Problem 24
Question
Find the unit vector \(\vec{u}\) in the direction of \(\vec{v} .\) \(\vec{v}=\langle 1,-2,2\rangle\)
Step-by-Step Solution
Verified Answer
The unit vector \(\vec{u}\) is \(\left\langle \frac{1}{3}, \frac{-2}{3}, \frac{2}{3} \right\rangle\).
1Step 1: Understanding the Problem
We are given a vector \(\vec{v} = \langle 1, -2, 2 \rangle\) and need to find a unit vector \(\vec{u}\) in the direction of \(\vec{v}\). A unit vector has a magnitude (length) of 1 and points in the direction of the given vector.
2Step 2: Compute the Magnitude of \(\vec{v}\)
To find the magnitude of \(\vec{v}\), use the formula for the magnitude of a vector: \[||\vec{v}|| = \sqrt{1^2 + (-2)^2 + 2^2} \]Calculate:\[||\vec{v}|| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\]
3Step 3: Calculate the Unit Vector \(\vec{u}\)
The unit vector \(\vec{u}\) in the direction of \(\vec{v}\) is given by scaling \(\vec{v}\) by the reciprocal of its magnitude. So, \[\vec{u} = \frac{1}{||\vec{v}||} \cdot \vec{v} = \frac{1}{3} \cdot \langle 1, -2, 2 \rangle\]Calculate the components:\[\vec{u} = \left\langle \frac{1}{3}, \frac{-2}{3}, \frac{2}{3} \right\rangle\]
Key Concepts
Vector MagnitudeVector DirectionScaling Vectors
Vector Magnitude
The concept of vector magnitude is essential in identifying not just how a vector directs, but how much it extends in space. Essentially, magnitude determines the size or length of a vector. When working with any vector, say \( \vec{v} = \langle a, b, c \rangle \), the magnitude is calculated using the Pythagorean formula:\[ ||\vec{v}|| = \sqrt{a^2 + b^2 + c^2} \]For example, consider the vector \( \vec{v} = \langle 1, -2, 2 \rangle \). Calculating its magnitude involves squaring each component, adding those squares, and taking the square root of the total.
- Square each component: \( 1^2 = 1 \), \((-2)^2 = 4 \), \(2^2 = 4 \)
- Add them up: \(1 + 4 + 4 = 9 \)
- Take the square root: \(\sqrt{9} = 3 \)
Vector Direction
Understanding vector direction involves recognizing where and in what direction a vector points in its space. Despite many vectors potentially sharing the same direction, each can differ in magnitude. To work with direction, it's helpful to use a unit vector, which simplifies the vector to its directional component by ensuring its length is 1.
The direction is maintained while the size or 'strength' of the vector is standardized. Take vector \( \vec{v} = \langle 1, -2, 2 \rangle \) again; by scaling down its magnitude to 1, you get a sense of where it points without affecting its magnitude.
For this, use the formula for the unit vector:\[ \vec{u} = \frac{1}{||\vec{v}||} \cdot \vec{v} \]Where \( \vec{u} \) represents the unit vector in the same direction as \( \vec{v} \). The unit vector retains the same direction as the original vector but is limited in magnitude, so it clearly depicts the direction.
The direction is maintained while the size or 'strength' of the vector is standardized. Take vector \( \vec{v} = \langle 1, -2, 2 \rangle \) again; by scaling down its magnitude to 1, you get a sense of where it points without affecting its magnitude.
For this, use the formula for the unit vector:\[ \vec{u} = \frac{1}{||\vec{v}||} \cdot \vec{v} \]Where \( \vec{u} \) represents the unit vector in the same direction as \( \vec{v} \). The unit vector retains the same direction as the original vector but is limited in magnitude, so it clearly depicts the direction.
Scaling Vectors
Scaling vectors is the key process to change a vector's magnitude without altering its direction. By multiplying a vector by a scalar — a real number — we effectively stretch or shrink its length. The original vector's direction remains fixed.
- If the scalar is greater than 1, the vector elongates.
- If the scalar is between 0 and 1, the vector contracts.
- If negative, the vector also reverses direction.
Other exercises in this chapter
Problem 24
Find the area of the parallelogram defined by the given vectors. \(\vec{u}=\langle-2,1,5\rangle, \quad \vec{v}=\langle-1,3,1\rangle\)
View solution Problem 24
Vectors \(\vec{u}\) and \(\vec{v}\) are given. Find \(\operatorname{proj}_{\vec{v}} \vec{u}\) the orthogonal projection of \(\vec{u}\) onto \(\vec{v},\) and ske
View solution Problem 25
Find the point of intersection between the line and the plane. line: \(\langle 1,2,3\rangle+t\langle 3,5,-1\rangle\) plane: \(3 x-2 y-z=4\)
View solution Problem 25
Find the distance from the point to the line. \(Q=(0,3), \quad \vec{\ell}(t)=\langle 2,0\rangle+t\langle 1,1\rangle\)
View solution