Answer on Question #65901 – Math – Geometry
Question
For what value of x x x will the angle between the lines with direction ratios ( x , 2 , 4 ) (x, 2, 4) ( x , 2 , 4 ) and ( 1 , 0 , 1 ) (1, 0, 1) ( 1 , 0 , 1 ) be 45 ∘ 45{}^\circ 45 ∘ ?
Solution
Denote a = ( a 1 , a 2 , a 3 ) = ( x , 2 , 4 ) a = (a_1, a_2, a_3) = (x, 2, 4) a = ( a 1 , a 2 , a 3 ) = ( x , 2 , 4 ) , b = ( b 1 , b 2 , b 3 ) = ( 1 , 0 , 1 ) b = (b_1, b_2, b_3) = (1, 0, 1) b = ( b 1 , b 2 , b 3 ) = ( 1 , 0 , 1 ) .
According to the algebraic definition of the dot product (see [1], p.4)
a b = a 1 b 1 + a 2 b 1 + a 3 b 3 . a b = a_1 b_1 + a_2 b_1 + a_3 b_3. ab = a 1 b 1 + a 2 b 1 + a 3 b 3 .
The angle between vectors a , b a, b a , b is defined by (see [1], p.6)
cos 45 ∘ = a ⋅ b ∣ ∣ a ∣ ∣ ⋅ ∣ ∣ b ∣ ∣ . \cos 45{}^\circ = \frac{a \cdot b}{||a|| \cdot ||b||}. cos 45 ∘ = ∣∣ a ∣∣ ⋅ ∣∣ b ∣∣ a ⋅ b .
Using that cos 45 ∘ = 2 / 2 \cos 45{}^\circ = \sqrt{2}/2 cos 45 ∘ = 2 /2 and the previous formulas, we get
a 1 b 1 + a 2 b 1 + a 3 b 3 = ∣ ∣ a ∣ ∣ ⋅ ∣ ∣ b ∣ ∣ ⋅ cos 45 ∘ , a_1 b_1 + a_2 b_1 + a_3 b_3 = ||a|| \cdot ||b|| \cdot \cos 45{}^\circ, a 1 b 1 + a 2 b 1 + a 3 b 3 = ∣∣ a ∣∣ ⋅ ∣∣ b ∣∣ ⋅ cos 45 ∘ , x ⋅ 1 + 2 ⋅ 0 + 4 ⋅ 1 = x 2 + 4 + 16 ⋅ 1 + 1 ⋅ 2 2 , x \cdot 1 + 2 \cdot 0 + 4 \cdot 1 = \sqrt{x^2 + 4 + 16} \cdot \sqrt{1 + 1} \cdot \frac{\sqrt{2}}{2}, x ⋅ 1 + 2 ⋅ 0 + 4 ⋅ 1 = x 2 + 4 + 16 ⋅ 1 + 1 ⋅ 2 2 , x + 4 = x 2 + 20 , x + 4 = \sqrt{x^2 + 20}, x + 4 = x 2 + 20 , ( x + 4 ) 2 = ( x 2 + 20 ) 2 , (x + 4)^2 = \left(\sqrt{x^2 + 20}\right)^2, ( x + 4 ) 2 = ( x 2 + 20 ) 2 , x 2 + 8 x + 16 = x 2 + 20 , x^2 + 8x + 16 = x^2 + 20, x 2 + 8 x + 16 = x 2 + 20 , 8 x = 20 − 16 , 8x = 20 - 16, 8 x = 20 − 16 , 8 x = 4 , 8x = 4, 8 x = 4 , x = 4 8 , x = \frac{4}{8}, x = 8 4 , x = 1 2 . x = \frac{1}{2}. x = 2 1 .
Answer: x = 1 2 x = \frac{1}{2} x = 2 1 .
References:
[1] S. Lipschutz; M. Lipson (2009). Linear Algebra (Schaum’s Outlines) (4th ed.). McGraw Hill. ISBN 978-0-07-154352-1.
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