Shortcut Methods

NEET:#

• 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:

$$\overrightarrow {AB} . \overrightarrow {BC}=x_1x_2 + y_1y_2$$

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

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

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

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

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 $$\overrightarrow {PQ} =(x_2 - x_1)\hat{i} + (y_2-y_1) \hat{j} + (z_2-z_1)\hat{k}$$ 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.