Prove that the line through points A(0, -1, -1) and B(4, 5, 1) intersects the line through points C(3, 9, 4) and D(-4, 4, 4). Hence, write the equation of line passing through the point of intersection of lines AB and CD as well as origin.
Prove that the line through points A(0, -1, -1) and B(4, 5, 1) intersects the line through points C(3, 9, 4) and D(-4, 4, 4). Hence, write the equation of line passing through the point of intersection of lines AB and CD as well as origin.
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Solution and Explanation
A line through two points has direction ratios equal to the difference of the points.
\( \overrightarrow{AB} = (4-0,\ 5+1,\ 1+1) = (4, 6, 2) \).
A general point on AB is \( (4\lambda,\ -1 + 6\lambda,\ -1 + 2\lambda) \).
Step 2: Write the line CD:
\( \overrightarrow{CD} = (-4-3,\ 4-9,\ 4-4) = (-7, -5, 0) \).
A general point on CD is \( (3 - 7\mu,\ 9 - 5\mu,\ 4) \).
Step 3: Check that the lines are not parallel:
The direction ratios \( (4, 6, 2) \) and \( (-7, -5, 0) \) are not proportional, because the last entry of one is zero and of the other is not.
So the lines can meet in at most one point.
Step 4: Equate the coordinates:
For a common point we need the three coordinates to match.
\[ 4\lambda = 3 - 7\mu, \qquad -1 + 6\lambda = 9 - 5\mu, \qquad -1 + 2\lambda = 4 \]
The third equation gives \( \lambda = \dfrac{5}{2} \).
The first equation then gives \( 10 = 3 - 7\mu \), so \( \mu = -1 \).
Step 5: Test the second equation:
Left side: \( -1 + 6\cdot\dfrac{5}{2} = 14 \). Right side: \( 9 - 5(-1) = 14 \).
Both sides are equal, so the same pair \( \lambda = \dfrac{5}{2},\ \mu = -1 \) satisfies all three equations.
Hence the lines AB and CD intersect.
Step 6: Point of intersection:
Put \( \lambda = \dfrac{5}{2} \) in the point on AB: \( P = (10, 14, 4) \).
Putting \( \mu = -1 \) in the point on CD gives the same point, \( (3 + 7,\ 9 + 5,\ 4) = (10, 14, 4) \).
Step 7: Line through the origin and P:
The direction ratios are \( (10, 14, 4) \), which reduce to \( (5, 7, 2) \).
The line passes through the origin, so its equation is
\[ \frac{x}{5} = \frac{y}{7} = \frac{z}{2} \]
In vector form this is \( \vec{r} = \lambda\,(5\hat{i} + 7\hat{j} + 2\hat{k}) \).
Final Answer:
The lines intersect at \( (10, 14, 4) \), and the required line is \[ \boxed{\frac{x}{5} = \frac{y}{7} = \frac{z}{2}} \]
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