Question

Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1} \ and \frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.

Answer 1

Answer:

Yes,these lines are perpendicular.

Step-by-step explanation:

To show that the lines

$\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1} \ and \\\\\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$

are perpendicular to each other.

We know that two lines given below

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \ and \\\\\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$

these two lines are perpendicular only if

$a_1a_2 + b_1b_2 + c_1c_2= 0$

in the given equations

$a_1 = 7 \\ \\ b_1 = - 5 \\ \\ c_1 = 1 \\ \\ a_2 = 1 \\ \\ b_2 = 2 \\ \\ c_2 = 3$

$= 7 \times 1 - 5 \times 2 + 1 \times 3 \\ \\ = 7 - 10 + 3 \\ \\ = 10 - 10 \\ \\ = 0$

it proves that both lines are perpendicular.

Hope it helps you.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

Two lines

$\displaystyle \sf{ \frac{x -x_1 }{ a_1 } = \frac{y -y_1 }{ b_1 } = \frac{z -z_1 }{ c_1 } \: \: } \: \: and \: \: \frac{x -x_2 }{ a_2 } = \frac{y -y_2 }{ b_2 } = \frac{z -z_2 }{ c_2 }$

are perpendicular if

$\displaystyle \sf{ {a_1a_2 + b_1b _2+ c_1c_2 } = 0 }$

CALCULATION

Comparing the given equation with with

$\displaystyle \sf{ \frac{x -x_1 }{ a_1 } = \frac{y -y_1 }{ b_1 } = \frac{z -z_1 }{ c_1 } \: \: } \: \: and \: \: \frac{x -x_2 }{ a_2 } = \frac{y -y_2 }{ b_2 } = \frac{z -z_2 }{ c_2 } \: \: we \: get$

$\displaystyle \sf{ {a_1 = 7 \: , b_1 = - 5 \: , \: c_1 } = 1 } \: \: and$

$\displaystyle \sf{ {a_2= 1 \: , b_2 = 2 \: , \: c_2 } = 3 }$

Now

$\displaystyle \sf{ {a_1a_2 + b_1b _2+ c_1c_2 } }$

$= \displaystyle \sf{ (7 \times 1) + ( - 5 \times 2) + (1 \times 3) }$

$= \displaystyle \sf{ 7 - 10 + 3 }$

$= \displaystyle \sf{ 0 }$

Hence the given two lines are perpendicular