Problem 27
Hard Vector Algebra

Let i , j and k be unit vectors of the coordinate axes of a…

Let $\vec{i}$, $\vec{j}$ and $\vec{k}$ be unit vectors of the coordinate axes of a spatial coordinate system. Decompose vector $\overrightarrow{OM}$ in terms of $\vec{i}$, $\vec{j}$ and $\vec{k}$ analytically and geometrically, where $M(x,y,z)$ is an arbitrary point in space and $O$ is the origin.

Explanation of Method

Decomposition of the position vector $\overrightarrow{OM}$ via $\vec i$, $\vec j$, $\vec k$.

Step 1: Setup

The unit vectors $\vec{i} = (1,0,0)$, $\vec{j} = (0,1,0)$, $\vec{k} = (0,0,1)$ are mutually orthogonal and form a right-handed orthonormal basis of space. The point $M$ has coordinates $(x,y,z)$.

Step 2: Geometric derivation

Step 1: project onto the $xy$-plane

Drop a perpendicular from $M(x,y,z)$ to the $xy$-plane; its foot is $M'(x,y,0)$. Then

\[ \overrightarrow{OM} = \overrightarrow{OM'} + \overrightarrow{M'M}, \qquad \overrightarrow{M'M} = z\,\vec{k}, \] because $M'M$ is parallel to the $z$-axis and has signed length $z$.

Step 2: decompose $\overrightarrow{OM'}$ in the $xy$-plane

Drop a perpendicular from $M'$ to the $x$-axis; its foot is $M_x(x,0,0)$. Then

\[ \overrightarrow{OM'} = \overrightarrow{OM_x} + \overrightarrow{M_xM'}, \qquad \overrightarrow{OM_x} = x\,\vec{i}, \qquad \overrightarrow{M_xM'} = y\,\vec{j}. \]

Step 3: combine

\[ \overrightarrow{OM} = \overrightarrow{OM_x} + \overrightarrow{M_xM'} + \overrightarrow{M'M} = x\,\vec{i} + y\,\vec{j} + z\,\vec{k} \]

\[\overrightarrow{OM} = x\,\vec{i} + y\,\vec{j} + z\,\vec{k}\]

The vector components are $x\vec i$, $y\vec j$, $z\vec k$; the scalar components (projections on the axes) are exactly the coordinates $x$, $y$, $z$.

x y z $\vec i$ $\vec j$ $\vec k$ $O$ $M_x$ $M'$ $M(x,y,z)$ $x\vec i$ $y\vec j$ $z\vec k$
  • $\overrightarrow{OM}=x\vec i+y\vec j+z\vec k$
  • $\overrightarrow{OM'}=x\vec i+y\vec j$
Path $O\to M_x\to M'\to M$ along the edges of the box; the unit vectors $\vec i,\vec j,\vec k$ are shown separately (top left).

Step 3: Magnitude

Applying the Pythagorean theorem twice (triangles $OM_xM'$ and $OM'M$ are right-angled): $|OM'| = \sqrt{x^2+y^2}, \qquad |OM| = \sqrt{|OM'|^2+z^2}$.

\[|\overrightarrow{OM}| = \sqrt{x^2+y^2+z^2}\]

Step 4: Direction cosines

The angles $\alpha,\beta,\gamma$ that $\overrightarrow{OM}$ makes with the $x$-, $y$-, $z$-axes satisfy (see Problem 29)

\[ \cos\alpha = \frac{x}{|\overrightarrow{OM}|}, \quad \cos\beta = \frac{y}{|\overrightarrow{OM}|}, \quad \cos\gamma = \frac{z}{|\overrightarrow{OM}|}, \qquad \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1. \]

Step 5: Numerical example

For $M(3, 2, 4)$: \[\overrightarrow{OM} = 3\vec{i}+2\vec{j}+4\vec{k}, \qquad |\overrightarrow{OM}| = \sqrt{9+4+16} = \sqrt{29},\]

\[ \cos\alpha = \tfrac{3}{\sqrt{29}},\quad \cos\beta = \tfrac{2}{\sqrt{29}},\quad \cos\gamma = \tfrac{4}{\sqrt{29}}, \qquad \tfrac{9+4+16}{29} = 1 \;\checkmark \]


Result

ItemResult
Decomposition$\overrightarrow{OM} = x\vec{i} + y\vec{j} + z\vec{k}$
Magnitude$\sqrt{x^2+y^2+z^2}$
Geometric path$O\xrightarrow{x\vec{i}} M_x \xrightarrow{y\vec{j}} M' \xrightarrow{z\vec{k}} M$