Question

If the point \( (3,6,k) \) lies on the line \( \dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3} \), then the value of \( k \) is

• A. \( 2 \)
• B. \( 3 \)
• C. \( 9 \)
• D. \( -2 \)
• E. \( -3 \)

Hint:

For a point lying on a line given in symmetric form, first set the common ratio equal to a parameter and convert the line into parametric equations. Then compare coordinates one by one.

Answer 1

The Correct Option is C

Step 1: Write the given line in symmetric form carefully.
The line is \[ \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3} \] Let the common value be \( \lambda \). Then, \[ \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}=\lambda \]

Step 2: Convert the symmetric form into parametric form.
From the above, \[ x-1=\lambda \quad \Rightarrow \quad x=1+\lambda \] \[ y-2=2\lambda \quad \Rightarrow \quad y=2+2\lambda \] \[ z-3=3\lambda \quad \Rightarrow \quad z=3+3\lambda \]

Step 3: Use the \( x \)-coordinate of the given point.
Since the point \( (3,6,k) \) lies on the line, its coordinates must satisfy the parametric equations.
Using \[ x=1+\lambda \] and \( x=3 \), we get \[ 3=1+\lambda \] \[ \lambda=2 \]

Step 4: Verify with the \( y \)-coordinate.
Using \[ y=2+2\lambda \] with \( \lambda=2 \), we get \[ y=2+2(2)=6 \] which matches the given \( y \)-coordinate.
So the value \( \lambda=2 \) is correct.

Step 5: Find the \( z \)-coordinate.
Now use \[ z=3+3\lambda \] Substituting \( \lambda=2 \), \[ z=3+3(2)=3+6=9 \] So, \[ k=9 \]

Step 6: Write the point explicitly.
Thus the point on the line corresponding to \( \lambda=2 \) is \[ (3,6,9) \] Hence the missing coordinate is \( 9 \).

Step 7: Final conclusion.
Therefore, the value of \( k \) is \[ \boxed{9} \] Hence, the correct option is \[ \boxed{(3)\ 9} \]