Answer :
Answer:
True: a, d
False: b, c
Explanation:
Vectors and Scalars
Vectors
Vectors are a quantity that includes magnitude (how big) and direction (location/"where is it pointing to").
Vectors can be drawn as arrows on a coordinate plane where the horizontal and vertical components are dictated by the numbers in the vector's form
Some common ways vectors are written are
- [tex]\left(\begin{array}{c}x\\y\end{array}\right)[/tex]
- < x, y >.
[tex]\dotfill[/tex]
Magnitude
Magnitude can be determined by using the initial and final points of a vector and plugging them into the distance formula.
[tex]\sqrt{(x_f-x_i)^2+(y_f-y_i)^2}[/tex]
If the vector originates from 0, the Pythagorean theorem formula where the vector is the hypotenuse.
[tex]\sqrt{x^2+y^2}[/tex]
[tex]\dotfill[/tex]
Scalars
Scalars are real numbers that act like factors: scalars multiply each component in the vector.
[tex]\hrulefill[/tex]
Solving the Problem
Part A
We can rephrase part A's claim as "a vector is equal to a scalar multiple of its opposite vector."
An opposite vector is a vector where it shares the same magnitude but an opposite direction. Their components are the negative version of the original vector.
If we want to prove part A's statement we must find a scalar value k to make the equation true.
[tex]\left(\begin{array}{c}x\\y\end{array}\right) = k\left(\begin{array}{c}-x\\-y\end{array}\right)[/tex]
If k = -1 then the negative signs on x and y cancel.
[tex]\left(\begin{array}{c}x\\y\end{array}\right) = -1\left(\begin{array}{c}-x\\-y\end{array}\right)[/tex]
[tex]\left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c}x\\y\end{array}\right)[/tex]
Thus, part A is true.
[tex]\dotfill[/tex]
Part B
Rephrasing part B's claim: "if a vector is multiplied by a positive scalar, then it must have a greater magnitude then the original vector itself."
To test the claim we use different positive values
Letting vector v have the components x and y where it originates from the origin, its magnitude is
[tex]|| \textbf{v}||= \sqrt{x^2+y^2[/tex].
(The notation is the convention of denoting the magnitude of a vector).
k = 100:
[tex]100\left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c}100x\\100y\end{array}\right)[/tex]
[tex]|| \textbf{v'}||= \sqrt{(100x)^2+(100y)^2} =\sqrt{10000x^2+10000y^2}[/tex]
[tex]= \sqrt{10000(x^2+y^2)} = 100\sqrt{x^2+y^2}[/tex]
[tex]|| \textbf{v'}|| > || \textbf{v}||[/tex]
It seems to hold truth. Let's try if k is between 0 and 1.
k = 1/2:
[tex]\dfrac{1}{2}\left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c}\frac{1}{2} x\\\frac{1}{2} y\end{array}\right)[/tex]
[tex]|| \textbf{v''}||= \sqrt{\left(\dfrac{1}{2} x\right)^2+\left(\dfrac{1}{2} y\right)^2} =\sqrt{\dfrac{1}{4} x^2+\dfrac{1}{4} y^2}[/tex]
[tex]=\sqrt{\dfrac{1}{4} (x^2+y^2)} =\dfrac{1}{2}\sqrt{x^2+ y^2}[/tex]
[tex]= \dfrac{\sqrt{x^2+ y^2} }{2}[/tex]
[tex]|| \textbf{v}|| > || \textbf{v''}||[/tex].
So, part B is false.
[tex]\dotfill[/tex]
Part C
Recalling the magnitude formulas, the components of the vector are squared.
[tex]||\textbf{v}|| = \sqrt{(x_f-x_i)^2+(y_f-y_i)^2} = \sqrt{x^2+y^2}[/tex]
Focusing on vector originating from the origin, let's consider two vectors
- the components of the vector v are x and y
- the components of vector w has the components -x and -y.
Determining their magnitudes,
[tex]||\textbf{v}|| = \sqrt{x^2+y^2}[/tex]
[tex]||\textbf{w}|| = \sqrt{(-x)^2+(-y)^2} = \sqrt{x^2+y^2}[/tex],
vectors v and w share the same magnitude.
Inspecting their components, vector v's would place its direction in the the first quadrant (up and to the right) of the coordinate plane whereas vector w's place its direction in the third quadrant (down and to the left).
Vectors v and w aren't going in the same direction
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Focusing on vectors not originating from the origin, let's consider two vectors and their initial/final points.
- Initial and final points of vector u are < 1, 1 > and < 5, 5 >
- Initial and final points of vector p are < -1, -1 > and < -5, -5 >
Determining their magnitudes using the distance formula,
[tex]|| \textbf{u}|| = \sqrt{(5-1)^2+(5-1)^2} = \sqrt{4^2+4^2} = \sqrt{32}[/tex]
[tex]|| \textbf{p}|| = \sqrt{(-1-(-5))^2+(-1-(-5))^2} = \sqrt{4^2+4^2} = \sqrt{32}[/tex],
vectors u and p share the same magnitude.
Inspecting the vectors closely,
- both points of vector u are located in the first quadrant, meaning that the vector is in the first quadrant.
- Vector p is located in the third quadrant.
Thus both vectors have different directions.
Thus, this claim is also false.
[tex]\dotfill[/tex]
Part D
For a vector to be parallel to another, the vector must be a scale factored version of the other.
[tex]k \textbf{a} = \textbf{b}[/tex],
where a and b are vectors.
In part D's case, a and b are the same vector, this leaves k to equal 1.
Since there's a known scalar quantity that multiplied to a vector equal itself, part D's claim is true.
