Application of dot products | Part-1

Vector Inequality and Angle Between Vectors

Question i)

Using vector prove the inequality that:
\( (a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3})^{2} \leq (a_{1}^{2} + a_{2}^{2} + a_{3}^{2})(b_{1}^{2} + b_{2}^{2} + b_{3}^{2}) \)

Solution:

Let \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \)

Then, \( \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \)

By Cauchy-Schwarz inequality:
\( |\vec{a} \cdot \vec{b}| \leq |\vec{a}| \cdot |\vec{b}| \)

Squaring both sides, we get:
\( (\vec{a} \cdot \vec{b})^2 \leq (|\vec{a}|^2)(|\vec{b}|^2) \)

Hence, \( (a_1 b_1 + a_2 b_2 + a_3 b_3)^2 \leq (a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2) \)

Question ii)

If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a}| = 4 \), \( |\vec{b}| = 3 \), and \( \vec{a} \cdot \vec{b} = 6 \), find the angle between \( \vec{a} \) and \( \vec{b} \).

Solution:

We know:
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \)

Substituting the values:
\( 6 = 4 \times 3 \times \cos \theta \Rightarrow \cos \theta = \frac{6}{12} = \frac{1}{2} \)

\( \theta = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ \)

Question iii)

If \( \vec{a} + \vec{b} + \vec{c} = 0 \), such that \( |\vec{a}| = 3 \), \( |\vec{b}| = 5 \), and \( |\vec{c}| = 7 \), then find the angle between \( \vec{a} \) and \( \vec{b} \).

Solution:

Given: \( \vec{a} + \vec{b} + \vec{c} = 0 \Rightarrow \vec{a} + \vec{b} = -\vec{c} \)

Taking modulus on both sides:
\( |\vec{a} + \vec{b}| = |\vec{c}| = 7 \)

Now, \( |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos \theta \)

\( 49 = 9 + 25 + 2 \times 3 \times 5 \cos \theta \Rightarrow 49 = 34 + 30 \cos \theta \)

\( 15 = 30 \cos \theta \Rightarrow \cos \theta = \frac{1}{2} \Rightarrow \theta = 60^\circ \)