Find two unit vectors orthogonal to both (3 , 2, 1) and (- 1, 1, 0 ).

Yulia

Yulia

Answered question

2021-09-16

Find two unit vectors orthogonal to both (3 , 2, 1) and (- 1, 1, 0 ).

Answer & Explanation

Laaibah Pitt

Laaibah Pitt

Skilled2021-09-17Added 98 answers

To find, explanation
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Nick Camelot

Nick Camelot

Skilled2023-05-10Added 164 answers

Step 1:
To find two unit vectors orthogonal to both (3,2,1) and (โˆ’1,1,0), we can use the cross product.
Let's denote the two given vectors as ๐š=(3,2,1) and ๐›=(โˆ’1,1,0).
First, we calculate the cross product ๐œ=๐šร—๐›:
๐œ=(321)ร—(โˆ’110)
Using the cross product formula, we can find the resulting vector:
๐œ=(2ยท0โˆ’1ยท13ยท0โˆ’1ยท03ยท1โˆ’2ยท(โˆ’1))=(โˆ’105)
Now, to obtain two unit vectors orthogonal to both ๐š and ๐›, we can normalize ๐œ by dividing it by its magnitude. The magnitude of ๐œ is given by:
โ€–๐œโ€–=(โˆ’1)2+02+52=26
Therefore, the unit vector u1 orthogonal to both ๐š and ๐› is:
u1=๐œโ€–๐œโ€–=126(โˆ’105)=(โˆ’1260526)
Step 2:
To find the second unit vector, we can take the cross product of ๐œ and either ๐š or ๐›. Let's use ๐š for this example. We calculate the cross product ๐=๐œร—๐š:
๐=(โˆ’105)ร—(321)
Using the cross product formula, we get:
๐=(0ยท1โˆ’5ยท2โˆ’1ยท1โˆ’5ยท3โˆ’1ยท2โˆ’0ยท3)=(โˆ’10โˆ’16โˆ’2)
Now, we normalize ๐ by dividing it by its magnitude:
โ€–๐โ€–=(โˆ’10)2+(โˆ’16)2+(โˆ’2)2=336
To obtain the second unit vector, we normalize ๐ by dividing it by its magnitude:
u2=๐โ€–๐โ€–=1336(โˆ’10โˆ’16โˆ’2)=(โˆ’10336โˆ’16336โˆ’2336)
Therefore, the two unit vectors orthogonal to both (3,2,1) and (โˆ’1,1,0) are:
u1=(โˆ’1260526)andu2=(โˆ’10336โˆ’16336โˆ’2336)
These vectors are orthogonal to both (3,2,1) and (โˆ’1,1,0), and they have unit length.
Mr Solver

Mr Solver

Skilled2023-05-10Added 147 answers

To find two unit vectors orthogonal to both ๐ฏ=(3,2,1) and ๐ฎ=(โˆ’1,1,0), we can use the cross product. The cross product of two vectors ๐š=(a1,a2,a3) and ๐›=(b1,b2,b3) is given by:
๐šร—๐›=(a2b3โˆ’a3b2,a3b1โˆ’a1b3,a1b2โˆ’a2b1)
Applying this formula, we have:
๐ฏร—๐ฎ=(2ยท0โˆ’1ยท1,1ยท(โˆ’1)โˆ’3ยท0,3ยท1โˆ’2ยท(โˆ’1))=(โˆ’1,โˆ’1,5)
Now, to obtain a unit vector, we divide ๐ฏร—๐ฎ by its magnitude, which is given by:
โ€–๐ฏร—๐ฎโ€–=(โˆ’1)2+(โˆ’1)2+52=27=33
Therefore, the unit vector in the direction of ๐ฏร—๐ฎ is:
u1=(โˆ’133,โˆ’133,533)
To find another orthogonal unit vector, we can take the cross product of ๐ฏร—๐ฎ and ๐ฏ:
(๐ฏร—๐ฎ)ร—๐ฏ
Calculating this cross product, we have:
(๐ฏร—๐ฎ)ร—๐ฏ=(โˆ’1ยท2โˆ’5ยท1,5ยท3โˆ’(โˆ’1)ยท1,(โˆ’1)ยท1โˆ’(โˆ’1)ยท3)=(โˆ’7,16,2)
Normalizing this vector, we get:
u2=(โˆ’772+162+22,1672+162+22,272+162+22)
Hence, two unit vectors orthogonal to both ๐ฏ=(3,2,1) and ๐ฎ=(โˆ’1,1,0) are u1=(โˆ’133,โˆ’133,533) and u2=(โˆ’772+162+22,1672+162+22,272+162+22).
To summarize, the two unit vectors orthogonal to both ๐ฏ=(3,2,1) and ๐ฎ=(โˆ’1,1,0) are:
u1=(โˆ’133,โˆ’133,533) and u2=(โˆ’772+162+22,1672+162+22,272+162+22).
These vectors are orthogonal to both ๐ฏ and ๐ฎ and have a magnitude of 1, making them unit vectors.
Jazz Frenia

Jazz Frenia

Skilled2023-05-10Added 106 answers

To identify two unit vectors that are orthogonal to both ๐ฏ1=(3,2,1) and ๐ฏ2=(โˆ’1,1,0), we can use the cross product.
The cross product of two vectors, ๐ฏ1 and ๐ฏ2, is orthogonal to both vectors. It can be calculated as follows:
๐ฏ1ร—๐ฏ2=|๐ข๐ฃ๐ค321โˆ’110|
Expanding this determinant, we get:
๐ฏ1ร—๐ฏ2=(2ยท0โˆ’1ยท1)๐ขโˆ’(3ยท0โˆ’1ยท(โˆ’1))๐ฃ+(3ยท1โˆ’2ยท(โˆ’1))๐ค
Simplifying further:
๐ฏ1ร—๐ฏ2=โˆ’๐ข+๐ฃ+5๐ค
Now, to find two unit vectors orthogonal to both ๐ฏ1 and ๐ฏ2, we can normalize the cross product vector. The magnitude of the cross product vector is (โˆ’1)2+12+52=27.
Hence, the unit vector in the direction of ๐ฏ1ร—๐ฏ2 is:
๐ฎ1=๐ฏ1ร—๐ฏ2โ€–๐ฏ1ร—๐ฏ2โ€–=โˆ’๐ข+๐ฃ+5๐ค27=(โˆ’127,127,527)
To find another orthogonal unit vector, we can take the cross product of ๐ฏ1ร—๐ฏ2 and ๐ฏ1.
๐ฎ2=๐ฏ1ร—(๐ฏ1ร—๐ฏ2)โ€–๐ฏ1ร—(๐ฏ1ร—๐ฏ2)โ€–
Calculating the cross product:
๐ฏ1ร—(๐ฏ1ร—๐ฏ2)=|๐ข๐ฃ๐ค321โˆ’110|
Expanding the determinant:
๐ฏ1ร—(๐ฏ1ร—๐ฏ2)=(2ยท0โˆ’1ยท1)๐ขโˆ’(3ยท0โˆ’1ยท(โˆ’1))๐ฃ+(3ยท1โˆ’2ยท(โˆ’1))๐ค
Simplifying further:
๐ฏ1ร—(๐ฏ1ร—๐ฏ2)=โˆ’๐ขโˆ’๐ฃ+5๐ค
Now, let's calculate the magnitude of ๐ฏ1ร—(๐ฏ1ร—๐ฏ2). Its magnitude is (โˆ’1)2+(โˆ’1)2+52=27.
Therefore, the unit vector in the direction of ๐ฏ1ร—(๐ฏ1ร—๐ฏ2) is:
๐ฎ2=๐ฏ1ร—(๐ฏ1ร—๐ฏ2)โ€–๐ฏ1ร—(๐ฏ1ร—๐ฏ2)โ€–=โˆ’๐ขโˆ’๐ฃ+5๐ค27=(โˆ’127,โˆ’127,527)
Hence, the two unit vectors orthogonal to both ๐ฏ1 and ๐ฏ2 are ๐ฎ1=(โˆ’127,127,527) and ๐ฎ2=(โˆ’127,โˆ’127,527), respectively.
xleb123

xleb123

Skilled2023-06-12Added 181 answers

To find two unit vectors orthogonal to both ๐ฏ=(3,2,1) and ๐ฐ=(โˆ’1,1,0), we can use the cross product of ๐ฏ and ๐ฐ. The cross product gives us a vector that is orthogonal to both ๐ฏ and ๐ฐ. Let's calculate it:
๐ฎ=๐ฏร—๐ฐ
The cross product of two vectors in three-dimensional space is given by:
๐ฎ=(v2w3โˆ’v3w2,v3w1โˆ’v1w3,v1w2โˆ’v2w1)
Substituting the values of ๐ฏ and ๐ฐ, we have:
๐ฎ=(2ยท0โˆ’1ยท1,1ยท(โˆ’1)โˆ’3ยท0,3ยท1โˆ’2ยท(โˆ’1))
Simplifying the expression:
๐ฎ=(โˆ’1,โˆ’1,5)
To obtain a unit vector, we divide ๐ฎ by its magnitude:
โ€–๐ฎโ€–=(โˆ’1)2+(โˆ’1)2+52=1+1+25=27=33
Dividing each component of ๐ฎ by 33, we get:
u1=(โˆ’133,โˆ’133,533)
Now, to find another unit vector orthogonal to both ๐ฏ and ๐ฐ, we can take the cross product of ๐ฎ and either ๐ฏ or ๐ฐ. Let's use ๐ฏ for this example:
u2=๐ฎร—๐ฏ
Using the same formula for the cross product as before, we substitute the values:
u2=(โˆ’1ยท1โˆ’(โˆ’1)ยท2,(โˆ’1)ยท3โˆ’(โˆ’1)ยท1,(โˆ’1)ยท2โˆ’(โˆ’1)ยท3)
Simplifying the expression:
u2=(โˆ’1+2,โˆ’3+1,โˆ’2+3)=(1,โˆ’2,1)
To obtain a unit vector, we divide u2 by its magnitude:
โ€–u2โ€–=12+(โˆ’2)2+12=1+4+1=6
Dividing each component of u2 by 6, we get:
u2=(16,โˆ’26,16)
Therefore, two unit vectors orthogonal to both ๐ฏ and ๐ฐ are:
u1=(โˆ’133,โˆ’133,533) and u2=(16,โˆ’26,16)

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