Vector and Matrix Quantities: Solving Problems with Vectors and Velocity (CCSS.N-VM.3)
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Common Core High School Number And Quantity › Vector and Matrix Quantities: Solving Problems with Vectors and Velocity (CCSS.N-VM.3)
What is the direction angle of the vector ⟨-7, -24⟩?
$\arctan(-7-24)$
$\arctan\left(\dfrac{-24}{-7}\right)$
$\arctan\left(\dfrac{-7}{-24}\right)$
$\arctan\left(\dfrac{-24}{7}\right)$
Explanation
Use $\theta=\arctan!\left(\dfrac{b}{a}\right)=\arctan!\left(\dfrac{-24}{-7}\right)$. Flipping the ratio, adding components, or dropping a negative sign leads to wrong expressions.
A velocity vector is $\langle -6, 8 \rangle$ m/s. What is its speed (magnitude)?
14
2
10
$\sqrt{2}$
Explanation
Magnitude is the length: $\sqrt{(-6)^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$. This comes from the Pythagorean theorem applied to the horizontal and vertical components.
What is the direction angle of the vector $\langle 3, 3\sqrt{3} \rangle$?
$\arctan!\left(\dfrac{3\sqrt{3}}{3}\right)$
$\arctan!\left(\dfrac{3}{3\sqrt{3}}\right)$
$\arctan(3\sqrt{3}+3)$
$\arctan(3\cdot 3\sqrt{3})$
Explanation
For a vector $\langle a,b\rangle$ in Quadrant I, the direction angle is $\theta=\arctan!\left(\dfrac{b}{a}\right)$. Here $a=3$, $b=3\sqrt{3}$, so $\theta=\arctan!\left(\dfrac{3\sqrt{3}}{3}\right)$.
Which statement describes the vector $\langle -5, 12 \rangle$?
It lies in Quadrant I.
Its magnitude is $13$.
Its magnitude is $7$.
Its magnitude is $\sqrt{(-5)+12}$.
Explanation
The magnitude is $\sqrt{(-5)^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$ by the Pythagorean theorem. The vector has negative $x$ and positive $y$, so it does not lie in Quadrant I.
What is the magnitude of the vector $\langle 5, 2 \rangle$?
$5+2$
$\sqrt{5^2-2^2}$
$\sqrt{5+2}$
$\sqrt{5^2+2^2}$
Explanation
Magnitude is length from the origin: $\sqrt{5^2+2^2}=\sqrt{25+4}=\sqrt{29}$. This follows from the Pythagorean theorem on the right triangle formed by the components.
What is the direction angle of the vector $\langle 4, 7 \rangle$?
$\arctan!\left(\dfrac{7}{4}\right)$
$\arctan!\left(\dfrac{4}{7}\right)$
$\arctan(7+4)$
$\arctan(7-4)$
Explanation
For $\langle a,b\rangle$ with $a>0$ and $b>0$, $\theta=\arctan!\left(\dfrac{b}{a}\right)$. Thus $\theta=\arctan!\left(\dfrac{7}{4}\right)$. The other options flip $a$ and $b$ or combine components incorrectly.
What is the magnitude of $\langle -3,4\rangle$?
5
7
$\sqrt{7}$
$\sqrt{13}$
Explanation
Use the Pythagorean theorem for vector length: $|\langle a,b\rangle|=\sqrt{a^2+b^2}$. Here $\sqrt{(-3)^2+4^2}=\sqrt{9+16}=5$.
What is the direction angle of $\langle -3,3\rangle$ (measured from the positive $x$-axis)?
45°
135°
-45°
315°
Explanation
Compute $\theta=\arctan!\left(\tfrac{b}{a}\right)=\arctan!\left(\tfrac{3}{-3}\right)=\arctan(-1)=-45°$. Since $a<0$ and $b>0$ (Quadrant II), adjust to $180°-45°=135°$.
Which statement describes the vector $\langle 6,8\rangle$?
It has magnitude $6+8=14$ and points opposite $\langle 3,4\rangle$.
It has magnitude $\sqrt{6+8}$ and points in Quadrant III.
It has magnitude 10 and the same direction as $\langle 3,4\rangle$.
It has magnitude 10 and is perpendicular to $\langle 3,4\rangle$.
Explanation
Magnitude uses the Pythagorean theorem: $\sqrt{6^2+8^2}=\sqrt{36+64}=10$. Also $\langle 6,8\rangle=2\langle 3,4\rangle$, so it has the same direction.
What is the direction angle of $\langle 5,-5\sqrt{3}\rangle$ (in degrees from the positive $x$-axis)?
60°
300°
120°
240°
Explanation
Compute $\theta=\arctan!\left(\tfrac{b}{a}\right)=\arctan!\left(\tfrac{-5\sqrt{3}}{5}\right)=\arctan(-\sqrt{3})=-60°$. With $a>0$, $b<0$ (Quadrant IV), the standard angle is $360°-60°=300°$.