Question

The line of intersection of the planes \(3x-6y-2z-15=0\) and \(2x=y-2z-5=0\) is ?

• A. \(\frac {x+3}{14}=\frac {y+1}{-2}=\frac {z}{15}\)
• B. \(\frac {(x+3)}{(-14)}=\frac {(y+1)}{2}=\frac {z}{15}\)
• C. \(\frac {(x-3)}{14}=\frac {(y+1)}{2}=\frac {z}{-15}\)
• D. \(\frac {(x+3)}{14}= \frac {(y-1)}{2}=\frac {(z+1)}{15}\)
• E. \(\frac {(x-3)}{14}=\frac {(y+1)}{2}=\frac {z}{15}\)

Answer 1

The Correct Option is B

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

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

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 \(x\) in terms of \(y\) and \(z\) :
\(x = 2y + (\frac 23)z + 5\)
Now, substitute this value of x into equation (2):
\(2(2y + (\frac 23)z + 5) - y + 2z = -5\)

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

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

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

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

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

\(x = -(\frac {60}{27})z + (\frac {18}{27})z - (\frac {135}{27})\)

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

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

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

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