Find TwoUnit Vectors Orthogonal to Both: A Step‑by‑Step Guide
When two vectors share a common perpendicular direction, the answer is not unique – there are exactly two unit vectors that satisfy the condition, pointing in opposite senses. This article walks you through the logic, the algebra, and a concrete example so you can solve similar problems with confidence.
Introduction
In three‑dimensional space, a vector that is orthogonal (perpendicular) to two given vectors must be parallel to their cross product. Because a direction and its opposite both qualify as orthogonal, there are precisely two unit vectors that meet the requirement: one pointing along the cross product and the other pointing in the exact reverse direction. This article explains why, shows how to compute them, and highlights typical pitfalls.
Understanding Orthogonality and the Cross Product
What Does “Orthogonal to Both” Mean?
A vector v is orthogonal to a vector a if their dot product is zero:
[ \mathbf{a}\cdot\mathbf{v}=0 ]
If v must be orthogonal to both a and b, it must satisfy
[\mathbf{a}\cdot\mathbf{v}=0 \quad\text{and}\quad \mathbf{b}\cdot\mathbf{v}=0 ]
The set of all vectors that meet these two equations forms a line through the origin. That line is aligned with the cross product a × b.
The Cross Product Gives the Direction
The cross product of two non‑parallel vectors a and b is defined as [ \mathbf{a}\times\mathbf{b}= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\ a_1&a_2&a_3\ b_1&b_2&b_3 \end{vmatrix} ]
Geometrically, a × b is perpendicular to the plane spanned by a and b, and its magnitude equals
[ |\mathbf{a}\times\mathbf{b}| = |\mathbf{a}|,|\mathbf{b}|\sin\theta ]
where (\theta) is the angle between a and b. Because the cross product is unique (up to sign), it provides the only line of vectors orthogonal to both.
How to Find the Two Unit Vectors
Step 1: Compute the Cross Product
Given a = ((a_1, a_2, a_3)) and b = ((b_1, b_2, b_3)), calculate
[ \mathbf{c}= \mathbf{a}\times\mathbf{b} ]
Write out the determinant explicitly or use the component formula:
[ \begin{aligned} c_1 &= a_2b_3 - a_3b_2\ c_2 &= a_3b_1 - a_1b_3\ c_3 &= a_1b_2 - a_2b_1 \end{aligned} ]
Step 2: Verify Non‑Zero Length If (\mathbf{c}) is the zero vector, the original vectors are parallel (or one is zero) and there is no unique perpendicular direction. In that degenerate case, any vector orthogonal to the common direction works, but the problem usually assumes non‑parallel inputs.
Step 3: Normalize to Obtain Unit Length
A unit vector has magnitude 1. To convert (\mathbf{c}) into a unit vector, divide by its norm:
[ \hat{\mathbf{u}} = \frac{\mathbf{c}}{|\mathbf{c}|} ]
The norm is
[ |\mathbf{c}| = \sqrt{c_1^{2}+c_2^{2}+c_3^{2}} ]
Step 4: Produce the Opposite Unit Vector
Because both (\hat{\mathbf{u}}) and (-\hat{\mathbf{u}}) are orthogonal to a and b, the two unit vectors you seek are
[ \boxed{\hat{\mathbf{u}} \quad\text{and}\quad -\hat{\mathbf{u}}} ]
Worked Example
Let
[\mathbf{a}= (1,,2,,3), \qquad \mathbf{b}= (4,,-5,,6) ]
1. Compute the cross product
[ \begin{aligned} c_1 &= 2\cdot6 - 3\cdot(-5) = 12 + 15 = 27\ c_2 &= 3\cdot4 - 1\cdot6 = 12 - 6 = 6\ c_3 &= 1\cdot(-5) - 2\cdot4 = -5 - 8 = -13 \end{aligned} ]
Thus (\mathbf{c}= (27,,6,,-13)).
2. Find its magnitude
[ |\mathbf{c}| = \sqrt{27^{2}+6^{2}+(-13)^{2}} = \sqrt{729+36+169} = \sqrt{934} \approx 30.56 ]
3. Normalize
[ \hat{\mathbf{u}} = \frac{1}{30.56}(27,,6,,-13) \approx (0.884,;0.196,;-0.426) ]
4. The opposite unit vector
[ -\hat{\mathbf{u}} \approx (-0.884,;-0.196,;0.426) ]
Both vectors have length 1 and satisfy
[\mathbf{a}\cdot\hat{\mathbf{u}} \approx 0,\qquad \mathbf{b}\cdot\hat{\mathbf{u}} \approx 0 ]
(and similarly for (-\hat{\mathbf{u}})).
Common Mistakes and How to Avoid Them
- Skipping the magnitude check – If (\mathbf{c}= \mathbf{0}), the original vectors are parallel; you cannot produce a unique perpendicular direction. Always verify that the cross product is non‑zero.
- Dividing by the wrong value – Remember to divide by the norm, not by any component. Using a component directly will not yield a unit vector.
- Confusing direction with magnitude – The cross product gives a direction; its magnitude is irrelevant for the final unit vectors except as a scaling factor.
- Neglecting the opposite sign – Many students stop after finding one unit vector, forgetting that the problem asks for two unit vectors. Both (\hat{\mathbf{u}}) and (-\hat{\mathbf{u}}) are valid answers.
Frequently Asked Questions (FAQ)
Q1: Can there be more than two unit vectors orthogonal to two given vectors?
A: No. In three‑dimensional Euclidean space, the set of vectors orthogonal to both a and b forms a one‑dimensional subspace (a line). The only unit vectors on that line are the two opposite directions.
**Q2: What if
