Unit Vectors — Question 5

Question 5

Normalize a spatial direction ratio

A cable points in the spatial direction ratio $2:-1:2$ relative to a right-handed orthonormal basis $\hat\imath,\hat\jmath,\hat k$. A displacement along that direction has length 18 m.

1. Construct a dimensionless unit vector along the cable.

2. Find the displacement components and verify their magnitude.

3. A second cable has ratio $-4:2:-4$. Is it in the same direction, the opposite direction, or neither?

Question 5 - Solution

Normalize a spatial direction ratio

(a) Normalize the ratio, not each entry separately.

$$\boxed{\hat u=\frac{(2,-1,2)}{\sqrt{4+1+4}}=\tfrac13(2,-1,2)}$$ The ratio vector has length $\sqrt{2^2+(-1)^2+2^2}=3$. Thus $\hat u=(2\hat\imath-\hat\jmath+2\hat k)/3$. Dividing by 3 gives squared length $4/9+1/9+4/9=1$.

(b) Multiply by the physical length.

$$\vec A=18\hat u=(12,-6,12)\,\mathrm m,\quad A=\sqrt{144+36+144}=18\,\mathrm m$$ $\vec A=(18\ \mathrm m)\hat u=(12\hat\imath-6\hat\jmath+12\hat k)\ \mathrm m$. Its magnitude is $\sqrt{144+36+144}=18$ m. The negative $y$ component specifies the sense along that axis.

(c) Compare normalized ratios.

$$\frac{(-4,2,-4)}{\sqrt{16+4+16}}=\tfrac13(-2,1,-2)=-\hat u$$ The second ratio is $-2$ times the first. Its norm is 6, so its unit direction is $(-2\hat\imath+\hat\jmath-2\hat k)/3=-\hat u$. It lies on a parallel line but points in the opposite sense. Direction ratios can be rescaled by a positive number without changing direction.