Application of dot product | Part -2

Vector Problems and Solutions

Problem i: Find the scalar projection of \(\overrightarrow{a}= \hat{i}+2\hat{j}+3\hat{k}\) on the line of support of \(\overrightarrow{b}=2 \hat{i}-3\hat{j}+6\).

Solution:

Scalar projection = \(\dfrac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|}\)

\(\overrightarrow{a} \cdot \overrightarrow{b} = (1)(2) + (2)(-3) + (3)(6) = 2 - 6 + 18 = 14\)

\(|\overrightarrow{b}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\)

\(Scalar\; projection = \dfrac{14}{7} = 2\)

Problem ii: If \(\overrightarrow{a}= \hat{i}+2\hat{j}-3\hat{k}\) and \(\overrightarrow{b}= 3\hat{i}-\hat{j}+2\hat{k}\), show that \(\overrightarrow{a}+\overrightarrow{b}\) is orthogonal to \(\overrightarrow{a}-\overrightarrow{b}\).

Solution:

\(\overrightarrow{a}+\overrightarrow{b} = (1+3)\hat{i} + (2+(-1))\hat{j} + (-3+2)\hat{k} = 4\hat{i} + 1\hat{j} - 1\hat{k}\)

\(\overrightarrow{a}-\overrightarrow{b} = (1-3)\hat{i} + (2-(-1))\hat{j} + (-3-2)\hat{k} = -2\hat{i} + 3\hat{j} - 5\hat{k}\)

Dot product: \((4)(-2) + (1)(3) + (-1)(-5) = -8 + 3 + 5 = 0\)

Since the dot product is 0, the vectors are orthogonal.

Problem iii: Prove by vector method, in a \(\triangle ABC\), \(c^2 = a^2 + b^2 - 2ab\cos C\)

Solution:

Let \(\vec{A} = \vec{0}, \vec{B} = \vec{b}, \vec{C} = \vec{c}\)

Then \(\vec{AB} = \vec{b}, \vec{AC} = \vec{c}\)

Let side opposite angle C be \(\vec{BC} = \vec{c} - \vec{b}\)

\(|\vec{BC}|^2 = (\vec{c} - \vec{b}) \cdot (\vec{c} - \vec{b}) = \vec{c} \cdot \vec{c} + \vec{b} \cdot \vec{b} - 2 \vec{b} \cdot \vec{c}\)

So, \(c^2 = a^2 + b^2 - 2ab \cos C\), where \(a = |\vec{b}|\), \(b = |\vec{c}|\), and \(\cos C = \dfrac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|}\)