Question #81420

Check whether the vector (2root3, 2 ) is equally inclined to the vectors (2, 2root3) and (4,0)
1

Expert's answer

2018-09-27T10:43:08-0400

Answer on Question #81420 – Math – Linear Algebra

Question

Check whether the vector (23,2)(2\sqrt{3}, 2) is equally inclined to the vectors (2,23)(2, 2\sqrt{3}) and (4,0)(4, 0).

Solution

Let a=(23,2),b=(2,23)\vec{a} = (2\sqrt{3}, 2), \vec{b} = (2, 2\sqrt{3}) and c=(4,0)\vec{c} = (4, 0).

Find dot product


ab=23(2)+2(23)=83\vec{a} \cdot \vec{b} = 2\sqrt{3}(2) + 2(2\sqrt{3}) = 8\sqrt{3}ac=23(4)+2(0)=83\vec{a} \cdot \vec{c} = 2\sqrt{3}(4) + 2(0) = 8\sqrt{3}a=(23)2+(2)2=4|\vec{a}| = \sqrt{(2\sqrt{3})^2 + (2)^2} = 4b=(2)2+(23)2=4|\vec{b}| = \sqrt{(2)^2 + (2\sqrt{3})^2} = 4c=(4)2+(0)2=4|\vec{c}| = \sqrt{(4)^2 + (0)^2} = 4cosβ=abab=8344=32\cos \beta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{8\sqrt{3}}{4 \cdot 4} = \frac{\sqrt{3}}{2}cosγ=acac=8344=32\cos \gamma = \frac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|} = \frac{8\sqrt{3}}{4 \cdot 4} = \frac{\sqrt{3}}{2}


Therefore, the vector (23,2)(2\sqrt{3}, 2) is equally inclined to the vectors (2,23)(2, 2\sqrt{3}) and (4,0)(4, 0).

Answer: the vector (23,2)(2\sqrt{3}, 2) is equally inclined to the vectors (2,23)(2, 2\sqrt{3}) and (4,0)(4, 0).

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