Mathematics
Grade 12
15 min
Find a threedimensional unit vector
Find a threedimensional unit vector
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1
Introduction & Learning Objectives
Learning Objectives
Define a unit vector in three dimensions and explain its purpose.
Accurately calculate the magnitude (or norm) of any three-dimensional vector.
State and apply the formula for normalizing a vector.
Find the unit vector in the same direction as a given non-zero vector.
Find the unit vector in the opposite direction of a given non-zero vector.
Verify that a calculated vector is a unit vector by checking if its magnitude is 1.
How does a GPS system tell your car to 'head northeast' without caring if you're going 5 mph or 70 mph? 🗺️ It's all about pure direction!
This tutorial focuses on one of the most fundamental operations in vector algebra: finding a unit vector. You will learn how to isolate a vector's direction by 'normali...
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Key Concepts & Vocabulary
TermDefinitionExample
Three-Dimensional VectorA mathematical object that has both magnitude (length) and direction in 3D space. It is typically represented by its components along the x, y, and z axes.The vector `v = <4, -2, 5>` represents a displacement of 4 units along the x-axis, -2 units along the y-axis, and 5 units along the z-axis.
Magnitude (or Norm)The length or size of a vector, calculated using the distance formula in three dimensions. It is always a non-negative scalar value.For vector `v = <3, 0, 4>`, its magnitude is `||v|| = \sqrt{3^2 + 0^2 + 4^2} = \sqrt{9 + 0 + 16} = \sqrt{25} = 5`.
Unit VectorA vector with a magnitude of exactly 1. It is used to represent a pure direction, without any associated length or scale.The standard basis vector `j = <0, 1, 0>`...
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Core Formulas
Magnitude of a 3D Vector
For a vector `v = <v_x, v_y, v_z>`, the magnitude `||v||` is given by: `||v|| = \sqrt{v_x^2 + v_y^2 + v_z^2}`
This formula is an extension of the Pythagorean theorem to three dimensions. Use it to find the length of any 3D vector. This is the first step in finding a unit vector.
Unit Vector Formula (Normalization)
For any non-zero vector `v`, the unit vector `u` in the same direction is: `u = \frac{v}{||v||} = \frac{1}{||v||} <v_x, v_y, v_z>`
This formula defines the process of normalization. To find the unit vector, simply divide the original vector by its own magnitude. The result is a new vector pointing in the same direction with a length of 1.
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Challenging
Find the positive value of `k` for which the vector `v = <1/2, k, \sqrt{11}/4>` is a unit vector.
A.1/4
B.1/16
C.3/4
D.1/2
Challenging
Let `u` be the unit vector in the direction of `v = <3, -4, 5>`. Find the vector `w` that has a magnitude of 10 and points in the opposite direction of `u`.
A.<-30/\sqrt{50}, 40/\sqrt{50}, -50/\sqrt{50}>
B.<-3\sqrt{2}, 4\sqrt{2}, -5\sqrt{2}>
C.<3\sqrt{2}, -4\sqrt{2}, 5\sqrt{2}>
D.<-6, 8, -10>
Challenging
Let `v = <a, a, a>` and `w = <2, -7, 5>`. If `a > 0` and `||v|| = ||w||`, what is the unit vector in the direction of `v`?
A.<\sqrt{26}, \sqrt{26}, \sqrt{26}>
B.<1, 1, 1>
C.<1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}>
D.<\sqrt{78}, \sqrt{78}, \sqrt{78}>
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