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Mathematics Question on Equation of a Line in Space

The line of intersection of the planes 3x−6y−2z−15=03x-6y-2z-15=0 and 2x=y−2z−5=02x=y-2z-5=0 is ?

A

x+314=y+1−2=z15\frac {x+3}{14}=\frac {y+1}{-2}=\frac {z}{15}

B

(x+3)(−14)=(y+1)2=z15\frac {(x+3)}{(-14)}=\frac {(y+1)}{2}=\frac {z}{15}

C

(x−3)14=(y+1)2=z−15\frac {(x-3)}{14}=\frac {(y+1)}{2}=\frac {z}{-15}

D

(x+3)14=(y−1)2=(z+1)15\frac {(x+3)}{14}= \frac {(y-1)}{2}=\frac {(z+1)}{15}

E

(x−3)14=(y+1)2=z15\frac {(x-3)}{14}=\frac {(y+1)}{2}=\frac {z}{15}

Answer

(x+3)(−14)=(y+1)2=z15\frac {(x+3)}{(-14)}=\frac {(y+1)}{2}=\frac {z}{15}

Explanation

Solution

Given that,
The planes 3x−6y−2z−15=03x - 6y - 2z - 15 = 0 and 2x=y−2z−5=0,2x = y - 2z - 5 = 0,
We first need to express both equations in standard form to find the line of intersection,
3x−6y−2z−15=03x - 6y - 2z - 15 = 0
⇒3x−6y−2z=15⇒3x - 6y - 2z = 15
Divide the equation by the common factor 3:3:
x−2y−(23)z=5x - 2y - (\frac 23)z = 5
2x=y−2z−52x = y - 2z - 5
⇒2x−y+2z=−5⇒2x - y + 2z = -5
Now, we have two equations in standard form:

  1. x−2y−(23)z=5x - 2y - (\frac 23)z = 5
  2. 2x−y+2z=52x - y + 2z = 5

To find the line of intersection, we need to solve these two equations simultaneously.
Let's use the method of substitution:
From equation (1), we can express xx in terms of yy and zz :
x=2y+(23)z+5x = 2y + (\frac 23)z + 5
Now, substitute this value of x into equation (2):
2(2y+(23)z+5)−y+2z=−52(2y + (\frac 23)z + 5) - y + 2z = -5

⇒4y+(43)z+10−y+2z=−5⇒4y + (\frac 43)z + 10 - y + 2z = -5
Combine the yy and zz terms:
3y+(103)z+10=−53y + (\frac {10}{3})z + 10 = -5
Now, let's isolate yy in terms of zz:
3y=−(103)z−153y = - (\frac {10}{3})z - 15

⇒y=−(109)z−5⇒y = - (\frac {10}{9})z - 5
Now, we have expressions for xx and yy in terms of zz:
x=2y+(23)z+5xx = 2y + (\frac 23)z + 5 x

=2(−(109)z−5)+(23)z+5x= 2(-(\frac {10}{9})z - 5) + (\frac 23)z + 5 x

=−(209)z−10+(23)z+5x= -(\frac {20}{9})z - 10 + (\frac 23)z + 5 x

=−(20/9)z+(23)z−5= -(20/9)z + (\frac 23)z - 5
To make it simpler, let's get a common denominator for the zz terms:

x=−(6027)z+(1827)z−(13527)x = -(\frac {60}{27})z + (\frac {18}{27})z - (\frac {135}{27})

⇒x=−(4227)z−(13527)⇒x = -(\frac {42}{27})z - (\frac {135}{27})

⇒x=−(149)z−(5)⇒x = -(\frac {14}{9})z - (5)
Now, we have the parametric equations for the line of intersection:

x=−(149)z−5y=−(109)z−5x = -(\frac {14}{9})z - 5 y = -(\frac {10}{9})z - 5
We can also express it in vector form:
r=(−(149)z−5)i−(109)z−5)j+zkr = (-(\frac {14}{9})z - 5) i - (\frac {10}{9})z - 5) j + z k
Hence, the line of intersection of the planes is,
(x+3)(−14)=(y+1)2=z15\frac {(x + 3)}{(-14)} = \frac {(y + 1)}{2} =\frac {z}{15}

So, the correct option is (B): (x+3)(−14)=(y+1)2=z15\frac {(x+3)}{(-14)}=\frac {(y+1)}{2}=\frac {z}{15}