Question

If \( A(1,0,2) \), \( B(2,1,0) \), \( C(2,-5,3) \), and \( D(0,3,2) \) are four points and the point of intersection of the lines \( AB \) and \( CD \) is \( P(a,b,c) \), then \( a + b + c = ? \) 

• A. \( 3 \)
• B. \( -5 \)
• C. \( 5 \)
• D. \( -3 \)

Hint:

To find the intersection of two lines in 3D, parameterize each line using direction vectors, equate their coordinates, and solve for the parameters.

Answer 1

The Correct Option is A

Step 1: Find the parametric equations of line \( AB \) 
The direction ratios of line \( AB \) are given by: \[ \overrightarrow{AB} = (2 - 1, 1 - 0, 0 - 2) = (1,1,-2). \] The parametric equations of line \( AB \) are: \[ x = 1 + \lambda, \quad y = 0 + \lambda, \quad z = 2 - 2\lambda. \] 
Step 2: Find the parametric equations of line \( CD \) 
The direction ratios of line \( CD \) are given by: \[ \overrightarrow{CD} = (0 - 2, 3 + 5, 2 - 3) = (-2, 8, -1). \] The parametric equations of line \( CD \) are: \[ x = 2 - 2\mu, \quad y = -5 + 8\mu, \quad z = 3 - \mu. \] 
Step 3: Find the intersection point 
Equating \( x, y, \) and \( z \) from both parameterized equations: \[ 1 + \lambda = 2 - 2\mu, \] \[ \lambda = 3 + 2\mu. \] \[ \lambda = -5 + 8\mu. \] \[ 2 - 2\lambda = 3 - \mu. \] Solving these equations simultaneously: 1. From \( \lambda = 3 + 2\mu \) and \( \lambda = -5 + 8\mu \): \[ 3 + 2\mu = -5 + 8\mu. \] \[ 3 + 5 = 8\mu - 2\mu. \] \[ 8 = 6\mu \Rightarrow \mu = \frac{4}{3}. \] 2. Substituting \( \mu = \frac{4}{3} \) into \( \lambda = 3 + 2\mu \): \[ \lambda = 3 + 2 \times \frac{4}{3} = 3 + \frac{8}{3} = \frac{17}{3}. \] 
Step 4: Find \( a, b, c \) using parametric equations 
\[ a = 1 + \lambda = 1 + \frac{17}{3} = \frac{20}{3}. \] \[ b = 0 + \lambda = \frac{17}{3}. \] \[ c = 2 - 2\lambda = 2 - 2 \times \frac{17}{3} = 2 - \frac{34}{3} = -\frac{28}{3}. \] \[ a + b + c = \frac{20}{3} + \frac{17}{3} - \frac{28}{3} = \frac{9}{3} = 3. \] 
Final Answer: \( \boxed{3} \)

Answer 2

To find the intersection point \( P(a,b,c) \) of the lines \( AB \) and \( CD \), we can use the parametric equation for each line. For line \( AB \), we represent any point \( P \) on the line as \( (1+t, t, 2-2t) \) where \( t \) is a parameter. Similarly, for line \( CD \), we represent any point \( P \) as \( (2s, 3-s, 2+s) \) where \( s \) is another parameter.

At the point of intersection, the coordinates must satisfy both parametric equations:

• \( 1+t = 2s \)
• \( t = 3-s \)
• \( 2-2t = 2+s \)

Let's solve these equations:

1. From \( t = 3-s \), express \( s \) in terms of \( t \): \( s = 3-t \).
2. Substitute \( s = 3-t \) into \( 1+t = 2s \):
   \( 1+t = 2(3-t) \)
   \( 1+t = 6-2t \)
   \( 3t = 5 \)
   \( t = \frac{5}{3} \).
3. Substitute \( t = \frac{5}{3} \) into \( s = 3-t \):
   \( s = 3-\frac{5}{3} \)
   \( s = \frac{4}{3} \).
4. Now, find the coordinates of \( P(a,b,c) \) using \( t = \frac{5}{3} \) in \( (1+t, t, 2-2t) \):
   \( a = 1+\frac{5}{3} = \frac{8}{3} \),
   \( b = \frac{5}{3} \),
   \( c = 2-2\left(\frac{5}{3}\right) = 2-\frac{10}{3} = -\frac{4}{3} \).

Finally, calculate \( a+b+c \):
\( a+b+c = \frac{8}{3}+\frac{5}{3}-\frac{4}{3} = \frac{9}{3} = 3 \).

Therefore, the value of \( a+b+c \) is \( 3 \).