Mv7 - Vectors in three dimensions | OCR A-Level Mathematics B (MEI)

Use three-dimensional vector language, operations and position calculations, including spatial directions described by perpendicular projections. Extend distance, ratio, geometric-proof and force-resultant methods by checking all three components.

Components and magnitude in space

A drone's straight displacement is $2\,\mathrm m$ east, $3\,\mathrm m$ south and $6\,\mathrm m$ upwards. How long is the flight, and how can you describe its direction without losing the height? A floor-plan arrow alone cannot distinguish a climb from a descent.

You already know

Cartesian components use perpendicular axes. In a plane, a vector's modulus comes from Pythagoras, and its direction comes from a reference angle with a quadrant check. Divide a non-zero vector by its modulus for a unit vector.

Add a vertical unit vector $\mathbf k$ to the horizontal $\mathbf i,\mathbf j$ . The drone's displacement is $\mathbf u=2\mathbf i-3\mathbf j+6\mathbf k=(2,-3,6)$ . The third component records vertical change; it cannot be swapped with another entry simply because the same three numbers remain.

Two right triangles used to calculate a three-dimensional magnitude
For the vector $(3,4,12)$ , the horizontal projection has length $5$ , and the second right triangle gives spatial length $13$ . The second panel's horizontal axis measures projection length, not the original $x$ -coordinate.

This example joins two integer right triangles: $3^2+4^2=5^2$ and $5^2+12^2=13^2$ . A three-dimensional diagonal can have an integer length even though it is not an edge of either horizontal component triangle.

Spatial modulus and unit vector

For $\mathbf a=(x,y,z)$ :

h=\sqrt{x^2+y^2},\qquad |\mathbf a|=\sqrt{h^2+z^2}=\sqrt{x^2+y^2+z^2}

For $\mathbf a\ne\mathbf0$ , $\hat{\mathbf a}=\mathbf a/|\mathbf a|$ is the unit vector in its direction.

Not in the formula sheet: learn it.

Addition, subtraction and scalar multiplication also extend to all three entries. For example, $(3,4,12)+2(-2,1,0)=(-1,6,12)$ . A zero component indicates no change along that axis; it does not make the whole vector zero.

Worked example 1: a drone's unit direction

Question. Find the magnitude of $\mathbf u=(2,-3,6)$ and a unit vector in its direction.

Plan. Use all three squared components, then divide every component by the resulting modulus.

\begin{aligned} |\mathbf u|&=\sqrt{4+9+36}=7 &&\text{spatial Pythagoras}\\ \hat{\mathbf u}&=\left(\frac27,-\frac37,\frac67\right) &&\text{normalise the complete vector} \end{aligned}

Why this step? Dividing by the floor-plan length $\sqrt{13}$ would ignore the climb and would not give modulus $1$ .

Answer. Magnitude $7$ , unit vector $(2/7,-3/7,6/7)$ .

Check. Its squared modulus is $(4+9+36)/49=1$ .

Change only the vertical sign, turning a climb into a descent.

Worked example 2: reflecting in the horizontal plane

Question. Find the magnitude and a unit vector of $\mathbf v=(2,-3,-6)$ .

Plan. Keep the component signs in the vector but square them when calculating length.

\begin{aligned} |\mathbf v|&=\sqrt{4+9+36}=7 &&\text{unchanged squared components}\\ \hat{\mathbf v}&=\left(\frac27,-\frac37,-\frac67\right) &&\text{retain the downward sign} \end{aligned}

Why this step? The reflected vector has the same length, but is not the negative of $\mathbf u$ : its horizontal components did not reverse.

Answer. Magnitude $7$ , unit vector $(2/7,-3/7,-6/7)$ .

Check. Multiplication by $7$ restores $\mathbf v$ , and the unit vector's modulus is $1$ .

A unit vector records direction without any angles. If a question instead supplies two angle conventions, use two perpendicular triangles. Let azimuth $\theta$ run anticlockwise from the positive $x$ -axis in the horizontal plane, and elevation $\phi$ run above that plane. State these meanings before using the words: another context may choose a different reference.

Spatial direction from two projections

For magnitude $r>0$ , azimuth $\theta$ in $0^\circ\le\theta<360^\circ$ , and elevation $-90^\circ\le\phi\le90^\circ$ :

h=r\cos\phi,\qquad x=h\cos\theta,\qquad y=h\sin\theta,\qquad z=r\sin\phi

For $h>0$ , recover elevation from $\tan\phi=z/h$ and azimuth from the signed horizontal components. A vertical vector has no unique azimuth; the zero vector has no direction.

Not in the formula sheet: learn it.

The intermediate horizontal length makes the conversion manageable. For magnitude $10$ , azimuth $90^\circ$ and elevation $30^\circ$ , $h=5\sqrt3$ , giving $(0,5\sqrt3,5)$ . In general substitute $h=r\cos\phi$ to obtain $x=r\cos\phi\cos\theta$ , $y=r\cos\phi\sin\theta$ .

Azimuth and elevation use different right triangles
The same vector $(3,4,12)$ gives azimuth from the horizontal projection and elevation from height over the full horizontal length.

Worked example 3: reading two spatial angles

Question. Find the azimuth and elevation of $(3,4,12)$ using the conventions above, to the nearest $0.1^\circ$ .

Plan. Find the horizontal projection, then calculate one angle in that plane and one above it.

\begin{aligned} h&=\sqrt{3^2+4^2}=5 &&\text{horizontal length}\\ \theta&=\tan^{-1}(4/3)=53.1301\ldots^\circ &&\text{first horizontal quadrant}\\ \phi&=\tan^{-1}(12/5)=67.3801\ldots^\circ &&\text{height over horizontal length} \end{aligned}

Why this step? Elevation compares vertical change with the whole horizontal projection, not just the $x$ -component.

Answer. Azimuth $53.1^\circ$ , elevation $67.4^\circ$ .

Check. With $r=13$ and unrounded angles, $r\cos\phi=5$ , then $5\cos\theta=3$ , $5\sin\theta=4$ and $r\sin\phi=12$ .

Change only the sign of the $y$ -component. Elevation stays fixed, but the floor-plan arrow crosses into the fourth quadrant.

Worked example 4: choosing the spatial quadrant

Question. Find the azimuth and elevation of $(3,-4,12)$ , to the nearest $0.1^\circ$ .

Plan. Keep the horizontal length and elevation, then correct the azimuth using the horizontal signs.

\begin{aligned} h&=5,\qquad \phi=\tan^{-1}(12/5)=67.3801\ldots^\circ &&\text{same height and projection}\\ \theta&=360^\circ-\tan^{-1}(4/3) &&\text{right and south}\\ &=306.8699\ldots^\circ &&\text{fourth horizontal quadrant} \end{aligned}

Why this step? The principal inverse-tangent result is negative, while our stated azimuth lies between $0^\circ$ and $360^\circ$ .

Answer. Azimuth $306.9^\circ$ , elevation $67.4^\circ$ .

Check. Using the unrounded azimuth, $5\cos\theta=3$ and $5\sin\theta=-4$ ; the height remains $12$ .

The opening drone has horizontal length $\sqrt{13}$ , azimuth $360^\circ-\tan^{-1}(3/2)=303.6901\ldots^\circ$ and elevation $\tan^{-1}(6/\sqrt{13})=58.9973\ldots^\circ$ . Thus one straight $7\,\mathrm m$ displacement records both the floor-plan direction and the climb.

Watch out

Wrong: one planar angle describes the whole spatial direction. Right: retain all three unit-vector components, or use two explicitly defined angles. Never normalise $\mathbf0$ , and never assign a unique azimuth to a vertical vector.

Change the sign of the remaining horizontal component from the last example.

Positions and distances in space

A spatial position vector locates a point from the origin. Its components are that point's coordinates, so finish minus start gives a three-dimensional displacement exactly as it did in a plane.

Spatial displacement and division of a segment

For positions $\mathbf a,\mathbf b$ :

\overrightarrow{AB}=\mathbf b-\mathbf a,\qquad AB=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+(b_3-a_3)^2}

A point at a specified fraction $t$ from $A$ has position $\mathbf a+t(\mathbf b-\mathbf a)$ . Internal ratio $m:n$ gives $t=m/(m+n)$ .

Not in the formula sheet: learn it.

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Mv7 - Vectors in three dimensions

Components and magnitude in space

Positions and distances in space

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