Unit Vectors — Question 6

Question 6

A missing unit-vector component has branches

A spatial unit vector has $u_x=1/3$, $u_y=2/3$, and unknown $u_z$. The basis is orthonormal.

1. Find all allowed values of $u_z$.

2. Select the direction if the vector points toward negative $z$, then form a 9 m displacement in that direction.

3. Can a unit vector have $u_x=0.8$ and $u_y=0.8$? Explain.

Question 6 - Solution

A missing unit-vector component has branches

(a) Enforce unit magnitude.

$$u_z^2=1-\frac19-\frac49=\frac49\Rightarrow\boxed{u_z=\pm\frac23}$$ $u_x^2+u_y^2+u_z^2=1$ gives $u_z^2=1-1/9-4/9=4/9$. Therefore $u_z=\pm2/3$. The first two components alone leave two directions, mirrored across the $xy$ plane.

(b) Use the directional condition.

$$u_z<0\Rightarrow\vec D=9(1,2,-2)/3=(3,6,-6)\,\mathrm m,\quad D=9\,\mathrm m$$ Negative $z$ selects $\hat u=(\hat\imath+2\hat\jmath-2\hat k)/3$. The requested displacement is $\vec D=(9\ \mathrm m)\hat u=(3\hat\imath+6\hat\jmath-6\hat k)\ \mathrm m$, whose length is $\sqrt{9+36+36}=9$ m.

(c) Reject inconsistent partial data.

$$u_z^2=1-.8^2-.8^2=-.28<0\Rightarrow\text{no real unit vector}$$ No: $u_x^2+u_y^2=1.28>1$ would require $u_z^2=-0.28$. Although each proposed component is individually less than one, they jointly exceed the available squared norm. Normalization is a constraint on the whole vector.