Name: Coplanar Equilibrium
Uploaded: 2019-07-19
Duration: 14 min 12 s
Description: In this video, we will learn the concept of condition for coplanarity of two lines and the equation of plane containing two non-parallel coplanar lines.

Question

Find the Cartesian equation of the line passing through the origin which is perpendicular to

x - 1 = y - 2 = z - 1

and intersects the

\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}

.

Accepted Answer

Given,

Line passing through the origin which is perpendicular to $x - 1 = y - 2 = z - 1$ and intersects the $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ ,

Now let the equation of line passing through the origin be $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ ,

Now given line $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ is perpendicular to $\frac{x - 1}{1} = \frac{y - 2}{1} = \frac{z - 1}{1}$ ,

So by perpendicular condition we get, $a + b + c = 0 . . . . . . . . . . (1)$

And line $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ intersect the given line $x - 1 = y - 2 = z - 1$ ,

Now we know that line intersect only if $|\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix}| = 0$

Now here $(x_{2}, y_{2}, z_{2}) \equiv (1, - 1, 1) & (x_{1}, y_{1}, z_{1}) \equiv (0, 0, 0)$ and $a_{1} = a, b_{1} = b, c_{1} = c & a_{2} = 2, b_{2} = 3, c_{2} = 4$

Now putting the values we get,

$|\begin{matrix} 1 & - 1 & 1 \\ a & b & c \\ 2 & 3 & 4 \end{matrix}| = 0$

$\Rightarrow 7 a + 2 b - 5 c = 0 . . . . . . . . . . (2)$

Now solving equation $(1) & (2)$ we get,

$\frac{a}{- 7} = \frac{b}{12} = \frac{c}{- 5}$

Hence direction ratio of line are $(7, - 12, 5)$

So the equation of line will be $\frac{x}{7} = \frac{y}{- 12} = \frac{z}{5}$

Find the Cartesian equation of the line passing through the origin which is perpendicular to $x - 1 = y - 2 = z - 1$ and intersects the $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ .

Important Questions on Line and Plane

Learn and Prepare for any exam you want !

Find the Cartesian equation of the line passing through the origin which is perpendicular to x-1=y-2=z-1 and intersects the x-12=y+13=z-14.

Important Questions on Line and Plane

Learn and Prepare for any exam you want !

Find the Cartesian equation of the line passing through the origin which is perpendicular to $x - 1 = y - 2 = z - 1$ and intersects the $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ .