Further vectors

Explanation

• A line through the point with position vector $\mathbf a$ and parallel to non-zero direction $\mathbf d$ has equation $\mathbf r=\mathbf a+\lambda\mathbf d$. Component equations follow immediately.
• When every direction component is non-zero, eliminating $\lambda$ gives Cartesian form $(x-a_1)/d_1=(y-a_2)/d_2=(z-a_3)/d_3$.
• A zero direction component instead gives a fixed coordinate, never a zero denominator.
• Through points $A$ and $B$, a suitable direction is $\overrightarrow{AB}=\mathbf b-\mathbf a$; any non-zero scalar multiple describes the same line.
• To intersect two lines, equate components and check that one common pair of parameters satisfies all three equations; parallel and skew lines do not meet.

Explanation

• A plane through position vector $\mathbf a$ and parallel to independent vectors $\mathbf b,\mathbf c$ has vector equation $\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c$. The spanning vectors must not be parallel.
• A Cartesian equation has form $px+qy+rz=d$, where $\mathbf n=(p,q,r)$ is normal to the plane and $d=\mathbf n\cdot\mathbf a$.
• To convert from vector form, find a non-zero vector perpendicular to both spanning directions by solving simultaneous scalar-product equations.
• To form a plane through three non-collinear points, subtract one point from the other two to obtain spanning directions.
• Substitution of each given point should confirm the final Cartesian equation.

Explanation

• The scalar product satisfies $\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta$.
• It gives the angle between lines from their direction vectors and the angle between planes from their normal vectors; use an absolute value when the acute angle is required.
• A plane through $\mathbf a$ with normal $\mathbf n$ has equation $\mathbf n\cdot(\mathbf r-\mathbf a)=0$, equivalently $\mathbf r\cdot\mathbf n=k$.
• For a line direction $\mathbf d$ and plane normal $\mathbf n$, the acute line-plane angle $\alpha$ satisfies $\sin\alpha=|\mathbf d\cdot\mathbf n|/(|\mathbf d||\mathbf n|)$ because it complements the direction-normal angle.
• The vector roles must therefore be identified before choosing sine or cosine.

Explanation

• Two non-zero vectors are perpendicular exactly when their scalar product is zero. In three dimensions, multiply corresponding components and add all three products.
• The vectors may originate at different points; only their directions matter, so displacement vectors must first be formed in a consistent order when points are given.
• An unknown component or parameter can be found by setting the scalar product equal to zero and solving the resulting equation.
• A diagram cannot prove perpendicularity, and one pair of components is insufficient.
• The zero vector has zero scalar product with every vector but has no direction, so it must not be described as perpendicular in the ordinary geometric sense.

Explanation

• To intersect a line and plane, substitute the line's parametric coordinates into the plane equation, solve for the parameter and recover all three coordinates. If the parameter equation is inconsistent the line is parallel to the plane; if it is an identity the line lies in the plane.
• The point-to-plane distance is the absolute value obtained by substituting the point into the plane expression, divided by the normal's magnitude.
• For point-to-line distance, choose a general point on the line and make its connector to the fixed point perpendicular to the line direction.
• For two lines, the shortest connector is perpendicular to both directions; this also handles skew lines.
• Every distance is non-negative.