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: $$OM={\sqrt{x_2-x_1}}^2+{\sqrt{y_2-y_1}}^2$$

• $$AM=\sqrt{(x_2-x_1{)}^2+(y2-y_1{)}^2+(z_2-z_1{)}^2}$$

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: $$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$$

$$\overrightarrow{AB}=\hat{i}(x_2-x_1)+\hat{j}(y_2-y_1)$$

$$\overrightarrow{AB}+\overrightarrow{BC}=\hat{i}[(x_3-x_2)+(x_2-x_1)]+\hat{j}[(y_2-y_3)+(y_3-y_1)]$$

Also $$\overrightarrow{AC}=\overrightarrow{OA}+\overrightarrow{OC}$$

$$\overrightarrow{AC}=\hat{i}(x_3-x_1)+\hat{j}(y_3-y_1)$$

Equating both expressions of (\overrightarrow {AC}) we get

$$\overrightarrow{AC}=\hat{i}(x_3-x_1)+\hat{j}(y_3-y_1)$$

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:

$$\overrightarrow{AB}.\overrightarrow{BC}=x_1{x}_2+y_1{y}_2$$

• 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 $$\overrightarrow{AB}\times \overrightarrow{BC}=\hat{k}(x_1{y}_2-x_2{y}_1)$$

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: $$|\overrightarrow{AB}|=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$$ the unit vector (\hat{AB}) is given by: $$\widehat{AB}=\frac{\overrightarrow{AB}}{|\overrightarrow{AB}|}=\frac{(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}}{\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}}$$

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

$$\overline{OA}.\overrightarrow{AB}=|\overline{OA}||\overrightarrow{AB}|\cos \theta =|\overrightarrow{AB}|\cos \theta$$

$$\overline{OD}=|\overline{OA}|\cos \theta$$

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 $$\overrightarrow{PQ}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$$ So parametric equations of the line passing through the points ((x_1, y_1, z_1)) and ((x_2, y_2, z_2)) are:

$$x=x_1+\lambda (x_2-x_1)$$

$$y=x_1+\lambda (y_2-x_1)$$

$$z=x_1+\lambda (z_2-x_1)$$

• 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: $$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$$ this is also known as the normal form of the equation of a plane with (\overrightarrow{n}) as normal vector.