Solveeit Logo
Login

Question

Question: Let the angle $\theta$; $0 < \theta < \frac{\pi}{2}$ between two unit vectors $\overline{a}$ & $\ove...

Let the angle θ\theta; 0<θ<π20 < \theta < \frac{\pi}{2} between two unit vectors a‾\overline{a} & b‾\overline{b} be Sin−1(3/5)Sin^{-1}(3/5). If the vector c‾=3a‾+6b‾+9(a‾×b‾)\overline{c} = 3\overline{a} + 6\overline{b} + 9(\overline{a} \times \overline{b}) then the value of 2(c‾.b‾)−(c‾.a‾)2(\overline{c}.\overline{b}) - (\overline{c}.\overline{a})

A

9

B

18

C

27

D

36

Answer

9

Explanation

Solution

The problem requires calculating a linear combination of dot products involving vector c‾\overline{c}. Given unit vectors a‾\overline{a} and b‾\overline{b} with an angle θ\theta such that 0<θ<π20 < \theta < \frac{\pi}{2} and sin⁡θ=35\sin \theta = \frac{3}{5}. From sin⁡θ=35\sin \theta = \frac{3}{5} and θ\theta being in the first quadrant, we find cos⁡θ=1−sin⁡2θ=1−(35)2=45\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}. Since a‾\overline{a} and b‾\overline{b} are unit vectors, ∣a‾∣=1|\overline{a}| = 1 and ∣b‾∣=1|\overline{b}| = 1. The dot product a‾⋅b‾=∣a‾∣∣b‾∣cos⁡θ=1⋅1⋅45=45\overline{a} \cdot \overline{b} = |\overline{a}| |\overline{b}| \cos \theta = 1 \cdot 1 \cdot \frac{4}{5} = \frac{4}{5}. Also, a‾⋅a‾=∣a‾∣2=1\overline{a} \cdot \overline{a} = |\overline{a}|^2 = 1 and b‾⋅b‾=∣b‾∣2=1\overline{b} \cdot \overline{b} = |\overline{b}|^2 = 1. The vector c‾\overline{c} is given by c‾=3a‾+6b‾+9(a‾×b‾)\overline{c} = 3\overline{a} + 6\overline{b} + 9(\overline{a} \times \overline{b}). We need to find 2(c‾⋅b‾)−(c‾⋅a‾)2(\overline{c} \cdot \overline{b}) - (\overline{c} \cdot \overline{a}).

First, calculate c‾⋅b‾\overline{c} \cdot \overline{b}: Using the distributive property of the dot product: c‾⋅b‾=(3a‾+6b‾+9(a‾×b‾))⋅b‾\overline{c} \cdot \overline{b} = (3\overline{a} + 6\overline{b} + 9(\overline{a} \times \overline{b})) \cdot \overline{b} c‾⋅b‾=3(a‾⋅b‾)+6(b‾⋅b‾)+9((a‾×b‾)⋅b‾)\overline{c} \cdot \overline{b} = 3(\overline{a} \cdot \overline{b}) + 6(\overline{b} \cdot \overline{b}) + 9((\overline{a} \times \overline{b}) \cdot \overline{b}) Since the cross product (a‾×b‾)(\overline{a} \times \overline{b}) is orthogonal to both a‾\overline{a} and b‾\overline{b}, (a‾×b‾)⋅b‾=0(\overline{a} \times \overline{b}) \cdot \overline{b} = 0. Substituting known values: c‾⋅b‾=3(45)+6(1)+9(0)=125+6=12+305=425\overline{c} \cdot \overline{b} = 3(\frac{4}{5}) + 6(1) + 9(0) = \frac{12}{5} + 6 = \frac{12 + 30}{5} = \frac{42}{5}.

Next, calculate c‾⋅a‾\overline{c} \cdot \overline{a}: Similarly, using the distributive property: c‾⋅a‾=(3a‾+6b‾+9(a‾×b‾))⋅a‾\overline{c} \cdot \overline{a} = (3\overline{a} + 6\overline{b} + 9(\overline{a} \times \overline{b})) \cdot \overline{a} c‾⋅a‾=3(a‾⋅a‾)+6(b‾⋅a‾)+9((a‾×b‾)⋅a‾)\overline{c} \cdot \overline{a} = 3(\overline{a} \cdot \overline{a}) + 6(\overline{b} \cdot \overline{a}) + 9((\overline{a} \times \overline{b}) \cdot \overline{a}) Since (a‾×b‾)(\overline{a} \times \overline{b}) is orthogonal to a‾\overline{a}, (a‾×b‾)⋅a‾=0(\overline{a} \times \overline{b}) \cdot \overline{a} = 0. Also, b‾⋅a‾=a‾⋅b‾\overline{b} \cdot \overline{a} = \overline{a} \cdot \overline{b}. Substituting known values: c‾⋅a‾=3(1)+6(45)+9(0)=3+245=15+245=395\overline{c} \cdot \overline{a} = 3(1) + 6(\frac{4}{5}) + 9(0) = 3 + \frac{24}{5} = \frac{15 + 24}{5} = \frac{39}{5}.

Finally, calculate the required expression: 2(c‾⋅b‾)−(c‾⋅a‾)=2(425)−(395)2(\overline{c} \cdot \overline{b}) - (\overline{c} \cdot \overline{a}) = 2(\frac{42}{5}) - (\frac{39}{5}) =845−395=84−395=455=9= \frac{84}{5} - \frac{39}{5} = \frac{84 - 39}{5} = \frac{45}{5} = 9.