The distance of the point $(1, -2, 3)$ from the plane $x - y + z = 5$ measured parallel to a line, whose direction ratios are $2, 3, -6$ is :
Show Hint
For distance measured parallel to a line with DRs $(a, b, c)$, the distance is $|r| \sqrt{a^2 + b^2 + c^2}$, where $r$ is the parameter value at the intersection point.
Step 1: Understanding the Concept:
This problem asks for the distance between a point and a plane, but not the perpendicular distance. It's measured along a specific direction (parallel to a given line). Step 2: Key Formula or Approach:
Equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $a, b, c$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = r$. Step 3: Detailed Explanation:
The line passing through $(1, -2, 3)$ parallel to the direction $(2, 3, -6)$ is:
\[ \frac{x-1}{2} = \frac{y+2}{3} = \frac{z-3}{-6} = r \]
Any point on this line is given by $Q(2r+1, 3r-2, -6r+3)$.
This point lies on the plane $x - y + z = 5$:
\[ (2r+1) - (3r-2) + (-6r+3) = 5 \]
\[ 2r + 1 - 3r + 2 - 6r + 3 = 5 \]
\[ -7r + 6 = 5 \implies -7r = -1 \implies r = \frac{1}{7} \]
The distance is given by:
\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \]
\[ d = \sqrt{(2r)^2 + (3r)^2 + (-6r)^2} = \sqrt{4r^2 + 9r^2 + 36r^2} \]
\[ d = \sqrt{49r^2} = 7|r| \]
Substituting $r = 1/7$:
\[ d = 7 \times \frac{1}{7} = 1 \] Step 4: Final Answer:
The distance is $1$.
