Question

The point of intersection of the lines \[ \frac{x-6}{-6}=\frac{y+4}{4}=\frac{z-4}{-8} \] and \[ \frac{x+1}{2}=\frac{y+2}{4}=\frac{z+3}{-2} \] is

• A. \((0,0,-4)\)
• B. \((1,0,0)\)
• C. \((0,2,0)\)
• D. \((1,2,0)\)

Hint:

To find the intersection of two 3D lines, convert both lines into parametric form and equate \(x\), \(y\), and \(z\) coordinates.

Answer 1

The Correct Option is A

Concept:
The equation of a line in symmetric form is: \[ \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \] To find the point of intersection of two lines, we convert both lines into parametric form and then equate the coordinates.

Step 1: Write the first line in parametric form.
Given first line: \[ \frac{x-6}{-6}=\frac{y+4}{4}=\frac{z-4}{-8} \] Let the common value be \(\lambda\). Then: \[ \frac{x-6}{-6}=\lambda \] \[ x-6=-6\lambda \] \[ x=6-6\lambda \] Also: \[ \frac{y+4}{4}=\lambda \] \[ y+4=4\lambda \] \[ y=-4+4\lambda \] And: \[ \frac{z-4}{-8}=\lambda \] \[ z-4=-8\lambda \] \[ z=4-8\lambda \] So, the first line is: \[ x=6-6\lambda,\quad y=-4+4\lambda,\quad z=4-8\lambda \]

Step 2: Write the second line in parametric form.
Given second line: \[ \frac{x+1}{2}=\frac{y+2}{4}=\frac{z+3}{-2} \] Let the common value be \(\mu\). Then: \[ \frac{x+1}{2}=\mu \] \[ x+1=2\mu \] \[ x=-1+2\mu \] Also: \[ \frac{y+2}{4}=\mu \] \[ y+2=4\mu \] \[ y=-2+4\mu \] And: \[ \frac{z+3}{-2}=\mu \] \[ z+3=-2\mu \] \[ z=-3-2\mu \] So, the second line is: \[ x=-1+2\mu,\quad y=-2+4\mu,\quad z=-3-2\mu \]

Step 3: Equate the corresponding coordinates.
At the point of intersection, the coordinates from both lines must be equal. From \(x\)-coordinates: \[ 6-6\lambda=-1+2\mu \] \[ 7=6\lambda+2\mu \] From \(y\)-coordinates: \[ -4+4\lambda=-2+4\mu \] \[ 4\lambda-4\mu=2 \] \[ \lambda-\mu=\frac{1}{2} \] So: \[ \lambda=\mu+\frac{1}{2} \]

Step 4: Find the values of \(\lambda\) and \(\mu\).
Substitute: \[ \lambda=\mu+\frac{1}{2} \] in: \[ 7=6\lambda+2\mu \] So: \[ 7=6\left(\mu+\frac{1}{2}\right)+2\mu \] \[ 7=6\mu+3+2\mu \] \[ 7=8\mu+3 \] \[ 8\mu=4 \] \[ \mu=\frac{1}{2} \] Now: \[ \lambda=\mu+\frac{1}{2} \] \[ \lambda=\frac{1}{2}+\frac{1}{2} \] \[ \lambda=1 \]

Step 5: Find the point of intersection.
Using the first line: \[ x=6-6\lambda \] \[ x=6-6(1)=0 \] \[ y=-4+4\lambda \] \[ y=-4+4(1)=0 \] \[ z=4-8\lambda \] \[ z=4-8(1)=-4 \] Therefore, the point of intersection is: \[ (0,0,-4) \] Hence, the correct answer is: \[ \boxed{(A)\ (0,0,-4)} \]