Cross Product Normalization (Unit Vector Perpendicular to Two Vectors)
A frequent job in vector algebra is: given two vectors, find a vector of length 1 that is perpendicular to both of them. The cross product does the perpendicular part; normalization does the length part. Putting them together is what this idea is about.
Step 1 — the cross product gives the direction
For two non-parallel vectors a and b, the cross product
a×b=∣a∣∣b∣sinθn^
is a vector that is perpendicular to botha and b. Its direction is fixed by the right-hand rule. So a×b already points the way we want — but its length is ∣a∣∣b∣sinθ, which is usually not1.
Step 2 — normalize to get unit length
To normalize any non-zero vector means to divide it by its own magnitude, producing a vector of length 1 in the same direction. Applying that to the cross product:
n^=±∣a×b∣a×b
This n^ is a unit vector perpendicular to botha and b. The ± matters: there are exactly two such unit normals, pointing in opposite directions. The plus sign gives the right-hand-rule direction of a×b; the minus sign gives the other side.
Why the division works
Dividing by the magnitude only rescales the vector — it never changes its direction. So n^ keeps the perpendicularity that the cross product built in, while its length becomes ∣a×b∣∣a×b∣=1.
d is parallel to a×b=32i^−j^−14k^; writing d=λ(a×b) and using c⋅d=15 gives λ=35, so d=3160i^−35j^−370k^.
The idea
Any vector perpendicular to both a and b must point along a×b, because the cross product is itself perpendicular to both, and in 3-D the perpendiculars to two non-parallel vectors form a single line. So d can only be a scalar multiple of a×b; the extra condition c⋅d=15 pins down that scalar.
Step-by-step
1. Compute a×b, with a=i^+4j^+2k^, b=3i^−2j^+7k^:
Mistake 1: Solving three scalar equations instead of using the cross product
Why it's wrong: setting d=(x,y,z) with d⋅a=0, d⋅b=0, c⋅d=15 works but is slow and error-prone. Correct approach: recognise d∥a×b and reduce to a single unknown λ.
Mistake 2: Assuming d is a unit vector or equals a×b itself
Why it's wrong: the condition c⋅d=15 fixes the length, generally giving a non-unit multiple. Correct approach: keep the free scalar λ and let the condition determine it. …
Q.The unit vector perpendicular to both vectors i^+k^ and i^−k^ is: (A) 2j^ (B) j^ (C) 2i^−k^ (D) 2i^+k^
›Reveal solutionSolution
To find a vector perpendicular to two given vectors, we use their cross product. Normalizing this resulting vector gives the unit vector. The unit vector perpendicular to i^+k^ and i^−k^ is j^.
Concept and Intuition
When you're asked to find a vector that is perpendicular to two other vectors simultaneously, the most direct and fundamental tool in vector algebra is the cross product.
Imagine two non-parallel vectors originating from the same point. They define a unique plane in space. The cross product of these two vectors yields a new vector that is perpendicular to this entire plane. This means the resulting vector is perpendicular to both of the original vectors.
The cross product of two vectors A and B is given by A×B=(AyBz−AzBy)i^+(AzBx−AxBz)j^+(AxBy−AyBx)k^.
A more convenient way to compute this is using a determinant:
A×B=i^AxBxj^AyByk^AzBz
Once we have a vector that is perpendicular to both, the problem asks for a unit vector. A unit vector is simply a vector with a magnitude of 1, pointing in the same direction as the original vector. To convert any non-zero vector V into a unit vector V^, we divide it by its own magnitude: V^=∣V∣V.
Step-by-step Solution
Identify the given vectors.
Let the two given vectors be A and B.
A=i^+k^
B=i^−k^
We can write these in component form as:
A=1i^+0j^+1k^
B=1i^+0j^−1k^
Calculate the cross product A×B.
This will give us a vector perpendicular to both A and B.
A×B=i^11j^00k^1−1
Expand the determinant:
$= \hat{i}((0)(-1) - (1)(0)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(0) - (0)(1))$
$= \hat{i}(0 - 0) - \hat{j}(-1 - 1) + \hat{k}(0 - 0)$
$= 0\hat{i} - \hat{j}(-2) + 0\hat{k}$
$= 2\hat{j}$
Let's call this resulting vector $\vec{P} = 2\hat{j}$. This vector $\vec{P}$ is perpendicular to both $\hat{i} + \hat{k}$ and $\hat{i} - \hat{k}$.
> [!TIP]
> You can quickly verify perpendicularity by checking the dot product. If $\vec{P} \cdot \vec{A} = 0$ and $\vec{P} \cdot \vec{B} = 0$, then $\vec{P}$ is indeed perpendicular to both.
> $\vec{P} \cdot \vec{A} = (0\hat{i} + 2\hat{j} + 0\hat{k}) \cdot (1\hat{i} + 0\hat{j} + 1\hat{k}) = (0)(1) + (2)(0) + (0)(1) = 0$. …