if u=(2,-3,4) & v=(-1,2,0)& w=(5,-1,2) find the angle b/n (u-2v) & (u+2w)
(u−2v)=(2,−3,4)−2⋅(−1,2,0)=(4,−7,4)(u-2v) = (2,-3,4) - 2\cdot(-1,2,0) = (4,-7,4)(u−2v)=(2,−3,4)−2⋅(−1,2,0)=(4,−7,4)
(u+2w)=(2,−3,4)+2⋅(5,−1,2)=(12,−5,8)(u+2w) = (2,-3,4) + 2\cdot(5,-1,2) = (12, -5, 8)(u+2w)=(2,−3,4)+2⋅(5,−1,2)=(12,−5,8)
The angle α\alphaα between two vectors a⃗\vec{a}a and b⃗\vec{b}b can be obtained as cosα=a⃗⋅b⃗∣a⃗∣∣b⃗∣.\cos\alpha = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}.cosα=∣a∣∣b∣a⋅b.
cosα=4⋅12+(−7)⋅(−5)+4⋅842+(−7)2+42122+(−5)2+82,cosα=1159⋅233,α=33.2∘.\cos\alpha = \dfrac{4\cdot12+(-7)\cdot(-5)+4\cdot8}{\sqrt{4^2+(-7)^2+4^2}\sqrt{12^2+(-5)^2+8^2}},\\ \cos\alpha = \dfrac{115}{9\cdot\sqrt{233}}, \\ \alpha = 33.2^\circ.cosα=42+(−7)2+42122+(−5)2+824⋅12+(−7)⋅(−5)+4⋅8,cosα=9⋅233115,α=33.2∘.
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