I need help with this problem 91+(-7)-5
1 answer:
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Answer:
angle of intersection: 5.2°
Step-by-step explanation:
The direction vector normal to the plane is ...
n = (1, 1, 3)
The direction vector of the line is ...
m = (1, -3, 1)
Then the angle θ between them can be found from the dot product:
n•m = |n|·|m|·cos(θ)
(1·1 +1(-3) +3·1) = 1 -3 +3 = 1 = √(1²+1²+3²)·√(1²+(-3)²+1²)·cos(θ)
1 = 11·cos(θ)
θ = arccos(1/11) ≈ 84.8°
This is the angle between the line and the normal to the plane, so the angle between the line and the plane will be the complement of this. Since this angle is not 90°, <em>the line and plane must intersect</em>.
acute angle = 90° -84.8° = 5.2°
_____
The attached graph shows the line and plane meet at a shallow angle, consistent with the above answer.
So for this, we will be using synthetic division. To set it up, have the equation so that the divisor is -10 (since that is the solution of k + 10 = 0) and the dividend are the coefficients. Our equation will look as such:
<em>(Note that synthetic division can only be used when the divisor is a 1st degree binomial)</em>
- -10 | 1 + 2 - 82 - 28
- ---------------------------
Now firstly, drop the 1:
- -10 | 1 + 2 - 82 - 28
- ↓
- -------------------------
- 1
Next, you are going to multiply -10 and 1, and then combine the product with 2.
- -10 | 1 + 2 - 82 - 28
- ↓ - 10
- -------------------------
- 1 - 8
Next, multiply -10 and -8, then combine the product with -82:
- -10 | 1 + 2 - 82 - 28
- ↓ -10 + 80
- -------------------------
- 1 - 8 - 2
Next, multiply -10 and -2, then combine the product with -28:
- -10 | 1 + 2 - 82 - 28
- ↓ -10 + 80 + 20
- -------------------------
- 1 - 8 - 2 - 8
Now, since we know that the degree of the dividend is 3, this means that the degree of the quotient is 2. Using this, the first 3 terms are k^2, k, and the constant, or in this case k² - 8k - 2. Now what about the last coefficient -8? Well this is our remainder, and will be written as -8/(k + 10).
<u>Putting it together, the quotient is </u>