The distance of the point $(7{,}10{,}11)$ from the line $\displaystyle\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ along the line $\displaystyle\frac{x-9}{2}=\frac{y-13}{-3}=\frac{z-17}{6}$ is:

Step 1: Identify the given lines and point

We are given a point $P(7{,}10{,}11)$ and two lines, $L_1$ and $L_2$. We need to find the distance from point $P$ to line $L_1$ measured along line $L_2$. This means we need to find the intersection point of line $L_2$ and line $L_1$, 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 $L_2$ using a parameter $\lambda$. The general coordinates of a point on $L_2$ are $(2\lambda + 9, -3\lambda + 13, 6\lambda + 17)$. This point will be the intersection point if it also lies on $L_1$.

Step 3: Substitute general point into line L1 equation

For the point on $L_2$ to also lie on $L_1$, its coordinates must satisfy the equation of $L_1$. We substitute the general coordinates of the point from $L_2$ into the equation for $L_1$.

Step 4: Solve for $\lambda$

From the second part of the equation for $L_1$, $\frac{y-4}{0}$, the numerator must be zero for the expression to be defined (representing a line parallel to the $y$ -axis in the $y$ -direction). So, $(-3\lambda + 13) - 4 = 0$, which simplifies to $-3\lambda + 9 = 0$. Solving for $\lambda$, we get $\lambda = 3$. 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 $\lambda$, we can substitute it back into the general coordinates of the point on $L_2$ 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 $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

More from this chapter

• Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square uni
• Let L_1: (x-1)/(1) = (y-2)/(1) = (z-1)/(2) and L_2: (x+1)/(1) = (y-2)/(2) = (z)/(4) be two lines. Let L_3 be a line passing through the point (α, β,) and be per
• The line L_1 is parallel to the vector a = -3 i + 2 j + 4 k and passes through the point (7, 6, 2) and the line L_2 is parallel to the vector b = 2 i + j + 3 k
• The distance of the line (x-2)/(2)=(y-6)/(3)=(z-3)/(4) from the point (1,4,0) along the line (x)/(4)=(y-2)/(2)=(z+3)/(3) is:
• Find the shortest distance between the skew lines: r⃗₁ = (i + j + k) + λ(2i + 3j + 4k) and r⃗₂ = (2i + 3j + 5k) + μ(i + 2j + 3k).
• Find the equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes x + 2y + 3z = 4 and 2x - 3y + 4z = 5.
• Let the line passing through the points (-1, 2, 1) and parallel to the line (x-1)/(2) = (y+1)/(3) = (z)/(4) intersect the line (x+2)/(3) = (y-3)/(2) = (z-4)/(3)
• Let the shortest distance between the lines (x-3)/(3) = (y-α)/(-1) = (z-3)/(1) and (x+3)/(-3) = (y+7)/(2) = (z-β)/(4) be 3√30. Then the positive value of 5α + β

All Three Dimensional Geometry solutions →