Vector Cross Product Let vectors A=(1,0,-3), B =(-2,5,1), and C =(3,1,1). Calculate the following, expressing your answers as ordered triples (three comma-separated numbers). (C) (2B―)(3C―) (D) (B―)(C―) (E) over→A(over→B×over→C) (F)If v―1 and v―2 are perpendicular, |v―1×v―2| (G) If v―1 and v―2 are parallel, |v―1×v―2|

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Vector Cross Product  Let vectors A=(1,0,-3), B =(-2,5,1), and C =(3,1,1). Calculate the following, expressing your answers as ordered triples (three comma-separated numbers).  (C) (2B―)(3C―)  (D) (B―)(C―)  (E)  over→A(over→B×over→C)  (F)If v―1  and  v―2 are perpendicular, |v―1×v―2|  (G) If v―1  and  v―2 are parallel, |v―1×v―2|

Viktor Wiley · Accepted Answer

Calculation:

alenahelenash · Accepted Answer

(C) To calculate (2B→)(3C→), we first find the cross product of B→ and C→:B→×C→=|i^j^k^−251311|=(−5−1)i^−(6−3)j^−(2+15)k^=−6i^−3j^−17k^.Multiplying this result by 2 and 3 respectively, we have:(2B→)(3C→)=2(−6i^−3j^−17k^)·3(3i^+j^+k^)=(−12i^−6j^−34k^)·(9i^+3j^+3k^).Expanding the dot product, we get:(−12i^−6j^−34k^)·(9i^+3j^+3k^)=(−12)(9)+(−6)(3)+(−34)(3)=(−108,−18,−102).(D) To calculate (B→)(C→), we simply take the dot product of B→ and C→:(B→)(C→)=(−2i^+5j^+1k^)·(3i^+1j^+1k^)=(−2)(3)+(5)(1)+(1)(1)=3.(E) To calculate A→(B→×C→), we substitute the cross product of B→ and C→ into the equation:A→(B→×C→)=(1i^+0j^−3k^)·(−6i^−3j^−17k^)=(1)(−6)+(0)(−3)+(−3)(−17)=51.(F) If v1→ and v2→ are perpendicular, the magnitude of their cross product is given by |v1→×v2→|.

star233 · Accepted Answer

Step 1:(C) (2B―)(3C―):Given vectors B―=(−2,5,1) and C―=(3,1,1), we can calculate the cross product as follows:2B―=2(−2,5,1)=(−4,10,2)3C―=3(3,1,1)=(9,3,3)Now, we&#039;ll calculate the cross product of 2B― and 3C―:(2B―)(3C―)=(−4,10,2)×(9,3,3)Using the cross product formula, we can evaluate this expression:(2B―)(3C―)=(10·3−2·3,2·9−(−4)·3,(−4)·3−10·9)(2B―)(3C―)=(24,30,−102)Therefore, the answer to (C) is (24,30,−102).Step 2:(D) (B―)(C―):Given vectors B―=(−2,5,1) and C―=(3,1,1), we&#039;ll calculate the dot product as follows:(B―)(C―)=(−2,5,1)·(3,1,1)Using the dot product formula, we get:(B―)(C―)=−2·3+5·1+1·1(B―)(C―)=−6+5+1(B―)(C―)=0Therefore, the answer to (D) is 0.Step 3:(E) A→(B→×C→):Given vectors A→=(1,0,−3), B→=(−2,5,1), and C→=(3,1,1), we&#039;ll calculate the cross product as follows:B→×C→=(−2,5,1)×(3,1,1)Using the cross product formula, we can evaluate this expression:B→×C→=(5·1−1·1,1·3−(−2)·1,(−2)·1−5·3)B→×C→=(4,5,−17)Now, we&#039;ll calculate A→(B→×C→):A→(B→×C→)=(1,0,−3)·(4,5,−17)Using the dot product formula, we get:A→(B→×C→)=1·4+0·5+(−3)·(−17)A→(B→×C→)=4+0+51A→(B→×C→)=55Therefore, the answer to (E) is 55.Step 4:(F) If v1→ and v2→ are perpendicular, |v1→×v2→|:If v1→ and v2→ are perpendicular, the magnitude of their cross product can be calculated using the formula:|v1→×v2→|=|v1→||v2→|sin(θ)Since v1→ and v2→ are perpendicular, sin(θ)=1, and the formula simplifies to:|v1→×v2→|=|v1→||v2→|×1|v1→×v2→|=|v1→||v2→|Therefore, the magnitude of the cross product of v1→ and v2→ is equal to the product of their magnitudes.Step 5:(G) If v1→ and v2→ are parallel, |v1→×v2→|:If v1→ and v2→ are parallel, the cross product of two parallel vectors is zero. Therefore:|v1→×v2→|=0Therefore, the magnitude of the cross product of v1→ and v2→ is zero if the vectors are parallel.

karton · Accepted Answer

Solution:(C) To find (2B→)(3C→), we first need to calculate 2B→ and 3C→ separately:2B→=2(−2,5,1)=(−4,10,2)3C→=3(3,1,1)=(9,3,3)Now, we can multiply these vectors:(2B→)(3C→)=(−4,10,2)×(9,3,3)Using the cross product formula, the resulting vector is:(2B→)(3C→)=|i^j^k^−4102933|=(−60,−6,66)Therefore, (2B→)(3C→)=(−60,−6,66)(D) To calculate (B→)(C→), we can simply use the dot product formula:(B→)(C→)=(−2,5,1)·(3,1,1)=(−2)(3)+(5)(1)+(1)(1)=−6+5+1=0Thus, (B→)(C→)=0(E) To find A→(B→×C→), we first need to calculate B→×C→:B→×C→=|i^j^k^−251311|=(4,−5,−13)Now, we can multiply this result with A→:A→(B→×C→)=(1,0,−3)×(4,−5,−13)Using the cross product formula, the resulting vector is:A→(B→×C→)=|i^j^k^10−34−5−13|=(−15,5,−5)Hence, A→(B→×C→)=(−15,5,−5)(F) If v1→ and v2→ are perpendicular, then the magnitude of their cross product can be calculated as:|v1→×v2→|=|v1→|·|v2→|Thus, if v1→ and v2→ are perpendicular, |v1→×v2→|=|v1→|·|v2→|(G) If v1→ and v2→ are parallel, then their cross product will be zero, i.e., v1→×v2→=0→.Thus, if v1→ and v2→ are parallel, |v1→×v2→|=|0→|=0

Vector Cross Product. Let vectors A=(1,0,-3), B =(-2,5,1), and C =(3,1,1).Calculate the following, expressing your answers as ordered triples

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