Ch 6Applications of Vector Algebra — Class 12 Mathematics, concept-first.

Vectors first appeared in your Class XI work as directed quantities — objects with both a magnitude and a direction, written or, in component form, .

A vector is represented as a directed straight-line segment in 3-dimensional space : it has an initial point and an end point , and is written .

Definition 6.1. For and :

Projections. If is any vector and a unit vector, then is the (signed) projection of onto the line along : it is positive when the angle between and is acute, and negative when that angle is obtuse.

Dot and cross products give slick, purely algebraic proofs of several plane-trigonometry results, with usual triangle notation ().

Placing a convenient vertex at the origin turns several classical Euclidean-geometry theorems into short vector computations.

Definition 6.2 (Work). If a constant force acts on a particle while it is displaced by (from one point to another), the work done by the force is Since , the work is positive when the force has an acu…

Definition 6.4. For three vectors , the scalar is called the scalar triple product of .

Theorem 6.2. For any three vectors : . Proof: both equal the same determinant , up to the row swaps then which cancel each other's sign flip.

Definition 6.5. For three vectors , the vector is called a vector triple product. Given any three vectors, the genuine vector triple products are and .

Theorem 6.9 (Jacobi's Identity). For any three vectors : Proof. Expand each of the three terms with Theorem 6.8: , , .

Vectors give an elegant, coordinate-free-until-you-need-it way to describe straight lines and planes in three dimensions.

A straight line is uniquely fixed by either of two kinds of data: - a point on the line, together with the direction of the line, or - two points on the line (the direction is then just the vector joi…

Theorem 6.11 (parametric vector equation). The line through the fixed point with position vector , parallel to a given vector , has vector equation Proof.

Theorem 6.12. The line through two given points with position vectors and has parametric vector equation This follows immediately from Theorem 6.11 by taking the direction to be (the vector from the f…

(a) Vector form. For two lines and , the acute angle between them equals the acute angle between their direction vectors : (The absolute value keeps the acute one, since a line has no preferred sense…

Given two lines and , every point on the first is of the form and every point on the second is , for parameters .

Definition 6.6. Two lines are coplanar if they lie in a common plane. Two lines that are either parallel or intersecting are automatically coplanar.

Definition 6.8. A vector perpendicular to a plane is called a normal to the plane.

Theorem 6.15 (Normal form). The plane at perpendicular distance from the origin, with unit normal vector , has vector equation Proof. Let be the foot of the perpendicular from to the plane, so .

(a) Vector form. Let the plane pass through a point (position vector ) with normal vector , and let (position vector ) be any point of the plane. Then is perpendicular to , so , i.e.

Suppose the plane meets the coordinate axes at with intercepts . Since lies on the plane, , i.e. ; similarly and . Substituting into gives , i.e. .

Theorem 6.17 (a) Parametric vector form. If (position vectors ) are three non-collinear points, the vector equation of the plane through them, in parametric form, is Proof idea: take on ray with and (…

Let the plane pass through a given point (position vector ) and be parallel to two given non-parallel vectors and .

Let the plane pass through two given distinct points (position vectors ) and be parallel to a non-zero vector that is NOT parallel to (otherwise all three directions would collapse to a line, not fix…

A straight line lies wholly in a plane precisely when two conditions both hold: every point of the line lies in the plane, AND the plane's normal is perpendicular to the line's direction (so the line…

(a) Vector form. Two non-parallel lines and are coplanar exactly when they lie in a single plane. Let (position vectors ) be points on the two lines.

Once two lines and are known to be coplanar (via §6.8.8), tells us the plane's normal direction, and either base point tells us where the plane sits.

The angle between two planes is defined to be the angle between their normals.

The angle between a line and a plane is the complement of the angle between the line's direction and the plane's normal — because a line lying flat IN the plane makes a angle with the plane while its…

Theorem 6.20 (vector form). The perpendicular distance from a point with position vector to the plane is Proof sketch.

Theorem 6.21. The distance between the two PARALLEL planes and (same normal direction ratios ) is Proof. Pick any point on the SECOND plane, so , i.e. .

Let and be two non-parallel planes, so are normal to them respectively. The line where they meet must be perpendicular to BOTH normals — hence it is parallel to .

Theorem 6.22. The vector equation of a plane passing through the line of intersection of the planes and is Proof. Regroup the left side as , i.e.

Let (position vector ) be a given point and a given plane. The mirror image (reflection) of in the plane is the point with position vector such that the plane is the perpendicular bisector of segment…

The image formula of §6.9 restates cleanly in pure coordinates. Let be the given point (so ) and let be the given plane (so , ). Writing the image as , matching -components of gives

Theorem 6.23. The point where the line meets the plane (assuming the line is not parallel to the plane, i.e. ) has position vector Proof.

Exercise 6.10 is a 25-question multiple-choice set (choose the correct/most suitable answer from four alternatives) that revises the entire chapter in one pass: scalar and vector triple products and t…

Sample papers and previous-year board questions for this subject.