Let L: $x + y + z + 4 = 0 = 2x -y-z+8$ be a line and P:$x+2y+3z + 1 = 0$ be a plane.
Finding the Intersection Point of Line L and Plane P
The problem asks for the coordinates of the point where a line L intersects a plane P.
Step 1: Define the Line L
The line L is defined by the intersection of two planes:
- Plane 1: \(x + y + z + 4 = 0\)
- Plane 2: \(2x - y - z + 8 = 0\)
To find the parametric equations for the line L, we can combine these two equations.
Step 2: Derive Parametric Equations for Line L
Add the two equations to eliminate \(y\) and \(z\):
\( (x + y + z + 4) + (2x - y - z + 8) = 0 \)
\( 3x + 12 = 0 \)
\( 3x = -12 \)
\( x = -4 \)
Now substitute \(x = -4\) back into the first equation (\(x + y + z + 4 = 0\)):
\( -4 + y + z + 4 = 0 \)
\( y + z = 0 \)
\( z = -y \)
Let's introduce a parameter, say \(t\), such that \(y = t\). Then, \(z = -t\).
So, the parametric equations for the line L are:
- \(x = -4\)
- \(y = t\)
- \(z = -t\)
Any point on the line L can be represented as \((-4, t, -t)\) for some real number \(t\).
Step 3: Substitute Line Equations into Plane P
The equation of the plane P is:
\( x + 2y + 3z + 1 = 0 \)
To find the intersection point, we substitute the parametric equations of the line L into the equation of the plane P:
\( (-4) + 2(t) + 3(-t) + 1 = 0 \)
Step 4: Solve for the Parameter t
Simplify and solve the equation for \(t\):
\( -4 + 2t - 3t + 1 = 0 \)
\( -3 - t = 0 \)
\( t = -3 \)
Step 5: Calculate the Coordinates of the Intersection Point
Substitute the value of \(t = -3\) back into the parametric equations of the line L:
- \(x = -4\)
- \(y = t = -3\)
- \(z = -t = -(-3) = 3\)
Therefore, the point of intersection of line L and plane P is \((-4, -3, 3)\).
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Important Questions from Equation in a Line
- What are the direction ratios of the line?
- What is the equation to the straight line passing through the point \((-\sin\theta, \cos\theta)\) and perpendicular to the line \(x\cos\theta + y\sin\theta = 9\)?
- Two points $P$ and $Q$ lie on line $y = 2x + 3$. These two points $P$ and $Q$ are at a distance $2$ units from another point $R(1, 5)$. What are the coordinates of the points $P$ and $Q$?
- If two sides of a square lie on the lines $2x+y-3=0$ and $4x + 2y + 5 = 0$, then what is the area of the square in square units ?
- \(ABC\) is an acute angled isosceles triangle. Two equal sides \(AB\) and \(AC\) lie on the lines \(7x - y - 3 = 0\) and \(x + y - 5 = 0\). If \(\theta\) is one of the equal angles, then what is \(\cot\theta\) equal to ?
