TN 12th Maths — Learn, Visualize, Practice

1. The scalar (dot) product

The dot product takes two vectors and returns a single number — a scalar — that measures how much the two vectors point in the same direction. It is the quiet workhorse behind angles, projections, work done by a force, and the test for perpendicularity.

DefinitionScalar (dot) product

For vectors $\vec a$ and $\vec b$ with angle $\theta$ between them $(0 \le \theta \le \pi)$ , the dot product is $\vec a \cdot \vec b = |\vec a|\,|\vec b|\cos\theta$ . In components, if $\vec a = a_1\vec i + a_2\vec j + a_3\vec k$ and $\vec b = b_1\vec i + b_2\vec j + b_3\vec k$ , then $\vec a \cdot \vec b = a_1 b_1 + a_2 b_2 + a_3 b_3$ .

The result is a scalar, not a vector. Geometrically, $|\vec b|\cos\theta$ is the length of the projection of $\vec b$ onto $\vec a$ , so $\vec a \cdot \vec b$ is $|\vec a|$ times that projection — a measure of shared direction.

Angle between two vectors

\cos\theta = \dfrac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|}

Projection of b on a (a scalar length)

\text{proj}_{\vec a}\vec b = \dfrac{\vec a \cdot \vec b}{|\vec a|}

Key idea: Two non-zero vectors are perpendicular exactly when $\vec a \cdot \vec b = 0$, because $\cos 90^\circ = 0$. Open the Visualize tab and rotate the vectors until the dot product hits zero.

Properties of the dot product

Commutative: $\vec a \cdot \vec b = \vec b \cdot \vec a$
Distributive: $\vec a \cdot (\vec b + \vec c) = \vec a \cdot \vec b + \vec a \cdot \vec c$
$\vec a \cdot \vec a = |\vec a|^2$
For the standard basis: $\vec i \cdot \vec i = \vec j \cdot \vec j = \vec k \cdot \vec k = 1$ and $\vec i \cdot \vec j = \vec j \cdot \vec k = \vec k \cdot \vec i = 0$

ExampleDot product and angle

Find

\vec a \cdot \vec b

and the angle between

\vec a = \vec i + 2\vec j + 2\vec k

and

\vec b = 2\vec i + \vec j - 2\vec k

.

ExampleA genuinely non-right angle

Find the angle between

\vec a = \vec i + \vec j

and

\vec b = \vec i + 2\vec j + 2\vec k

.

ExampleProjection (the vector shadow)

Find the length of the projection of

\vec b = 3\vec i + \vec j - \vec k

onto

\vec a = 2\vec i - \vec j + 2\vec k

.

ExampleWork done by a force

A constant force

\vec F = 2\vec i + \vec j + \vec k

moves a particle from

A(1,0,1)

to

B(3,2,0)

. Find the work done.

2. The vector (cross) product

The cross product of two vectors in 3D returns a new vector — one that is perpendicular to both inputs. Its length records the area they span, and its direction records the orientation. Unlike the dot product, order matters.

DefinitionVector (cross) product

The magnitude $|\vec a \times \vec b|$ equals the area of the parallelogram with $\vec a$ and $\vec b$ as adjacent sides. Half of it is the area of the triangle formed by the two vectors.

Determinant form

\vec a \times \vec b = \begin{vmatrix} \vec i & \vec j & \vec k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\vec i - (a_1 b_3 - a_3 b_1)\vec j + (a_1 b_2 - a_2 b_1)\vec k

Areas & unit normal

\text{Area}_{\parallel} = |\vec a \times \vec b|,\quad \text{Area}_{\triangle} = \tfrac{1}{2}|\vec a \times \vec b|,\quad \hat n = \dfrac{\vec a \times \vec b}{|\vec a \times \vec b|}

Note: The cross product is anti-commutative: $\vec a \times \vec b = -(\vec b \times \vec a)$. Reversing the order flips the resulting vector to the opposite direction.

Key idea: Two non-zero vectors are parallel exactly when $\vec a \times \vec b = \vec 0$, because $\sin 0^\circ = 0$. In the explorer this shows up as the $z$-component (and the parallelogram area) collapsing to zero.

Standard basis cross products

$\vec i \times \vec j = \vec k,\quad \vec j \times \vec k = \vec i,\quad \vec k \times \vec i = \vec j$ (cyclic)
$\vec i \times \vec i = \vec j \times \vec j = \vec k \times \vec k = \vec 0$
Reversing any of the above negates the result, e.g. $\vec j \times \vec i = -\vec k$

ExampleCross product and area

Find

\vec a \times \vec b

and the area of the parallelogram for

\vec a = \vec i + \vec j + \vec k

and

\vec b = \vec i - \vec j + \vec k

.

ExampleArea of a triangle from its vertices

Find the area of the triangle with vertices

A(1,1,1)

,

B(2,3,1)

,

C(1,2,3)

.

ExampleUnit vector perpendicular to two vectors

Find a unit vector perpendicular to both

\vec a = 2\vec i - \vec j + \vec k

and

\vec b = 3\vec i + 4\vec j - \vec k

.

3. The scalar triple product

Combine three vectors by taking the cross product of two, then the dot product with the third. The single number you get measures the volume of the box they span — and its sign tells you the orientation.

DefinitionScalar triple product

$[\vec a\ \vec b\ \vec c] = \vec a \cdot (\vec b \times \vec c)$ . In components it is the $3\times 3$ determinant of the rows of coordinates.

Determinant form

[\vec a\ \vec b\ \vec c] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}

The absolute value $\big|[\vec a\ \vec b\ \vec c]\big|$ is the volume of the parallelepiped with $\vec a$ , $\vec b$ , $\vec c$ as adjacent edges. The tetrahedron on the same three edges has one-sixth of that volume.

Key idea: Three vectors are coplanar exactly when $[\vec a\ \vec b\ \vec c] = 0$ — the box they would span is flat, so it has zero volume.

Properties

Cyclic invariance: $[\vec a\ \vec b\ \vec c] = [\vec b\ \vec c\ \vec a] = [\vec c\ \vec a\ \vec b]$
Swapping any two vectors changes the sign
If any two of the three vectors are equal, the product is $0$
$\vec a \cdot (\vec b \times \vec c) = (\vec a \times \vec b) \cdot \vec c$ — dot and cross may be interchanged

ExampleVolume of a parallelepiped

Find the volume of the parallelepiped whose edges are

\vec a = \vec i + 2\vec j + 3\vec k

,

\vec b = 2\vec i + \vec j + \vec k

,

\vec c = \vec j + 2\vec k

.

ExampleCoplanarity — solve for a parameter

Find

\lambda

so that

\vec a = \vec i + \vec j + \vec k

,

\vec b = 2\vec i + 3\vec j + \vec k

,

\vec c = \vec i + \lambda\vec j + 3\vec k

are coplanar.

4. The vector triple product

Cross two vectors and cross the result again with a third, and you land back inside the plane of the last two vectors. The expansion below — sometimes called the BAC–CAB rule — lets you avoid computing nested cross products.

Vector triple product identity

\vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c)\,\vec b - (\vec a \cdot \vec b)\,\vec c

Note: The result lies in the plane of $\vec b$ and $\vec c$. Note the cross product is not associative: in general $\vec a \times (\vec b \times \vec c) \ne (\vec a \times \vec b) \times \vec c$, so the brackets matter.

ExampleApplying the identity abstractly

Simplify

\vec a \times (\vec b \times \vec c)

given

\vec a \cdot \vec b = 4

and

\vec a \cdot \vec c = -1

.

ExampleWith explicit vectors

Evaluate

\vec a \times (\vec b \times \vec c)

for

\vec a = \vec i + \vec j

,

\vec b = \vec i + \vec k

,

\vec c = \vec j + \vec k

.

5. Equation of a straight line

A line is fixed either by a point plus a direction, or by two points. Every form below is a repackaging of one idea: from a base point $\vec a$ , slide along a direction $\vec b$ by any amount $t$ .

Vector form (parametric)

\vec r = \vec a + t\,\vec b,\qquad t \in \mathbb{R}

Cartesian (symmetric) form

\dfrac{x - x_1}{b_1} = \dfrac{y - y_1}{b_2} = \dfrac{z - z_1}{b_3}

Through two points $A$ and $B$ , the direction is just $\vec b = \overrightarrow{AB} = \vec b_{\text{pos}} - \vec a$ , so $\vec r = \vec a + t(\vec b_{\text{pos}} - \vec a)$ .

Note: The denominators $b_1, b_2, b_3$ in the Cartesian form are direction ratios, not coordinates. If one direction ratio is $0$, that fraction is written separately, e.g. $\dfrac{x-1}{2} = \dfrac{y-3}{-1},\ z = 4$.

ExampleLine: all three forms

A line passes through

(1, 2, -3)

and is parallel to

4\vec i + 5\vec j - 7\vec k

. Give its vector and Cartesian equations.

ExampleLine through two points, and where it meets a plane

Find the line through

(-5, 7, -4)

and

(13, -5, 2)

, and the point where it crosses the

xy

-plane.

ExampleAngle between two lines

Find the acute angle between lines with directions

\vec b = 2\vec i + 2\vec j - \vec k

and

\vec d = 6\vec i + 3\vec j + 2\vec k

.

6. Equation of a plane

A plane in space is pinned down by a point on it and a direction perpendicular to it — the normal vector $\vec n$ . Every description below is really the single condition: the vector from a fixed point to any point on the plane is perpendicular to $\vec n$ .

Vector / normal form

(\vec r - \vec a) \cdot \vec n = 0 \quad\Longleftrightarrow\quad \vec r \cdot \vec n = \vec a \cdot \vec n

Cartesian (general) form

ax + by + cz = d, \quad\text{where } \vec n = a\vec i + b\vec j + c\vec k

Normal form & intercept form

\vec r \cdot \hat n = p \quad(p = \text{distance from origin}),\qquad \dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1

The coefficients $a, b, c$ are exactly the components of the normal vector. To find the plane through three points $A, B, C$ , take $\vec n = \overrightarrow{AB} \times \overrightarrow{AC}$ and then use one of the points.

ExamplePlane in normal form

Reduce the plane

6x + 2y + 3z = 12

to normal form and read off its distance from the origin.

ExamplePlane through three points

Find the Cartesian equation of the plane through

A(2,1,1)

,

B(3,2,2)

,

C(1,3,2)

.

ExamplePlane from intercepts

A plane cuts the axes at

(2,0,0)

,

(0,3,0)

,

(0,0,4)

. Find its equation.

7. Angles: planes and lines

Two planes meet at the same angle as their normals. A line and a plane, however, use the sine — because the natural angle to measure is between the line and the normal, and the line-to-plane angle is its complement.

Angle between two planes

\cos\theta = \dfrac{\vec n_1 \cdot \vec n_2}{|\vec n_1|\,|\vec n_2|}

Angle between a line and a plane

\sin\theta = \dfrac{|\vec b \cdot \vec n|}{|\vec b|\,|\vec n|}

Special cases

Planes parallel $\iff \vec n_1 \parallel \vec n_2$ ; planes perpendicular $\iff \vec n_1 \cdot \vec n_2 = 0$
Line parallel to a plane $\iff \vec b \cdot \vec n = 0$ (line direction perpendicular to the normal)
Line perpendicular to a plane $\iff \vec b \parallel \vec n$

ExampleAngle between two planes

Find the angle between the planes

x + y + z = 1

and

2x - y + 2z = 5

.

ExampleAngle between a line and a plane

Find the angle between the line

\vec r = \vec a + t(2\vec i - \vec j + 2\vec k)

and the plane

x + y + z = 5

.

8. Distances: point to plane and point to line

Distance is always measured along the perpendicular. To a plane, project onto the normal; to a line, use the cross product to isolate the perpendicular part.

Point to plane

D = \dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}

Point to line

D = \dfrac{|(\vec u - \vec a) \times \vec b|}{|\vec b|}

ExampleDistance from a point to a plane

Find the distance from the point

(1, 2, 0)

to the plane

2x - y + 2z = 5

.

ExampleDistance from a point to a line

Find the distance from

P(1, 0, 0)

to the line

\vec r = t(\vec i + \vec j + \vec k)

.

9. Shortest distance between skew lines

Two lines in space are skew if they are neither parallel nor intersecting. The shortest distance between them is the length of the unique common perpendicular — captured by a scalar triple product.

DefinitionSkew lines

Lines that do not lie in a common plane. For $\vec r = \vec a + t\vec b$ and $\vec r = \vec c + s\vec d$ , they are skew when $\vec b \times \vec d \ne \vec 0$ and $(\vec c - \vec a)\cdot(\vec b\times\vec d) \ne 0$ .

Shortest distance (skew)

d = \dfrac{\big|(\vec c - \vec a) \cdot (\vec b \times \vec d)\big|}{|\vec b \times \vec d|}

Shortest distance (parallel lines)

d = \dfrac{|(\vec c - \vec a) \times \vec b|}{|\vec b|}

Note: If the numerator of the skew-line formula comes out $0$ (while $\vec b \times \vec d \ne \vec 0$), the lines are not skew — they intersect. That vanishing triple product is exactly the coplanarity condition.

ExampleShortest distance between two skew lines

Find the shortest distance between

\vec r = (\vec i + \vec j) + t(2\vec i - \vec j + \vec k)

and

\vec r = (2\vec i + \vec j - \vec k) + s(3\vec i - 5\vec j + 2\vec k)

.

10. Image of a point in a plane

The mirror image of a point in a plane sits on the opposite side, the same perpendicular distance away, along the normal. The foot of the perpendicular is the midpoint between the point and its image.

Image of point A across plane r·n̂ = p

\vec{A'} = \vec a - 2\left(\dfrac{\vec a \cdot \vec n - d}{|\vec n|^2}\right)\vec n

In coordinates, to reflect $(x_0, y_0, z_0)$ in $ax + by + cz = d$ : compute $k = \dfrac{ax_0 + by_0 + cz_0 - d}{a^2 + b^2 + c^2}$ , then the image is $(x_0 - 2ak,\ y_0 - 2bk,\ z_0 - 2ck)$ .

ExampleImage of a point in a plane

Find the image of the point

(1, 2, 3)

in the plane

x + 2y + 2z = 3

.

Watch out: Common exam slips: using $\cos$ instead of $\sin$ for the line–plane angle; forgetting the absolute value in distance formulas (distance is never negative); and mixing up direction ratios with coordinates in the Cartesian line form. Always confirm $\vec b \times \vec d \ne \vec 0$ before calling two lines skew.