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

Question

Laaibah Pitt · Accepted Answer

To find, explanation

Nick Camelot · Accepted Answer

Step 1:To find two unit vectors orthogonal to both (3,2,1) and (−1,1,0), we can use the cross product.Let&#039;s denote the two given vectors as 𝐚=(3,2,1) and 𝐛=(−1,1,0).First, we calculate the cross product 𝐜=𝐚×𝐛:𝐜=(321)×(−110)Using the cross product formula, we can find the resulting vector:𝐜=(2·0−1·13·0−1·03·1−2·(−1))=(−105)Now, to obtain two unit vectors orthogonal to both 𝐚 and 𝐛, we can normalize 𝐜 by dividing it by its magnitude. The magnitude of 𝐜 is given by:‖𝐜‖=(−1)2+02+52=26Therefore, the unit vector u1 orthogonal to both 𝐚 and 𝐛 is:u1=𝐜‖𝐜‖=126(−105)=(−1260526)Step 2:To find the second unit vector, we can take the cross product of 𝐜 and either 𝐚 or 𝐛. Let&#039;s use 𝐚 for this example. We calculate the cross product 𝐝=𝐜×𝐚:𝐝=(−105)×(321)Using the cross product formula, we get:𝐝=(0·1−5·2−1·1−5·3−1·2−0·3)=(−10−16−2)Now, we normalize 𝐝 by dividing it by its magnitude:‖𝐝‖=(−10)2+(−16)2+(−2)2=336To obtain the second unit vector, we normalize 𝐝 by dividing it by its magnitude:u2=𝐝‖𝐝‖=1336(−10−16−2)=(−10336−16336−2336)Therefore, the two unit vectors orthogonal to both (3,2,1) and (−1,1,0) are:u1=(−1260526)andu2=(−10336−16336−2336)These vectors are orthogonal to both (3,2,1) and (−1,1,0), and they have unit length.

Mr Solver · Accepted Answer

To find two unit vectors orthogonal to both

𝐯 = (3, 2, 1)

and

𝐰 = (- 1, 1, 0)

, we can use the cross product of

𝐯

and

𝐰

. The cross product gives us a vector that is orthogonal to both

𝐯

and

𝐰

. Let's calculate it:

𝐮 = 𝐯 \times 𝐰

The cross product of two vectors in three-dimensional space is given by:

𝐮 = (v_{2} w_{3} - v_{3} w_{2}, v_{3} w_{1} - v_{1} w_{3}, v_{1} w_{2} - v_{2} w_{1})

Substituting the values of

𝐯

and

𝐰

, we have:

𝐮 = (2 \cdot 0 - 1 \cdot 1, 1 \cdot (- 1) - 3 \cdot 0, 3 \cdot 1 - 2 \cdot (- 1))

Simplifying the expression:

𝐮 = (- 1, - 1, 5)

To obtain a unit vector, we divide

𝐮

by its magnitude:

‖ 𝐮 ‖ = \sqrt{(- 1)^{2} + (- 1)^{2} + 5^{2}} = \sqrt{1 + 1 + 25} = \sqrt{27} = 3 \sqrt{3}

Dividing each component of

𝐮

by

3 \sqrt{3}

, we get:

u_{1} = (\frac{- 1}{3 \sqrt{3}}, \frac{- 1}{3 \sqrt{3}}, \frac{5}{3 \sqrt{3}})

Now, to find another unit vector orthogonal to both

𝐯

and

𝐰

, we can take the cross product of

𝐮

and either

𝐯

or

𝐰

. Let's use

𝐯

for this example:

u_{2} = 𝐮 \times 𝐯

Using the same formula for the cross product as before, we substitute the values:

u_{2} = (- 1 \cdot 1 - (- 1) \cdot 2, (- 1) \cdot 3 - (- 1) \cdot 1, (- 1) \cdot 2 - (- 1) \cdot 3)

Simplifying the expression:

u_{2} = (- 1 + 2, - 3 + 1, - 2 + 3) = (1, - 2, 1)

To obtain a unit vector, we divide

u_{2}

by its magnitude:

‖ u_{2} ‖ = \sqrt{1^{2} + (- 2)^{2} + 1^{2}} = \sqrt{1 + 4 + 1} = \sqrt{6}

Dividing each component of

u_{2}

by

\sqrt{6}

, we get:

u_{2} = (\frac{1}{\sqrt{6}}, \frac{- 2}{\sqrt{6}}, \frac{1}{\sqrt{6}})

Therefore, two unit vectors orthogonal to both

𝐯

and

𝐰

are:

u_{1} = (\frac{- 1}{3 \sqrt{3}}, \frac{- 1}{3 \sqrt{3}}, \frac{5}{3 \sqrt{3}})

and

u_{2} = (\frac{1}{\sqrt{6}}, \frac{- 2}{\sqrt{6}}, \frac{1}{\sqrt{6}})

Jazz Frenia · Accepted Answer

To identify two unit vectors that are orthogonal to both 𝐯1=(3,2,1) and 𝐯2=(−1,1,0), we can use the cross product.The cross product of two vectors, 𝐯1 and 𝐯2, is orthogonal to both vectors. It can be calculated as follows:𝐯1×𝐯2=|𝐢𝐣𝐤321−110|Expanding this determinant, we get:𝐯1×𝐯2=(2·0−1·1)𝐢−(3·0−1·(−1))𝐣+(3·1−2·(−1))𝐤Simplifying further:𝐯1×𝐯2=−𝐢+𝐣+5𝐤Now, to find two unit vectors orthogonal to both 𝐯1 and 𝐯2, we can normalize the cross product vector. The magnitude of the cross product vector is (−1)2+12+52=27.Hence, the unit vector in the direction of 𝐯1×𝐯2 is:𝐮1=𝐯1×𝐯2‖𝐯1×𝐯2‖=−𝐢+𝐣+5𝐤27=(−127,127,527)To find another orthogonal unit vector, we can take the cross product of 𝐯1×𝐯2 and 𝐯1.𝐮2=𝐯1×(𝐯1×𝐯2)‖𝐯1×(𝐯1×𝐯2)‖Calculating the cross product:𝐯1×(𝐯1×𝐯2)=|𝐢𝐣𝐤321−110|Expanding the determinant:𝐯1×(𝐯1×𝐯2)=(2·0−1·1)𝐢−(3·0−1·(−1))𝐣+(3·1−2·(−1))𝐤Simplifying further:𝐯1×(𝐯1×𝐯2)=−𝐢−𝐣+5𝐤Now, let&#039;s calculate the magnitude of 𝐯1×(𝐯1×𝐯2). Its magnitude is (−1)2+(−1)2+52=27.Therefore, the unit vector in the direction of 𝐯1×(𝐯1×𝐯2) is:𝐮2=𝐯1×(𝐯1×𝐯2)‖𝐯1×(𝐯1×𝐯2)‖=−𝐢−𝐣+5𝐤27=(−127,−127,527)Hence, the two unit vectors orthogonal to both 𝐯1 and 𝐯2 are 𝐮1=(−127,127,527) and 𝐮2=(−127,−127,527), respectively.

xleb123 · Accepted Answer

To find two unit vectors orthogonal to both

𝐯 = (3, 2, 1)

and

𝐰 = (- 1, 1, 0)

, we can use the cross product of

𝐯

and

𝐰

. The cross product gives us a vector that is orthogonal to both

𝐯

and

𝐰

. Let's calculate it:

𝐮 = 𝐯 \times 𝐰

The cross product of two vectors in three-dimensional space is given by:

𝐮 = (v_{2} w_{3} - v_{3} w_{2}, v_{3} w_{1} - v_{1} w_{3}, v_{1} w_{2} - v_{2} w_{1})

Substituting the values of

𝐯

and

𝐰

, we have:

𝐮 = (2 \cdot 0 - 1 \cdot 1, 1 \cdot (- 1) - 3 \cdot 0, 3 \cdot 1 - 2 \cdot (- 1))

Simplifying the expression:

𝐮 = (- 1, - 1, 5)

To obtain a unit vector, we divide

𝐮

by its magnitude:

‖ 𝐮 ‖ = \sqrt{(- 1)^{2} + (- 1)^{2} + 5^{2}} = \sqrt{1 + 1 + 25} = \sqrt{27} = 3 \sqrt{3}

Dividing each component of

𝐮

by

3 \sqrt{3}

, we get:

u_{1} = (\frac{- 1}{3 \sqrt{3}}, \frac{- 1}{3 \sqrt{3}}, \frac{5}{3 \sqrt{3}})

Now, to find another unit vector orthogonal to both

𝐯

and

𝐰

, we can take the cross product of

𝐮

and either

𝐯

or

𝐰

. Let's use

𝐯

for this example:

u_{2} = 𝐮 \times 𝐯

Using the same formula for the cross product as before, we substitute the values:

u_{2} = (- 1 \cdot 1 - (- 1) \cdot 2, (- 1) \cdot 3 - (- 1) \cdot 1, (- 1) \cdot 2 - (- 1) \cdot 3)

Simplifying the expression:

u_{2} = (- 1 + 2, - 3 + 1, - 2 + 3) = (1, - 2, 1)

To obtain a unit vector, we divide

u_{2}

by its magnitude:

‖ u_{2} ‖ = \sqrt{1^{2} + (- 2)^{2} + 1^{2}} = \sqrt{1 + 4 + 1} = \sqrt{6}

Dividing each component of

u_{2}

by

\sqrt{6}

, we get:

u_{2} = (\frac{1}{\sqrt{6}}, \frac{- 2}{\sqrt{6}}, \frac{1}{\sqrt{6}})

Therefore, two unit vectors orthogonal to both

𝐯

and

𝐰

are:

u_{1} = (\frac{- 1}{3 \sqrt{3}}, \frac{- 1}{3 \sqrt{3}}, \frac{5}{3 \sqrt{3}})

and

u_{2} = (\frac{1}{\sqrt{6}}, \frac{- 2}{\sqrt{6}}, \frac{1}{\sqrt{6}})

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

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