(a) Is the vector (i + j + i) a unit vector? Justify your answer.
(b) Can a unit vector have any components with magnitude greater than unity?
Can it have any negative components? In each case justify your answer.
(c) If A = a (3.0i + 4.0j), where a is a constant, determine the value of a that makes 1 a unit vector.
The Correct Answer and Explanation is :
Solution:
(a) Is the vector ((\mathbf{i} + \mathbf{j} + \mathbf{i})) a unit vector?
To determine if ((\mathbf{i} + \mathbf{j} + \mathbf{i})) is a unit vector, we first simplify the vector:
[
\mathbf{i} + \mathbf{j} + \mathbf{i} = 2\mathbf{i} + \mathbf{j}.
]
Next, calculate its magnitude:
[
|\mathbf{v}| = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}.
]
A unit vector has a magnitude of 1. Since (\sqrt{5} \neq 1), ((2\mathbf{i} + \mathbf{j})) is not a unit vector.
(b) Can a unit vector have any components with magnitude greater than unity?
No, a unit vector cannot have any component with a magnitude greater than 1. This is because the magnitude of a vector is given by:
[
|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}.
]
If any single component ((x), (y), or (z)) exceeds 1 in magnitude, the sum of the squares will be greater than 1, making the magnitude of the vector greater than 1, which contradicts the definition of a unit vector.
Can it have negative components?
Yes, a unit vector can have negative components. Negative components only indicate direction and do not affect the magnitude. For example, the vector ((-1/\sqrt{2})\mathbf{i} + (-1/\sqrt{2})\mathbf{j}) is a unit vector because:
[
|\mathbf{v}| = \sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1.
]
(c) If (\mathbf{A} = a(3.0\mathbf{i} + 4.0\mathbf{j})), determine the value of (a) that makes (\mathbf{A}) a unit vector.
Calculate the magnitude of (\mathbf{A}):
[
|\mathbf{A}| = |a(3.0\mathbf{i} + 4.0\mathbf{j})| = a \sqrt{(3.0)^2 + (4.0)^2} = a \sqrt{9 + 16} = a \sqrt{25} = 5a.
]
For (\mathbf{A}) to be a unit vector, (|\mathbf{A}| = 1):
[
5a = 1 \implies a = \frac{1}{5}.
]
Explanation:
A unit vector has a magnitude of exactly 1. In (a), the magnitude of the given vector is (\sqrt{5}), which exceeds 1, proving it is not a unit vector. In (b), components exceeding 1 in magnitude would make the total magnitude larger than 1, violating the unit vector condition. However, negative components are permissible as they merely change direction, not magnitude. For (c), scaling the vector by (a = 1/5) normalizes it to a unit vector because the constant (a) adjusts the magnitude to precisely 1.