Q .12. - Calculus

Question

. If a, b, and c are nonzero, find the points where the line intersects each of the coordinate planes. If$a = 0,$ explain why the line is parallel to the y z-plane. 

Step-by-Step Solution

Answer

$\text{ The required answer is, }$

$\left(0,y_0-\frac{b{x}_0}{a},z_0-\frac{c{x}_0}{a}\right)\left(x_0-\frac{a{y}_0}{b},0,z_0-\frac{c{y}_0}{b}\right)\left(x_0-\frac{a{z}_0}{c},y_0-\frac{b{z}_0}{c},0\right).$

Step 1:Given information

A line  and $a=0$

Step 2:Calculation

$\text{ Consider the line }\mathcal{L}\text{ determined by the equations }r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)\text{, }$

Then,$r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)$

$\text{ So, }x=x_0+at,y=y_0+bt,z=z_0+ct$

When the line intersect in yz-plane the value of $x=0$

$\text{ Now substitute }x=0\text{ in }x=x_0+at\text{. }$

$0=x_0+at$

$-x_0=at$

$\Rightarrow \frac{-x_0}{a}=t$

$\text{ Substitute }t=\frac{-x_0}{a}\text{ in }r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)\text{. }$

$r\left(\frac{-x_0}{a}\right)=\left(x_0+a·\frac{-x_0}{a},y_0+b·\frac{-x_0}{a},z_0+c·\frac{-x_0}{a}\right)$

$r\left(\frac{-x_0}{a}\right)=\left(x_0-x_0,y_0+b·\frac{-x_0}{a},z_0+c·\frac{-x_0}{a}\right)$

$r\left(\frac{-x_0}{a}\right)=\left(0,y_0-\frac{b{x}_0}{a},z_0-\frac{c{x}_0}{a}\right)$

$\text{ Thus the point in }yz\text{-plane is }\left(0,y_0-\frac{b{x}_0}{a},z_0-\frac{c{x}_0}{a}\right)\text{. }$

When the line intersect in zx -plane the value of $y=0$

$\text{ Now substitute }y=0\text{ in }y=y_0+bt\text{. }$

$0=y_0+bt$

$0-y_0=y_0+bt-y_0$

$-y_0=bt$

$\Rightarrow t=\frac{-y_0}{b}$

$\text{ Substitute }t=\frac{-y_0}{b}\text{ in }r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)\text{. }$

$r\left(\frac{-y_0}{b}\right)=\left(x_0+a·\frac{-y_0}{b},y_0+b·\frac{-y_0}{b},z_0+c·\frac{-y_0}{b}\right)$

$r\left(\frac{-y_0}{b}\right)=\left(x_0-\frac{a{y}_0}{b},0,z_0-\frac{c{y}_0}{b}\right)$

$\text{ Thus the point in }zx\text{-plane is }\left(x_0-\frac{a{y}_0}{b},0,z_0-\frac{c{y}_0}{b}\right)\text{. }$

 When the line intersects in  xy -plane the value of $z=0$

$\text{ Now substitute }z=0\text{ in }z=z_0+ct$

$0=z_0+ct$

$-z_0=ct$

$\Rightarrow t=\frac{-z_0}{c}$

$\text{ Substitute }t=\frac{-z_0}{c}\text{ in }r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)\text{. }$

$r\left(\frac{-z_0}{c}\right)=\left(x_0+a·\frac{-z_0}{c},y_0+b·\frac{-z_0}{c},z_0+c·\frac{-z_0}{c}\right)$

$r\left(\frac{-z_0}{c}\right)=\left(x_0+\frac{-a{z}_0}{c},y_0+\frac{-b{z}_0}{c},z_0-z_0\right)$

$r\left(\frac{-z_0}{c}\right)=\left(x_0-\frac{a{z}_0}{c},y_0-\frac{b{z}_0}{c},0\right)$

$\text{ Thus the point in }xy\text{-plane is }\left(x_0-\frac{a{z}_0}{c},y_0-\frac{b{z}_0}{c},0\right)\text{. }$

 Thus the points where the line intersects the coordinate planes $yz, zx, xy$  are,

$\left(0,y_0-\frac{b{x}_0}{a},z_0-\frac{c{x}_0}{a}\right)\left(x_0-\frac{a{y}_0}{b},0,z_0-\frac{c{y}_0}{b}\right)\left(x_0-\frac{a{z}_0}{c},y_0-\frac{b{z}_0}{c},0\right)$

$\text{ When }a=0\text{, the equation is }r\left(t\right)=\left(x_0+0·t,y_0+bt,z_0+ct\right)\text{. }$

$r\left(t\right)=\left(x_0,y_0+bt,z_0+ct\right)$

The x component becomes zero in yz-plane.

Thus, when $a=0$ the line is parallel to yz-plane.

Therefore, the required answer is,

$\left(0,y_0-\frac{b{x}_0}{a},z_0-\frac{c{x}_0}{a}\right)\left(x_0-\frac{a{y}_0}{b},0,z_0-\frac{c{y}_0}{b}\right)\left(x_0-\frac{a{z}_0}{c},y_0-\frac{b{z}_0}{c},0\right).$