B}\)

⇒ \(\frac{(3 \hat{i}+4 \hat{j}) \cdot(12 \hat{i}+5 \hat{j})}{\sqrt{3^2+4^2} \sqrt{12^2+5^2}}\)

⇒ \(\cos \theta=\frac{36+20}{5 \times 13}=\frac{56}{65}\)

⇒ \(\theta=\cos ^{-1} \frac{56}{65}\)

2. Vector Product

The vector product or cross product of any two vectors \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) A and B, denoted as \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}(\text { read } \overrightarrow{\mathrm{A}} \text { cross } \overrightarrow{\mathrm{B}})\) is defined as:

⇒ \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=\mathrm{AB} \sin \theta \hat{n}\)

Here θ is the angle between the vectors and the direction \(\hat{n}\) is given by the right-hand-thumb rule.

Right-Hand-Thumb Rule:

To find the direction of \(\hat{n}\) draw the two vectors \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) with both the tails coinciding.

Now place your stretched right palm perpendicular to the plane of \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) in such a way that the fingers are along the vector \(\overrightarrow{\mathrm{A}}\) and when the fingers are closed they go towards \(\overrightarrow{\mathrm{B}}\). The direction of the thumb gives the direction of \(\hat{n}\)

Vector Product Properties

The vector product of two vectors is always a vector perpendicular to the plane containing the two vectors i.e. orthogonal to both vectors.

A and B, though the vectors \(\vec{A} \text { and } \vec{B}\) though the vector \(\vec{A} \text { and } \vec{B}\) may or may not be orthogonal.

Vector product of two vectors is not commutative i.e. \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}} \neq \overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}\)

But \(|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|=|\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}|\) = AB sin θ

The vector product is distributive when the order of the vectors is strictly maintained i.e.

⇒ \(\mathrm{A} \times(\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}})=\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{C}}\)

The magnitude of the vector product of two vectors will be maximum when sinθ = max = 1, i.e, θ = 90º

⇒ \(|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|_{\max }=\mathrm{AB}\) i.e., the magnitude of the vector product is maximum if the vectors are orthogonal.

The magnitude of the vector product of two non–zero vectors will be minimum when |sinθ| = minimum = 0,i.e., θ = 0º or 180º, and vectors are collinear.
min | A B | 0 × =i.e., if the vector product of two non–zero vectors vanishes, the vectors are collinear.

Note: When θ = 0º then vectors may be called like vectors or parallel vectors and when θ = 180º then vectors may be called unlike vectors or antiparallel vectors.

The self-cross product i.e. product of a vector by itself vanishes i.e. is a null vector.

Note: Null vector or zero vector: A vector of zero magnitude is called a zero vector. The direction of a zero vector is determinate (unspecified).

⇒ \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{A}}=\mathrm{AA} \sin 0^{\circ} \hat{\mathrm{n}}=\overrightarrow{0}\)

Note: Geometrical meaning of vector product of two vectors

1. Consider two vectors \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) which are represented by \(\overrightarrow{\mathrm{OP}} \text { and } \overrightarrow{\mathrm{QP}} \text { and } \angle \mathrm{POQ}=\theta\)
2. Complete the parallelogram OPRQ. Join P with Q. Here OP = A and OQ = B. Draw QN ⊥OP
3. Magnitude of the cross product of \(\vec{A} \text { and } \vec{B}\)

⇒ \(|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|=\mathrm{AB} \sin \theta\)

= (OP)(OQ sin θ)

= (OP)(NQ) (∵ NQ = OQ sin θ)

= base × height

= Area of parallelogram OPRQ

Area of ΔPOQ = ΔPOQ = \(\frac{\text { base } \times \text { height }}{2}=\frac{(\mathrm{OP})(\mathrm{NQ})}{2}\)

⇒ \(\frac{1}{2}|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|\)

∴ Area of parallelogram OPRQ = 2 [area of Δ OPQ] = \(|\vec{A} \times \vec{B}|\)

Formulae to find Area

If \(\vec{A} \text { and } \vec{B}\) are two adjacent sides of a triangle, then its area = \(\frac{1}{2}|\vec{A} \times \vec{B}|\)

If \(\vec{A} \text { and } \vec{B}\) are two adjacent sides of a parallelogram, then its area = \(|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|\)

If \(\vec{A} \text { and } \vec{B}\) are diagonals of a parallelogram then its area = \(\frac{1}{2}|\vec{A} \times \vec{B}|\)

In case of unit vector \(\hat{n}, \quad \hat{n} \times \hat{n}=\overrightarrow{0} \Rightarrow \hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\overrightarrow{0}\)

In case of orthogonal unit vectors \(\hat{i}, \hat{j} \text { and } \hat{k}\) in accordance with right-hand-thumb-rule,

In terms of compound, \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)

⇒ \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
A_x & A_y & A_z \\
B_x & B_y & B_z
\end{array}\right|=\hat{i}\left|\begin{array}{cc}
A_y & A_z \\
B_y & B_z
\end{array}\right|-\hat{j}\left|\begin{array}{cc}
A_x & A_z \\
B_x & B_z
\end{array}\right|+\hat{k}\left|\begin{array}{cc}
A_x & A_y \\
B_x & B_y
\end{array}\right|\)

The magnitude of the area of the parallelogram formed by the adjacent sides of vectors \(\vec{A} \text { and } \vec{B}\) equal to

Question 4. \(\overrightarrow{\mathrm{A}}\) is Eastwards and \(\overrightarrow{\mathrm{B}}\) is downwards. Find the direction of \(\overrightarrow{\mathrm{A}} \times \vec{B}\)?

Answer:

Applying the right-hand thumb rule we find that \(\overrightarrow{\mathrm{A}} \times \vec{B}\) is along North.

Question 5. If \(\vec{A} \cdot \vec{B}=|\vec{A} \times \vec{B}|\), find angle between \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\)

Answer:

⇒ \(\vec{A} \cdot \vec{B}=|\vec{A} \times \vec{B}|\)

AB cos θ = AB sin θ

tan θ = 1

⇒ θ = 45º

Question 6. Two vectors \(\vec{A} \text { and } \vec{B}\) are inclined to each other at an angle θ. Find a unit vector that is perpendicular to both \(\vec{A} \text { and } \vec{B}\)

Answer:

⇒ \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=\mathrm{AB} \sin \theta \hat{n}\)

⇒ \(\hat{n}=\frac{\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}}{\mathrm{AB} \sin \theta}\) here \(\hat{n}\) is perpendicular to both \(\vec{A} \text { and } \vec{B}\)

Question 7. Find \(\vec{A} \times \vec{B}\) if \(\vec{A}=\hat{i}-2 \hat{j}+4 \hat{k}\) and \(\vec{B}=2 \hat{i}-\hat{j}+2 \hat{k}\)

Answer:

⇒ \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)

⇒ \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 4 \\
3 & -1 & 2
\end{array}\right|\)

⇒ \(\hat{i}(-4-\hat{j}(-4))-(2-12)+\hat{k}(-1-(-6))=10 \hat{j}+5 \hat{k}\)

Question 8. Find the value of

1. sin (− θ)
2. cos (− θ)
3. tan (− θ)
4. cos (\(\frac{\pi}{2}\)− θ)
5. sin (\(\frac{\pi}{2}\)+ θ)
6. cos (\(\frac{\pi}{2}\)+ θ) 2
7. sin (π − θ)
8. cos (π − θ)
9. sin (\(\frac{3 \pi}{2}\)− θ)
10. cos (\(\frac{3 \pi}{2}\)− θ)
11. sin (\(\frac{3 \pi}{2}\)+ θ)
12. cos (\(\frac{3 \pi}{2}\)+ θ)
13. tan (\(\frac{\pi}{2}\)− θ)
14. cot (\(\frac{\pi}{2}\)− θ) 2

Answers :

1. – sin θ
2. cos θ
3. – tan θ
4. sin θ
5. cos θ
6. – sin θ
7. sin θ
8. – cos θ
9. – cos θ
10. – sin θ
11. – cos θ
12. sin θ
13. cot θ
14. tan θ

Question 9.

1. For what value of m the vector \(\vec{A}=2 \hat{i}+3 \hat{j}-6 \hat{k}\) is perpendicular to \(\vec{B}=3 \hat{i}-m \hat{j}+6 \hat{k}\)
2. Find the components of vector \(\vec{A}=2 \hat{i}+3 \hat{j}\) along the direction of \(\hat{i}+\hat{j}\)?

Answers :

1. m = –10
2. \(\frac{5}{\sqrt{2}}\)

Question 10.

1. \(\overrightarrow{\mathrm{A}}\) is North–East and \(\overrightarrow{\mathrm{B}}\) is downwards, find the direction of \(\vec{A} \times \vec{B}\)
2. Find \(\vec{B} \times \vec{A} \text { if } \vec{A}=3 \hat{i}-2 \hat{j}+6 \hat{k}\) and \(\vec{B}=\hat{i}-\hat{j}+\hat{k}\)

Answers :

1. North-West.
2. \(-4 \hat{i}-3 \hat{j}+\hat{k}\)