Show Scalar Multiplication Visually - Precalculus

1. Identify the scalar $c$ if $c(6,9)=(2,3)$.

   $c=\frac{1}{3}$. Since $\frac{1}{3}(6)=2$ and $\frac{1}{3}(9)=3$, $c=\frac{1}{3}$.

2. Identify the scalar multiple: Compute $-2\langle 4, -3 \rangle$ component-wise.

   $\langle -8, 6 \rangle$. Multiply each component: $-2(4)=-8$ and $-2(-3)=6$.

3. Which option describes $\frac{1}{3}\vec{v}$ geometrically for nonzero $\vec{v}$?

   Same direction, one-third the magnitude. Positive fraction less than 1 shrinks but doesn't reverse.

4. What happens to a vector's direction when it is multiplied by a scalar $c>0$?

   Direction stays the same. Positive scalars preserve the vector's direction.

5. Compute the scaled endpoint: If $\vec{v}=\langle 2, -1 \rangle$, what is $4\vec{v}$?

   $\langle 8, -4 \rangle$. Multiply each component by 4: $4(2)=8$ and $4(-1)=-4$.

6. Identify the correct statement: If $c=-5$, how does $c\vec{v}$ compare to $\vec{v}$?

   Opposite direction and $5$ times as long. Negative scalar reverses direction; $|-5|=5$ scales magnitude.

7. What happens to a vector's direction when it is multiplied by a negative scalar $c<0$?

   Direction reverses (points opposite). Negative scalars flip the vector to point the opposite way.

8. What is the result of scalar multiplication $0\vec{v}$ for any vector $\vec{v}$?

   $\vec{0}$. Multiplying any vector by zero gives the zero vector.

9. What is the geometric effect on magnitude when multiplying a vector by a scalar $c$?

   Magnitude is multiplied by $|c|$. The length scales by the absolute value of the scalar.

10. What is the scalar $c$ if $c(-5,1)=(15,-3)$?

    $c=-3$. Since $-3(-5)=15$ and $-3(1)=-3$, $c=-3$.

11. Find the missing vector: If $c=3$ and $c\vec{v}=\langle 9, -12 \rangle$, what is $\vec{v}$?

    $\langle 3, -4 \rangle$. Divide each component by 3: $\vec{v}=\langle 9/3, -12/3 \rangle$.

12. What is the scalar multiple $3(2,-5)$?

    $(6,-15)$. Multiply each component: $3(2)=6$, $3(-5)=-15$.

13. What is the scalar multiple $-\frac{3}{4}(12,-8)$?

    $(-9,6)$. Multiply: $-\frac{3}{4}(12)=-9$, $-\frac{3}{4}(-8)=6$.

14. What is the scalar $c$ if $c(3,-2)=(12,-8)$?

    $c=4$. Since $4(3)=12$ and $4(-2)=-8$, $c=4$.

15. State the component-wise formula for multiplying a vector $\langle v_x, v_y \rangle$ by a scalar $c$.

    $c\langle v_x, v_y \rangle = \langle cv_x, cv_y \rangle$. Multiply each component by the scalar separately.

16. Compute the scalar: If $\vec{v}=\langle 3, -2 \rangle$ and $c\vec{v}=\langle 12, -8 \rangle$, what is $c$?

    $c=4$. Since $12=4(3)$ and $-8=4(-2)$, the scalar is $c=4$.

17. Identify the error: A student claims $2\langle 1, -4 \rangle=\langle 1, -8 \rangle$. What is the correct result?

    $\langle 2, -8 \rangle$. Student forgot to multiply the first component by 2.

18. What happens to a vector's magnitude when it is multiplied by a scalar $c$?

    Magnitude is multiplied by $|c|$. The scalar's absolute value scales the length.

19. What happens to a vector's direction when it is multiplied by a scalar $c<0$?

    Direction reverses. Negative scalars flip the vector to point opposite.

20. What happens to a vector's direction when it is multiplied by a positive scalar $c>0$?

    Direction stays the same. Positive scalars preserve the vector's original direction.

21. What is the magnitude of $c\vec{v}$ in terms of $|c|$ and $\|\vec{v}\|$?

    $\|c\vec{v}\|=|c|\,\|\vec{v}\|$. Magnitude scales by the absolute value of the scalar.

22. What is the component-wise rule for scalar multiplication $c(v_x, v_y)$?

    $c(v_x, v_y) = (cv_x, cv_y)$. Multiply each component by the scalar $c$.

23. On a coordinate plane, if $\vec{v}$ goes from $(0,0)$ to $(2,3)$, where does $-2\vec{v}$ end?

    $(-4,-6)$. Scale and reverse: $-2(2)=-4$ and $-2(3)=-6$.

24. Identify the scalar multiple: Compute $-\frac{3}{4}\langle 8, 12 \rangle$ component-wise.

    $\langle -6, -9 \rangle$. Multiply: $-\frac{3}{4}(8)=-6$ and $-\frac{3}{4}(12)=-9$.

25. Identify the scalar multiple: Compute $\frac{1}{2}\langle 6, -10 \rangle$ component-wise.

    $\langle 3, -5 \rangle$. Multiply each component: $\frac{1}{2}(6)=3$ and $\frac{1}{2}(-10)=-5$.

26. Find the missing vector: If $c=-2$ and $c\vec{v}=\langle -10, 14 \rangle$, what is $\vec{v}$?

    $\langle 5, -7 \rangle$. Divide by $-2$: $\vec{v}=\langle -10/(-2), 14/(-2) \rangle$.

27. Identify the scalar multiple: Compute $3\langle -2, 5 \rangle$ component-wise.

    $\langle -6, 15 \rangle$. Multiply each component: $3(-2)=-6$ and $3(5)=15$.

28. What is the scalar multiple $-2(4,1)$?

    $(-8,-2)$. Multiply each component: $-2(4)=-8$, $-2(1)=-2$.

29. On a coordinate plane, what does multiplying by $c$ do to each component of $(v_x,v_y)$?

    Multiply both components by $c$. Scalar multiplication applies to each component.

30. Compute the scaled endpoint: If $\vec{v}=\langle -3, 7 \rangle$, what is $-1\vec{v}$?

    $\langle 3, -7 \rangle$. Multiplying by $-1$ reverses the vector: $-1(-3)=3$, $-1(7)=-7$.

31. Compute the scalar: If $\vec{v}=\langle -4, 6 \rangle$ and $c\vec{v}=\langle 2, -3 \rangle$, what is $c$?

    $c=-\frac{1}{2}$. Since $2=-\frac{1}{2}(-4)$ and $-3=-\frac{1}{2}(6)$, $c=-\frac{1}{2}$.

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