The distance of the point (7,10,11) from the line 1x−4=0y−4=3z−2 along the line 2x−9=−3y−13=6z−17 is:
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Step-by-Step Solution
Step 1: Identify the given lines and point
We are given a point P(7,10,11) and two lines, L1 and L2. We need to find the distance from point P to line L1 measured along line L2. This means we need to find the intersection point of line L2 and line L1, and then calculate the distance between this intersection point and point P.
Step 2: Find the general point on line L2
Let's represent any point on line L2 using a parameter λ. The general coordinates of a point on L2 are (2λ+9,−3λ+13,6λ+17). This point will be the intersection point if it also lies on L1.
Step 3: Substitute general point into line L1 equation
For the point on L2 to also lie on L1, its coordinates must satisfy the equation of L1. We substitute the general coordinates of the point from L2 into the equation for L1.
Step 4: Solve for λ
From the second part of the equation for L1, 0y−4, the numerator must be zero for the expression to be defined (representing a line parallel to the y -axis in the y -direction). So, (−3λ+13)−4=0, which simplifies to −3λ+9=0. Solving for λ, we get λ=3. We can verify this with the other parts of the equation.
Step 5: Find the intersection point Q
Now that we have the value of λ, we can substitute it back into the general coordinates of the point on L2 to find the intersection point Q. This point Q is (15,4,35).
Step 6: Calculate the distance PQ
Finally, we calculate the distance between the given point P(7,10,11) and the intersection point Q(15,4,35) using the distance formula in 3D. The distance is (x2−x1)2+(y2−y1)2+(z2−z1)2.