8. Let α, β, γ be the angles made by a vector u ∈ R 3 with positive x, y, z−axis respectively. Then, the numbers l = cos α, m =
cos β, n = cos γ are called the direction cosines of u.
(a) Find the formulas for the direction cosines: l, m, n of the vector u = hu1, u2, u3i.
(b) Prove that l 2 + m2 + n 2 = 1.
(c) Find the direction cosines l, m, n of the vector u = h1, −1, 1i.
(a) Find the formulas for the direction cosines: l, m, n of the vector u = hu1, u2, u3i.
(b) Prove that l 2 + m2 + n 2 = 1.
(c) Find the direction cosines l, m, n of the vector u = h1, −1, 1i.
1 answer:
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For function k(x, y) = -x² - y² + 4x + 4y,
the absolute minimum is 0 and the absolute maximum is 6
For given question,
We have been given a function k(x, y) = -x² - y² + 4x + 4y
We need to find the absolute maximum and minimum values of the function, subject to the constraints 0 ≤ x ≤ 3, y ≥ 0, and x + y ≤ 6
First we find the partial derivative of function k(x, y) with respect to x.
⇒
Now, we find the partial derivative of function k(x, y) with respect to y.
To find the critical point:
consider and
⇒ -2x + 4 = 0 and -2y + 4 = 0
⇒ x = 2 and y = 2
This means, the critical point of function is (2, 2)
We have been given constraints 0 ≤ x ≤ 3, y ≥ 0, and x + y ≤ 6
Consider k(0, 0)
⇒ k(0, 0) = -0² - 0² + 4(0) + 4(0)
⇒ k(0, 0) = 0
Consider k(3, 3)
⇒ k(3, 3) = -3² - 3² + 4(3) + 4(3)
⇒ k(3, 3) = -9 - 9 + 12 + 12
⇒ k(3, 3) = -18 + 24
⇒ k(3, 3) = 6
Therefore, for function k(x, y) = -x² - y² + 4x + 4y,
the absolute minimum is 0 and the absolute maximum is 6
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