Question #266369

For the vectors ๐‘Ÿโƒ—1 = 2๐‘–ฬ‚โˆ’ 3 ๐‘—ฬ‚+ 4๐‘˜ฬ‚ ๐‘Ž๐‘›๐‘‘ ๐‘Ÿ2ฬ‚ = 4๐‘–ฬ‚+ 2๐‘˜ฬ‚


(a) Find the angle between them.


(b) Find the cross-product of the vectors.

1
Expert's answer
2021-11-15T17:54:44-0500

(a) Find the dot product and magnitudes:

rโƒ—1โ‹…rโƒ—2=2โ‹…4+(โˆ’3)โ‹…2=2.โˆฃrโƒ—1โˆฃ=22+(โˆ’3)2=3.6โˆฃrโƒ—2โˆฃ=42+22=4.5.\vec r_1ยท\vec r_2=2ยท4+(-3)ยท2=2.\\ |\vec r_1|=\sqrt{2^2+(-3)^2}=3.6\\ |\vec r_2|=\sqrt{4^2+2^2}=4.5.

Find the angle between them:


cosโกฮธ=rโƒ—1โ‹…rโƒ—2โˆฃrโƒ—1โˆฃโ‹…โˆฃrโƒ—2โˆฃ=23.6โ‹…4.5=0.12, ฮธ=83.1ยฐ\cos\theta=\frac{\vec r_1ยท\vec r_2}{|\vec r_1|ยท|\vec r_2|}=\frac{2}{3.6ยท4.5}=0.12,\\\space\\ \theta=83.1ยฐ

(b) Find the cross-product of the vectors:

rโƒ—1ร—rโƒ—2=โˆฃrโƒ—1โˆฃโ‹…โˆฃrโƒ—2โˆฃsinโกฮธ=1.95\vec r_1\times\vec r_2=|\vec r_1|ยท|\vec r_2|\sin\theta=1.95


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