# Three Dimensional Coordinate Geometry II

• ### Lines

A line is uniquely specified, when coordinates of 2 distinct points are known. In other words, only 1 line passes through 2 distinct points. Let $P (x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ be those 2 points.

Let $R(x,y,z)$ be any point on that line. Then,

${\vec {PR} = t (\vec {PQ})}$

since both the vectors are collinear. $t$ is any real number (or a scalar).

In other words,

${x-x_1 = t(x_2 - x_1),\ y-y_1 = t (y_2 - y_1), \ z-z_1 = t (z_2 - z_1)}$

$(x_2-x_1)$ , $(y_2-y_1)$ and $(z_2-z_1)$ are known as the direction ratios of the line.

If each of the direction ratios is divided by the distance between $P$ and $Q$, we get the direction cosines of the line. These are the cosines of the angles made by the line with $X$, $Y$ and $Z$ axes respectively. Let them be $l,m$ and $n$.

Having studied vectors, we know that the direction cosines of a line are the components of the unit vector along that direction. Hence,

${l^2 + m^2 + n^2 = 1}$

(To verify the fact that these represent the cosines of the angles, we have to use the dot product of vectors. For example, consider $\hat a = l \hat i + m \hat j + n \hat k$. Unit vector along $X$ axis is $\hat i$. Taking the dot product $\hat a \cdot \hat i$ we get $cos (\theta_x) = l$)

• #### Angle between any 2 lines is given by

${cos^{-1} (l_1 l_2 + m_1 m_2 + n_1 n_2)}$

• #### Two lines are parallel if their direction cosines have a constant ratio. i.e.

${\frac {l_1}{l_2} = \frac {m_1}{m_2} = \frac {n_1}{n_2}}$

• #### 2 lines are coincident if

${l_1l_2 + m_1 m_2 + n_1 n_2 = 1}$

• #### 2 lines are perpendicular if

${l_1 l_2 + m_1 m_2 + n_1 n_2 = 0}$

• #### Projection of a segment on a line

Let $P (x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ be the endpoints of a segment and $l,m,n$ be direction cosines of a line $L$. The projection of segment $PQ$ on line $L$ is given by

${l(x_2-x_1) + m(y_2 -y_1)+ n(z_2-z_1)}$

• ### Planes

A plane is an entity formed by 3 non-collinear points. The general equation of plane is

${ax+by+cz + d = 0}$

$a,b,c$ are the direction ratios of the vector normal to the plane. This can be easily verified by using the fact that dot product of any 2 perpendicular vectors is 0.

• #### If the plane passes through $(x_1,y_1,z_1)$ and has a normal with direction ratios $a,b,c$, then the equation is

${a(x-x_1)+ b(y-y_1)+c (z-z_1)= 0}$

• #### If $A,B,C$ are the intercepts made by the plane with $X,Y$ and $Z$ axes respectively, then

${\frac x A + \frac y B + \frac z C = 1}$

This is known as the slope-intercept form of the equation of plane.

• #### If a plane is parallel to $ax+by+cz + d =0$, its equation is

${ax+ by+cz + d_1 = 0}$

• #### Length of perpendicular from a point $(x,y,z)$ on the plane $ax+by+cz+d=0$ is given by

${\Big| \frac {ax_1 + by_1 + cz_1 + d}{\sqrt {a^2 + b^2 + c^2}} \Big|}$

• #### Equation of a plane, passing through the intersection of 2 planes $P_1 =0$ and $P_2 =0$ is given by

${P_1 + \lambda P_2 =0}$

$\lambda$ is a parameter $\in \mathbb R$.

• #### Equation of plane passing through $P_1, P_2$ and $P_3$ is given by

${\begin {vmatrix} x& y & z & 1 \\ x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \end {vmatrix} = 0}$