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Snowcat [4.5K]
3 years ago
10

Find the three angles of the triangle formed using the position vectors 2i hat − j + 5k and i hat + 2j + 4k and the line segment

connecting their endpoints. Give your answers in degrees to two decimal places.
Mathematics
1 answer:
victus00 [196]3 years ago
8 0

Given

\vec{a}=2\vec{i}-\vec{j}+5\vec{k},

\vec{b}=\vec{i}+2\vec{j}+4\vec{k},

you can find

\vec{a}-\vec{b}=2\vec{i}-\vec{j}+5\vec{k}-(\vec{i}+2\vec{j}+4\vec{k})=\vec{i}-3\vec{j}+\vec{k}.

Three vectors \vec{a}, \vec{b}, \vec{a}-\vec{b} form a triangle.

1.

\cos\angle 1=\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|}=\dfrac{2\cdot 1+(-1)\cdot 2+5\cdot 4}{\sqrt{2^2+(-1)^2+5^2}\cdot \sqrt{1^2+2^2+4^2} }=\dfrac{20}{\sqrt{30} \cdot \sqrt{21} }.

2.

\cos\angle 2=\dfrac{\vec{a}\cdot (\vec{a}-\vec{b})}{|\vec{a}|\cdot |\vec{a}-\vec{b}|}=\dfrac{2\cdot 1+(-1)\cdot (-3)+5\cdot 1}{\sqrt{2^2+(-1)^2+5^2}\cdot \sqrt{1^2+(-3)^2+1^2} }=\dfrac{10}{\sqrt{30} \cdot \sqrt{11} }.

3.

\cos\angle 3=\dfrac{\vec{b}\cdot (\vec{a}-\vec{b})}{|\vec{b}|\cdot |\vec{a}-\vec{b}|}=\dfrac{1\cdot 1+2\cdot (-3)+4\cdot 1}{\sqrt{1^2+2^2+4^2}\cdot \sqrt{1^2+(-3)^2+1^2} }=\dfrac{-1}{\sqrt{21} \cdot \sqrt{11} }.

Then

  • \angle 1=\arccos \left(\dfrac{20}{\sqrt{30} \cdot \sqrt{21} }\right)\approx 37.17^{\circ};
  • \angle 2=\arccos \left(\dfrac{10}{\sqrt{30} \cdot \sqrt{11} }\right)\approx 56.60^{\circ};
  • \angle 3=\arccos \left(\dfrac{-1}{\sqrt{21} \cdot \sqrt{11} }\right)\approx 93.77^{\circ}.
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