Shortcut Methods
JEE Mains:

Magnitude of vectors:

For vectors in 2 dimensions, the magnitude is given by $\sqrt{x^2 + y^2}$.

For vectors in 3 dimensions, the magnitude is given by $\sqrt{x^2 + y^2 + z^2}$.

Addition and subtraction of vectors:

To add two vectors, simply add their corresponding components.

To subtract two vectors, subtract the corresponding components of the second vector from the corresponding components of the first vector.

Scalar and vector products of vectors:

The scalar product of two vectors is given by $$\overrightarrow{A}\cdot\overrightarrow{B}=AB\cos\theta$$ where $A$ and $B$ are the magnitudes of the two vectors and $\theta$ is the angle between them.

The vector product of two vectors is given by $$\overrightarrow{A}\times\overrightarrow{B}=AB\sin\theta$$ where $A$ and $B$ are the magnitudes of the two vectors and $\theta$ is the angle between them.

Unit vectors:

To find the unit vector in the direction of a given vector, divide the vector by its magnitude.

Projection of vectors:

The projection of a vector $\overrightarrow{A}$ onto a vector $\overrightarrow{B}$ is given by $$\overrightarrow{A}\cdot\hat{B}$$ where $\hat{B}$ is the unit vector in the direction of $\overrightarrow{B}$.

Vector equations of lines and planes:

The vector equation of a line is given by $\overrightarrow{r} = \overrightarrow{r}_0 + t\overrightarrow{v}$ where $\overrightarrow{r}_0$ is the position vector of a point on the line, $\overrightarrow{v}$ is the direction vector of the line, and $t$ is a scalar parameter.

The vector equation of a plane is given by $$ \overrightarrow{r} \cdot \overrightarrow{n} = d $$ where $\overrightarrow{n}$ is the normal vector to the plane and $d$ is the distance from the origin to the plane.
## **CBSE Boards:**
Magnitude of vectors

Length of the magnitude of the vector: $$OM= \sqrt{x_2 x_1}^2 + \sqrt{y_2y_1}^2$$

$$AM= \sqrt{(x_2  x_1)^2 + (y2  y_1)^2 + (z_2 z_1)^2}$$
Addition and subtraction of vectors
 If the initial and terminal points of vectors (\overrightarrow {AB}) and (\overrightarrow {BC}) are (A(x_1, y_1)), (B(x_2, y_2)) and (C(x_3, y_3)) respectively, then by the triangle law of vector addition, we have: $$\overrightarrow {AC}=\overrightarrow {AB}+\overrightarrow {BC}$$
$$\overrightarrow {AB}= \hat{i}(x_2  x_1)+\hat{j}(y_2y_1)$$
$$\overrightarrow {AB}+\overrightarrow {BC}=\hat{i}[(x_3x_2)+(x_2 x_1)] + \hat{j}[(y_2 y_3)+(y_3y_1)]$$
Also $$\overrightarrow {AC}=\overrightarrow {OA}+ \overrightarrow {OC}$$
$$\overrightarrow {AC}= \hat{i}(x_3 x_1) +\hat{j}(y_3y_1)$$
Equating both expressions of (\overrightarrow {AC}) we get
$$\overrightarrow {AC}= \hat{i}(x_3 x_1) +\hat{j}(y_3y_1)$$
Scalar and vector products
 The scalar product of two vectors (\overrightarrow {AB}) and (\overrightarrow {BC}) having components (x_1, y_1) and (x_2, y_2) is defined as:
$$\overrightarrow {AB} . \overrightarrow {BC}=x_1x_2 + y_1y_2$$
 We can also determine (\overrightarrow {AB}) and (\overrightarrow {BC}) and the angle (\theta) between them to evaluate their dot product as
$$\overrightarrow {AB} . \overrightarrow {BC}= \overrightarrow {AB}. \overrightarrow {BC} . \cos\theta$$
 The vector product of two vectors (\overrightarrow {AB}) and (\overrightarrow {BC}) is defined as $$\overrightarrow {AB} \times \overrightarrow {BC} = \hat{k}(x_1y_2  x_2y_1) $$
Unit vectors
The unit vector in a direction is a vector whose magnitude is 1 and it points in the same direction as the given vector.
 If (\overrightarrow {AB}) is the vector determined by A(x_1, y_1) and (B(x_2, y_2)) its magnitude is: $$\overrightarrow {AB}=\sqrt{(x_2x_1)^2+(y_2y_1)^2}$$ the unit vector (\hat{AB}) is given by: $$\hat{AB}= \frac{\overrightarrow{AB}}{\overrightarrow{AB}} = \frac{(x_2x_1)\hat{i}+(y_2y_1)\hat{j}}{\sqrt{(x_2x_1)^2+(y_2y_1)^2}}$$
Projection of vectors
The projection of (\overrightarrow {AB}) onto (\overrightarrow {OA}) is given by: $$\overline {OA} \cdot \overrightarrow {AB} = \overrightarrow {OA}.\overrightarrow {AB} \cos \theta$$
Also it is given by
$$\overline{OA}.\overrightarrow {AB}= \overline {OA}\overrightarrow {AB} \cos \space\theta = \overrightarrow {AB} \cos \theta$$
$$\overline {OD}= \overline {OA}  \cos \theta$$
Therefore the projection of (\overrightarrow {AB}) on (\overrightarrow {OA}) is the adjacent side.
Vector equations of lines and planes
 Two points ((x_1, y_1, z_1)) and ((x_2, y_2, z_2)) is (\overrightarrow {PQ} = \overrightarrow {OP}+\overrightarrow {OP} ) where (\overrightarrow {OP} = x_1 \hat{i} +y_1\hat{j} + z_1 \hat{k} ) and ( \overrightarrow {PQ} = x_2 \hat{i} + y_2 \hat{j} + z_2\hat{k} )
Hence (\overrightarrow {PQ}) is given by $$\overrightarrow {PQ} =(x_2  x_1)\hat{i} + (y_2y_1) \hat{j} + (z_2z_1)\hat{k}$$ So parametric equations of the line passing through the points ((x_1, y_1, z_1)) and ((x_2, y_2, z_2)) are:
$$x= x_1 + \lambda (x_2x_1)$$
$$y=x_1 + \lambda(y_2x_1)$$
$$z=x_1 + \lambda(z_2x_1)$$
 The vector equation of a plane through a given point ((x_1, y_1, z_1)) and perpendicular to a given vector (\overrightarrow{n} =a\hat{i} + b\hat{j} + c\hat{k}) is given by: $$a(xx_1) + b(yy_1)+ c(zz_1) = 0$$ this is also known as the normal form of the equation of a plane with (\overrightarrow{n}) as normal vector.