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lutik1710 [3]
3 years ago
11

Which of the following vectors are orthogonal to (1,5)? Check all that apply

Mathematics
1 answer:
Andre45 [30]3 years ago
7 0
We know that
If the scalar product of two vectors<span> is zero, both vectors are </span><span>orthogonal

</span><span>A. (-2,5)
</span>(-2,5)*(1,5)-------> -2*1+5*5=23-----------> <span>are not orthogonal 

</span><span>B. (10,-2)
</span>(10,-2)*(1,5)-------> 10*1-2*5=0-----------> are  orthogonal 

<span>C. (-1,-5)
</span>(-1,-5)*(1,5)-------> -1*1-5*5=-26-----------> are not orthogonal 

<span>D. (-5,1)
</span>(-5,1)*(1,5)-------> -5*1+1*5=0-----------> are  orthogonal 

the answer is 
B. (10,-2) and D. (-5,1)  are  orthogonal  to  (1,5)
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Originally A(16,0) so now it would be A1(8,0)

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I'm having a hard time with this question so someone help please
bulgar [2K]

Answer:

  7.7 km

Step-by-step explanation:

This question is asking you to find the third side of a triangle in which two sides and an angle not between them are given. The law of sines provides the necessary relationships.

<h3>Law of Sines</h3>

The Law of Sines tells you that triangle side lengths are proportional to the sine of the opposite angle. For this triangle TPJ, the relation is ...

  t/sin(T) = p/sin(P) = j/sin(J)

We are given T=68°, p=5.2 km, t=7.5 km, and we are asked to find the length of j, the TP distance.

<h3>Setup</h3>

We know the sum of angles in a triangle is 180°, and the sine of an angle is equal to the sine of its supplement. This lets us fill in the Law of Sines equation like this:

  7.5/sin(68°) = 5.2/sin(P) = j/sin(68°+P) . . . . . solve for j

<h3>Solution</h3>

This can be solved in two steps. First, we find angle P, then we use that to find length j.

  sin(P) = 5.2/7.5 × sin(68°) ≈ 0.642847

  P = arcsin(0.642847) ≈ 40.0045°

Now, we can find j:

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The distance from the tower to the plane is about 7.7 km.

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<em>Additional comment</em>

We have used T, P, and J to refer to the vertices at the tower, plane, and jet.

The triangle can also be solved using the Law of Cosines. In that case, the missing side will be the solution to a quadratic equation with irrational coefficients.

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