2.2 Vectorial form of a line passing through two given points

2.1 Vectorial form of a line passing through a given point and parallel to a given vector
2.2 Vectorial form of a line passing through two given points
2.3 Cartesian form of a line passing through a given point and parallel to a given vector
2.4 Cartesian form of a line passing through two given points

3.1 Condition for perpendicularity
3.2 Condition for parallelism
3.3 Co-planarity of two lines
3.4 Shortest distance between two lines
3.5 Distance between two skew lines
3.6 Distance between parallel lines
3.7 Perpendicular distance of a point from a line
3.8 Image of a point in a straight line

4.1 Equation of plane in normal form
4.2 Equation of a plane perpendicular to a given vector and passing through a given point
4.3 Equation of plane passing through a given point and parallel to the two given vectors
4.4 Equation of plane passing through three non-collinear points
4.5 Intercept form of plane
4.6 Equation of plane passing through the intersection of two given planes

5.1 Angle between two planes
5.2 Angle between a line and a plane
5.3 Distance of a point from a plane
5.4 Distance between two parallel planes

Vectorial form of a line passing through two given points $A$ and $B$ whose position vectors are $\overrightarrow a $ and $\overrightarrow b $ respectively is $$\overrightarrow r = \overrightarrow a + \lambda \left( {\overrightarrow b - \overrightarrow a } \right)$$

Proof: Let us assume any arbitrary point $P(x,y,z)$ on a line whose position vector is $\overrightarrow r $ which passes through two given points $A$ and $B$ whose position vectors are $\overrightarrow a $ and $\overrightarrow b $ respectively and $O$ be the origin. Then, $$\overrightarrow {OP} = \overrightarrow r {\text{ }}\overrightarrow {OA} = \overrightarrow a {\text{ and }}\overrightarrow {OB} = \overrightarrow b $$

Since, $A$, $B$ and $P$ lie on a same line means $\overrightarrow {AP} $ is collinear with $\overrightarrow {AB} $ i.e., $$\overrightarrow {AP} = \lambda \overrightarrow {AB} $$ for any scalar $\lambda $.

$$\begin{equation} \begin{aligned} \overrightarrow {OP} - \overrightarrow {OA} = \lambda \left( {\overrightarrow {OB} - \overrightarrow {OA} } \right) \\ \Rightarrow \overrightarrow r - \overrightarrow a = \lambda \left( {\overrightarrow b - \overrightarrow a } \right) \\ \Rightarrow \overrightarrow r = \overrightarrow a + \lambda \left( {\overrightarrow b - \overrightarrow a } \right) \\\end{aligned} \end{equation} $$

Note: When two points are given and we have to find the equation of the line passing through them, there is no need to remember the above-derived formulae.

Just remember the basic concept to find the equation of line i.e., we need one given point and one parallel vector. From the two given points take any one point and in order to find the parallel vector, we have the position vectors of two given points and by subtracting one from another we get the collinear vector which is also the parallel vector to the line.

The basic concept to find the equation of the line is very useful when other conditions than the above mentioned two conditions in the question are given. Just remember one thing that we need a point through which line passes and parallel vector to the line. [explained in Question 5.]

Question 4.Find the vectorial form of the equation of straight line parallel to $2\widehat i + 4\widehat j - 3\widehat k$ and passing through the point $(5,-2,4)$.

Solution: The line is passing through $(5,-2,4)$. Therefore, the position vector of the point through which line passes is $$\overrightarrow a = 5\hat i - 2\hat j + 4\hat k$$ and the vector parallel to the line is $$\overrightarrow b = 2\hat i + 4\hat j - 3\hat k$$ Therefore, the vectorial form of equation of straight line is $$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b $$ i.e., $$\overrightarrow r = 5\hat i - 2\hat j + 4\hat k + \lambda \left( {2\hat i + 4\hat j - 3\hat k} \right)$$

Question 5. Find the equation of a line passes through $(2,-1,3)$ and is perpendicular to the lines given by $\overrightarrow r = \left( {\widehat i + \widehat j - \widehat k} \right) + \lambda \left( {2\widehat i - 2\widehat j + \widehat k} \right)$ and $\overrightarrow r = \left( {2\widehat i - \widehat j + 3\widehat k} \right) + \mu \left( {\widehat i + 2\widehat j + 2\widehat k} \right)$.

Solution: Let us assume the equation of line be $$\overrightarrow r = \overrightarrow a + \phi \overrightarrow b $$

From the basic concept, we know that to find the equation of a line we need a point through which a line pass which is given $(2,-1,3)$ i.e.,

$\overrightarrow a = 2\widehat i - \widehat j + 3\widehat k$ and a parallel vector which is not directly given in the question.

But it is given that the line is perpendicular to the given two lines i.e.,

$${L_1} \equiv \overrightarrow r = \left( {\widehat i + \widehat j - \widehat k} \right) + \lambda \left( {2\widehat i - 2\widehat j + \widehat k} \right)$$ and

$${L_2} \equiv \overrightarrow r = \left( {2\widehat i - \widehat j + 3\widehat k} \right) + \mu \left( {\widehat i + 2\widehat j + 2\widehat k} \right)$$

If a line is perpendicular to two different lines means the parallel vector to that line $\overrightarrow b $ is also perpendicular to the parallel vectors of two given lines ${\overrightarrow {{b_1}} = 2\widehat i - 2\widehat j + \widehat k}$ and ${\overrightarrow {{b_2}} = \widehat i + 2\widehat j + 2\widehat k}$. Therefore, the cross product of ${\overrightarrow {{b_1}} }$ and ${\overrightarrow {{b_2}} }$ gives the parallel vector to the line $\overrightarrow b $ i.e., $$\overrightarrow b = \overrightarrow {{b_1}} \times \overrightarrow {{b_2}} $$

\[\overrightarrow b = \left| {\begin{array}{c} {\widehat i}&{\widehat j}&{\widehat k} \\ 2&{ - 2}&1 \\ 1&2&2 \end{array}} \right|\]

$$\overrightarrow b = - 6\widehat i - 3\widehat j + 6\widehat k$$ Therefore, the equation of line is $$\overrightarrow r = \left( {2\widehat i - \widehat j + 3\widehat k} \right) + \phi \left( { - 6\widehat i - 3\widehat j + 6\widehat k} \right)$$