Find two unit vectors orthogonal to both u=<4,-3,1> and v=<0,1,8> . Give exact values (no decimals). Separate the vectors with a comma and do not pull any numbers or roots outside of the vector.
The Correct Answer and Explanation is:
To find two unit vectors orthogonal to both vectors u = <4, -3, 1> and v = <0, 1, 8>, we can follow these steps:
Step 1: Find the Cross Product
The cross product of two vectors u and v gives a vector that is orthogonal to both. The formula for the cross product of two vectors u = <u₁, u₂, u₃> and v = <v₁, v₂, v₃> is: u×v=∣i^j^k^u1u2u3v1v2v3∣u \times v = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ u₁ & u₂ & u₃ \\ v₁ & v₂ & v₃ \end{matrix} \right|u×v=i^u1v1j^u2v2k^u3v3
Now, applying the formula for the vectors u = <4, -3, 1> and v = <0, 1, 8>: u×v=i^∣−3118∣−j^∣4108∣+k^∣4−301∣u \times v = \hat{i} \left| \begin{matrix} -3 & 1 \\ 1 & 8 \end{matrix} \right| – \hat{j} \left| \begin{matrix} 4 & 1 \\ 0 & 8 \end{matrix} \right| + \hat{k} \left| \begin{matrix} 4 & -3 \\ 0 & 1 \end{matrix} \right|u×v=i^−3118−j^4018+k^40−31
Calculating each of the 2×2 determinants: i^:(−3)(8)−(1)(1)=−24−1=−25\hat{i}: (-3)(8) – (1)(1) = -24 – 1 = -25i^:(−3)(8)−(1)(1)=−24−1=−25 j^:(4)(8)−(0)(1)=32−0=32\hat{j}: (4)(8) – (0)(1) = 32 – 0 = 32j^:(4)(8)−(0)(1)=32−0=32 k^:(4)(1)−(−3)(0)=4−0=4\hat{k}: (4)(1) – (-3)(0) = 4 – 0 = 4k^:(4)(1)−(−3)(0)=4−0=4
Thus, the cross product is: u×v=−25i^−32j^+4k^=<−25,−32,4>u \times v = -25\hat{i} – 32\hat{j} + 4\hat{k} = <-25, -32, 4>u×v=−25i^−32j^+4k^=<−25,−32,4>
Step 2: Find the Magnitude of the Cross Product
Next, we calculate the magnitude (or length) of the vector u × v. The magnitude of a vector <x, y, z> is given by: ∥u×v∥=x2+y2+z2\| u \times v \| = \sqrt{x^2 + y^2 + z^2}∥u×v∥=x2+y2+z2
For <-25, -32, 4>, we have: ∥u×v∥=(−25)2+(−32)2+42=625+1024+16=1665\| u \times v \| = \sqrt{(-25)^2 + (-32)^2 + 4^2} = \sqrt{625 + 1024 + 16} = \sqrt{1665}∥u×v∥=(−25)2+(−32)2+42=625+1024+16=1665
Step 3: Normalize the Cross Product
To find a unit vector, we divide the vector u × v by its magnitude. The unit vector is: u^=11665⟨−25,−32,4⟩=⟨−251665,−321665,41665⟩\hat{u} = \frac{1}{\sqrt{1665}} \langle -25, -32, 4 \rangle = \left\langle \frac{-25}{\sqrt{1665}}, \frac{-32}{\sqrt{1665}}, \frac{4}{\sqrt{1665}} \right\rangleu^=16651⟨−25,−32,4⟩=⟨1665−25,1665−32,16654⟩
Step 4: Find the Second Unit Vector
The second unit vector orthogonal to both u and v is simply the negative of the first. Thus, it is: −u^=⟨251665,321665,−41665⟩-\hat{u} = \left\langle \frac{25}{\sqrt{1665}}, \frac{32}{\sqrt{1665}}, \frac{-4}{\sqrt{1665}} \right\rangle−u^=⟨166525,166532,1665−4⟩
Final Answer:
The two unit vectors orthogonal to both u and v are: ⟨−251665,−321665,41665⟩,⟨251665,321665,−41665⟩\left\langle \frac{-25}{\sqrt{1665}}, \frac{-32}{\sqrt{1665}}, \frac{4}{\sqrt{1665}} \right\rangle, \left\langle \frac{25}{\sqrt{1665}}, \frac{32}{\sqrt{1665}}, \frac{-4}{\sqrt{1665}} \right\rangle⟨1665−25,1665−32,16654⟩,⟨166525,166532,1665−4⟩
