FP3.5 - Vectors

The vector product $\mathbf a\times\mathbf b$ is a vector perpendicular to both $\mathbf a$ and $\mathbf b$. Its direction follows the right-hand orientation, while its magnitude records the area scale generated by the two vectors:$$|\mathbf a\times\mathbf b|=|\mathbf a|\,|\mathbf b|\sin\theta,$$where $0\le\theta\le\pi$ is the angle between them.

Order matters: $\mathbf b\times\mathbf a=-(\mathbf a\times\mathbf b)$, whereas cyclic changes leave a scalar triple product unchanged. A zero cross product means the two non-zero vectors are parallel. A zero scalar triple product means the three vectors are coplanar, because the third vector has no component perpendicular to the plane of the first two.

Let $\mathbf a=(1,2,-1)$ and $\mathbf b=(3,0,2)$. Then$$\mathbf a\times\mathbf b=(4,-5,-6),\qquad |\mathbf a\times\mathbf b|=\sqrt{77}.$$The triangle with adjacent sides $\mathbf a$ and $\mathbf b$ has area $\sqrt{77}/2$. If $\mathbf c=(0,1,1)$, then$$(\mathbf a\times\mathbf b)\cdot\mathbf c=-11,$$so the parallelepiped volume is $11$; the sign records orientation, not a negative physical volume.

A calculated cross product should satisfy $(\mathbf a\times\mathbf b)\cdot\mathbf a=0$ and $(\mathbf a\times\mathbf b)\cdot\mathbf b=0$. These two dot-product checks catch most component or sign errors before the result is used as a normal or area vector.

A vector equation separates position from direction. A line through the point with position vector $\mathbf a$ and non-zero direction $\mathbf b$ may be written$$\mathbf r=\mathbf a+\lambda\mathbf b$$or equivalently$$(\mathbf r-\mathbf a)\times\mathbf b=\mathbf0.$$The second form says exactly that the displacement from the fixed point is parallel to the line direction.

Each distance formula is a projection. For a point and plane, the displacement is projected onto the unit normal. For skew lines, $\mathbf b_1\times\mathbf b_2$ is perpendicular to both directions, so projecting any joining displacement onto this common normal gives the fixed perpendicular separation.

For the plane $2x-y+2z=6$ and point $P=(1,1,1)$, the normal is $\mathbf n=(2,-1,2)$. Substitution gives the signed residual $2-1+2-6=-3$, so$$d=\frac{|-3|}{\sqrt{2^2+(-1)^2+2^2}}=1.$$The absolute value is essential because distance is non-negative.

For$$\ell_1:\mathbf r=\lambda(1,0,1),\qquad \ell_2:\mathbf r=(0,1,0)+\mu(0,1,1),$$the common normal is $(1,0,1)\times(0,1,1)=(-1,-1,1)$. With joining vector $(0,1,0)$,$$d=\frac{|(0,1,0)\cdot(-1,-1,1)|}{\sqrt3}=\frac1{\sqrt3}.$$Solving the coordinate equations confirms the lines do not intersect, so the skew formula applies.

For two plane equations, first cross their normals to obtain the intersection-line direction. Then choose one coordinate conveniently and solve the two simultaneous linear equations for a point. Substitute the final parametric line into both plane equations: both identities must hold for every parameter value.

A plane is fixed by one point and two independent in-plane directions, or by one point and a non-zero normal. These descriptions produce equivalent parametric, vector-normal and Cartesian equations; converting between them reveals the geometry needed for intersections and angles.

From a parametric form, calculate $\mathbf n=\mathbf b\times\mathbf c$ and then $p=\mathbf a\cdot\mathbf n$. From a Cartesian form, read off the normal and find any convenient point satisfying the equation; two independent vectors perpendicular to the normal may then serve as parametric directions.

A plane through $A=(1,-1,2)$ with directions $\mathbf b=(1,2,0)$ and $\mathbf c=(0,1,1)$ has$$\mathbf n=\mathbf b\times\mathbf c=(2,-1,1).$$Since $\mathbf a\cdot\mathbf n=2+1+2=5$, its normal and Cartesian forms are$$\mathbf r\cdot(2,-1,1)=5,qquad 2x-y+z=5.$$Substituting $A$ and both direction displacements verifies the conversion.

If three non-collinear points $A,B,C$ define a plane, use $\overrightarrow{AB}$ and $\overrightarrow{AC}$ as the two directions. Their cross product is a normal. If the cross product is zero, the points are collinear and do not determine a unique plane.

The acute angle $\theta$ between two planes is the acute angle between their normals:$$\cos\theta=\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1|\,|\mathbf n_2|},\qquad 0\le\theta\le\frac\pi2.$$The absolute value selects the acute angle because reversing a normal does not change its plane.