Question

Two vectors a and b are at an angle of 23 if a= 3 i cap 4 j cap +12 k cap and b is recorded as 2 i cap + j cap ( )k cap where coefficient of k is missing, calculate the coefficient given that cos 23=12/13

Answer 1

Given:

Two vectors a and b are at an angle of 23° ; a= 3 i cap 4 j cap +12 k cap and b is recorded as 2 i cap + j cap ( )k cap where coefficient of k is missing.

To find:

Coefficient of k cap ?

Calculation:

Let the coefficient required be x:

$\vec{a} = 3 \hat{i} + 4 \hat{j} + 12 \hat{k}$

$\vec{b} = 2 \hat{i} + \hat{j} + x \hat{k}$

$\therefore \: \vec{a}. \vec{b} = | a| \times |b| \times \cos( \theta)$

$\implies \: (2 \times 3) + (4 \times 1) + (12 \times x) = ( \sqrt{ {3}^{2} + {4}^{2} + {12}^{2} } ) \times (\sqrt{ {2}^{2} + {1}^{2} + {x}^{2} } ) \times \cos( {23}^{ \circ} )$

$\implies \: 10 + 12x = ( \sqrt{169 } ) \times (\sqrt{5 + {x}^{2} }) \times \dfrac{12}{13}$

$\implies \: 10 + 12x = 13 \times (\sqrt{5 + {x}^{2} }) \times \dfrac{12}{13}$

$\implies \: 10 + 12x = (\sqrt{5 + {x}^{2} }) \times 12$

$\implies \: \dfrac{10}{12} + x = \sqrt{5 + {x}^{2} }$

$\implies \: \dfrac{5}{6} + x = \sqrt{5 + {x}^{2} }$

Squaring Both Sides:

$\implies \: \dfrac{25}{36} + {x}^{2} + \dfrac{5x}{3} = 5 + {x}^{2}$

$\implies \: \dfrac{25}{36} + \dfrac{5x}{3} = 5$

$\implies \: \dfrac{5x}{3} = 5 - \dfrac{25}{36}$

$\implies \: \dfrac{5x}{3} = \dfrac{180 - 25}{36}$

$\implies \: \dfrac{5x}{3} = \dfrac{155}{36}$

$\implies \: x= \dfrac{31}{12}$

So, value of x = 31/12.

Answer 2

Explanation:

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