1.6 Vectors
- Syllabus
- 9231–2028–2029
- Topic
- 1.6
- Level
- AS
Learning objectives
1.6.1Plane equations• use the equation of a plane in any of the forms ax + by + cz = d or r.n = p or r = a + λb + μc and convert equations of planes from one form to another as necessary in solving problems1.6.2Vector product• recall that the vector product ab# of two vectors can be expressed either as sinab ni t, where nt is a unit vector, or in component form as ab ab ab abij23 32 31 13-- +`` jj ab ab k12 21-+`j1.6.3Lines and planes• use equations of lines and planes, together with scalar and vector products where appropriate, to solve problems concerning distances, angles and intersections, including - determining whether a line lies in a plane, is parallel to a plane or intersects a plane, and finding the point of intersection of a line and a plane when it exists - finding the foot of the perpendicular from a point to a plane - finding the angle between a line and a plane, and the angle between two planes - finding an equation for the line of intersection of two planes - calculating the shortest distance between two skew lines - finding an equation for the common perpendicular to two skew lines.
A plane has three equivalent equations with different useful data
| Form | Meaning | Best use |
|---|---|---|
| ax+by+cz=d | Normal n=(a,b,c) | Membership, distances and intersections |
| r⋅n=p | Same normal form, with p=d | Compact vector reasoning |
| r=a+λb+μc | Point a and two non-parallel in-plane directions b,c | Generating points and lines in the plane |
Parametric to normal: compute n=b×c, then p=a⋅n. Normal to parametric: choose one point satisfying the equation and two independent vectors b,c satisfying b⋅n=c⋅n=0. Scale any equation or normal by a non-zero constant without changing the plane.
For\mathbf r=\begin{pmatrix}1\0\2\end{pmatrix}+\lambda\begin{pmatrix}1\1\0\end{pmatrix}+\mu\begin{pmatrix}0\1\1\end{pmatrix},\mathbf n=\begin{pmatrix}1\1\0\end{pmatrix}\times\begin{pmatrix}0\1\1\end{pmatrix}=\begin{pmatrix}1\-1\1\end{pmatrix},\qquad p=3.Hence $\mathbf r\cdot(1,-1,1)=3$, orx-y+z=3.
Verify that the base point satisfies the scalar equation and both direction vectors dot to zero with the normal. One point alone does not determine a plane; two parallel directions also fail to span one.
The vector product encodes perpendicular direction and swept area
For the angle $0\le\theta\le\pi$ between non-zero vectors a and b,\mathbf a\times\mathbf b=|\mathbf a|,|\mathbf b|\sin\theta,\hat{\mathbf n},where $\hat{\mathbf n}$ is the unit normal selected by the right-hand rule. Thus $|\mathbf a\times\mathbf b|$ is the parallelogram area and half of it is the triangle area.
If $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf b=(b_1,b_2,b_3)$,\mathbf a\times\mathbf b=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1).
For $\mathbf a=(1,2,-1)$ and $\mathbf b=(3,0,2)$,\mathbf a\times\mathbf b=(4,-5,-6).Check: $(4,-5,-6)\cdot(1,2,-1)=0$ and $(4,-5,-6)\cdot(3,0,2)=0$. Its magnitude is $\sqrt{77}$, so a unit normal is $\pm(4,-5,-6)/\sqrt{77}$.
b×a=−(a×b), and the product is zero exactly when the vectors are parallel (or one is zero). It is a vector, unlike the scalar dot product; verify both perpendicular dot products to catch component-sign errors.
Reduce every line-plane problem to membership or perpendicularity
For line r=a+λd and plane r⋅n=p:
| Question | Decisive calculation |
|---|---|
| Line versus plane | If d⋅n=0, substitute and solve one intersection. If it is 0, the line lies in the plane when a⋅n=p and is otherwise parallel/disjoint. |
| Foot H from point q | Set H=q+tn, where t=(p−q⋅n)/∣n∣2. |
| Point-plane distance | ∣q⋅n−p∣/∣n∣. |
| Acute line-plane angle alpha | sinα=∣d⋅n∣/(∣d∣∣n∣). |
| Acute plane-plane angle theta | cosθ=∣n1⋅n2∣/(∣n1∣∣n2∣). |
For two non-parallel planes, their intersection direction is n1×n2. Find one point satisfying both scalar equations (choose a convenient coordinate, then solve), and write the line as point plus a parameter times that direction.
For skew lines $L_1:\mathbf r=\mathbf a+\lambda\mathbf u$ and $L_2:\mathbf r=\mathbf b+\mu\mathbf v$, let $\mathbf w=\mathbf u\times\mathbf v$. Their shortest distance is\frac{|(\mathbf b-\mathbf a)\cdot\mathbf w|}{|\mathbf w|}. Tofindthecommonperpendicularitself,solve[\mathbf b+\mu\mathbf v-(\mathbf a+\lambda\mathbf u)]\cdot\mathbf u=0,[\mathbf b+\mu\mathbf v-(\mathbf a+\lambda\mathbf u)]\cdot\mathbf v=0.The resulting points P and Q determine $\mathbf r=\mathbf P+t(\mathbf Q-\mathbf P)$.
Example: from Q=(4,2,9) to 6x−y+8z=−7, use H=Q+t(6,−1,8). Substitution gives 94+101t=−7, so t=−1 and H=(−2,3,1). The displacement H−Q=(−6,1,−8) is parallel to the normal, and H satisfies the plane equation.
A line-plane angle is complementary to the direction-normal angle, which is why its formula uses sine. For skew lines, the shortest segment must be perpendicular to both directions; the distance between arbitrary selected points is not the answer.