Vector Projection Equality: Finding the Unit Vector
The problem asks for a unit vector $\vec{a}$ where its projections onto three specific vectors are equal. Let the vector $\vec{a}$ be represented as $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$. We are given that $\vec{a}$ is a nonzero vector.
Vector Definitions and Magnitudes
Let the three vectors be:
- $\vec{v_1} = 2\hat{i}-\hat{j}+2\hat{k}$
- $\vec{v_2} = \hat{i}+2\hat{j}-2\hat{k}$
- $\vec{v_3} = \hat{k}$
Calculate the magnitudes of these vectors:
- $|\vec{v_1}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$
- $|\vec{v_2}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1+4+4} = \sqrt{9} = 3$
- $|\vec{v_3}| = \sqrt{0^2 + 0^2 + 1^2} = \sqrt{1} = 1$
Projection Calculations
The projection of vector $\vec{a}$ onto a vector $\vec{v}$ is given by the formula $\frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}$.
The projections are:
- Projection onto $\vec{v_1}$: $\frac{\vec{a} \cdot \vec{v_1}}{|\vec{v_1}|} = \frac{(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i}-\hat{j}+2\hat{k})}{3} = \frac{2x - y + 2z}{3}$
- Projection onto $\vec{v_2}$: $\frac{\vec{a} \cdot \vec{v_2}}{|\vec{v_2}|} = \frac{(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i}+2\hat{j}-2\hat{k})}{3} = \frac{x + 2y - 2z}{3}$
- Projection onto $\vec{v_3}$: $\frac{\vec{a} \cdot \vec{v_3}}{|\vec{v_3}|} = \frac{(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{k})}{1} = z$
Equating Projections
We are given that these projections are equal. Setting the projections equal to each other:
- $\frac{2x - y + 2z}{3} = z \implies 2x - y + 2z = 3z \implies 2x - y - z = 0 \quad (1)$
- $\frac{x + 2y - 2z}{3} = z \implies x + 2y - 2z = 3z \implies x + 2y - 5z = 0 \quad (2)$
Solving the System of Equations
Solve the system of linear equations (1) and (2) for $x, y, z$. From equation (1), express $y$ in terms of $x$ and $z$: $y = 2x - z$. Substitute this into equation (2):
$x + 2(2x - z) - 5z = 0$
$x + 4x - 2z - 5z = 0$
$5x - 7z = 0 \implies 5x = 7z$
Let $x = 7k$ for some non-zero scalar $k$. Then $z = 5k$. Substitute these values back into the expression for $y$: $y = 2(7k) - 5k = 14k - 5k = 9k$.
Thus, the vector $\vec{a}$ is proportional to $7k\hat{i} + 9k\hat{j} + 5k\hat{k}$, or $k(7\hat{i} + 9\hat{j} + 5\hat{k})$.
Calculating the Unit Vector
A unit vector along $\vec{a}$ has the same direction but a magnitude of 1. The direction vector is $7\hat{i} + 9\hat{j} + 5\hat{k}$.
Calculate the magnitude of this direction vector:
$|\text{direction vector}| = \sqrt{7^2 + 9^2 + 5^2} = \sqrt{49 + 81 + 25} = \sqrt{155}$
The unit vector is the direction vector divided by its magnitude:
Unit vector = $\frac{7\hat{i} + 9\hat{j} + 5\hat{k}}{\sqrt{155}} = \frac{1}{\sqrt{155}}(7\hat{i} + 9\hat{j} + 5\hat{k})$
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