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Coplanarity of Two Lines is considered one of the most asked concept.
17 Questions around this concept.
If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then k is equal to
The lines $L_1: y-x=0$ and $L_2: 2 x+y=0$ intersect the line $L_3: y+2=0$ at $P$ and $Q$ respectively. The bisector of the acute angle between $L_1$ and $L_2$ intersects $L_3$ at $R$.
Statement-1 : The ratio $P R: R Q$ equals $2 \sqrt{2}: \sqrt{5}$.
Statement 2: In any triangle, the bisector of an angle divides the triangle into two similar triangles.
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We studied in the previous concept that a line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ to lie in the plane $\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0$ iff $a l+b m+c n=0$ and $a x_1+b y_1+c z_1+d=0$ Thus, the general equation of the plane containing a straight line
$
\begin{aligned}
& \frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n} \text { is } \\
& a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0
\end{aligned}
$
where, $\quad a l+b m+c n=0$
The equation of the plane containing a straight line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ and parallel to the straight line $\frac{x-x_2}{l_1}=\frac{y-y_2}{m_1}=\frac{z-z_2}{n_1}$ is
$
\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
l & m & n \\
l_1 & m_1 & n_1
\end{array}\right|=0
$
Hence, the equation of the plane containing two given straight lines
$\begin{array}{ll} & \frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n} \\ \text { and } \quad & \frac{x-x_2}{l_1}=\frac{y-y_2}{m_1}=\frac{z-z_2}{n_1} \\ \text { or } & \left|\begin{array}{ccc}x-x_1 & y-y_1 & z-z_1 \\ l & m & n \\ l_1 & m_1 & n_1\end{array}\right|=0 \\ \left|\begin{array}{ccc}x-x_2 & y-y_2 & z-z_2 \\ l & m & n \\ l_1 & m_1 & n_1\end{array}\right|=0\end{array}$
And, the condition of coplanarity of the given straight lines is given by:
$
\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
l & m & n \\
l_1 & m_1 & n_1
\end{array}\right|=0
$
In Vector Form:
If the line $\mathrm{Ł}_1: \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_1+\lambda \overrightarrow{\mathbf{b}}_1$ and $\mathrm{Ł}_2: \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_2+\lambda \overrightarrow{\mathbf{b}}_2$ are coplanar then,
$
\left[\begin{array}{lll}
\overrightarrow{\mathbf{r}}_1 & \overrightarrow{\mathbf{b}}_1 & \overrightarrow{\mathbf{b}}_2
\end{array}\right]=\left[\begin{array}{lll}
\overrightarrow{\mathbf{r}}_2 & \overrightarrow{\mathbf{b}}_1 & \overrightarrow{\mathbf{b}}_2
\end{array}\right]
$
and the equation of the plane containing them is
$
\begin{aligned}
& {\left[\begin{array}{lll}
\overrightarrow{\mathbf{r}} & \overrightarrow{\mathbf{b}}_1 & \overrightarrow{\mathbf{b}}_2
\end{array}\right] } \\
&=\left[\begin{array}{lll}
\overrightarrow{\mathbf{r}}_1 & \overrightarrow{\mathbf{b}}_1 & \overrightarrow{\mathbf{b}}_2
\end{array}\right] \\
& \text { or } {\left[\begin{array}{lll}
\overrightarrow{\mathbf{r}} & \overrightarrow{\mathbf{b}}_1 & \overrightarrow{\mathbf{b}}_2
\end{array}\right] }
\end{aligned}=\left[\begin{array}{lll}
\overrightarrow{\mathbf{r}}_2 & \overrightarrow{\mathbf{b}}_1 & \overrightarrow{\mathbf{b}}_2
\end{array}\right] .
$
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