Let u= (5, -12) and c= -3. What is || cu ||? A- -39 B- 39 C- 21 D- 51
The Correct Answer and Explanation is:
We are given:
- A vector u = (5, -12)
- A scalar c = -3
- We are asked to find the magnitude of the vector cu, which is written as ∥cu∥\| c\mathbf{u} \|.
Step-by-Step Solution:
1. Understand scalar multiplication:
When we multiply a vector u by a scalar c, we multiply each component of the vector by the scalar: cu=c⋅(5,−12)=(−3)(5,−12)=(−15,36)c\mathbf{u} = c \cdot (5, -12) = (-3)(5, -12) = (-15, 36)
2. Find the magnitude of the new vector ∥cu∥\| cu \|:
The magnitude of a 2D vector (x,y)(x, y) is given by: ∥(x,y)∥=x2+y2\| (x, y) \| = \sqrt{x^2 + y^2}
So for the vector (−15,36)(-15, 36): ∥(−15,36)∥=(−15)2+(36)2=225+1296=1521=39\| (-15, 36) \| = \sqrt{(-15)^2 + (36)^2} = \sqrt{225 + 1296} = \sqrt{1521} = 39
Final Answer: B – 39
Explanation
The magnitude of a vector represents its length or size regardless of direction. Given a vector u = (5, -12) and a scalar c = -3, we are to find the magnitude of the scaled vector cu.
To begin, scalar multiplication involves multiplying each component of the vector by the scalar. Here, multiplying each component of u by -3 gives us a new vector cu = (-15, 36). Notice that this changes the direction of the vector (because the scalar is negative), but not its magnitude in absolute terms.
Next, we find the magnitude of the new vector. The formula for the magnitude of a 2D vector (x, y) is: ∥(x,y)∥=x2+y2\| (x, y) \| = \sqrt{x^2 + y^2}
Applying this to (-15, 36): ∥(−15,36)∥=(−15)2+362=225+1296=1521=39\| (-15, 36) \| = \sqrt{(-15)^2 + 36^2} = \sqrt{225 + 1296} = \sqrt{1521} = 39
The magnitude remains the same as if we scaled the original magnitude directly. Indeed, an alternative method confirms this: ∥cu∥=∣c∣⋅∥u∥\| c\mathbf{u} \| = |c| \cdot \| \mathbf{u} \| ∥u∥=52+(−12)2=25+144=169=13\| \mathbf{u} \| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13
So: ∥−3⋅u∥=∣−3∣⋅13=3⋅13=39\| -3 \cdot \mathbf{u} \| = |-3| \cdot 13 = 3 \cdot 13 = 39
Thus, the correct answer is B – 39.
