Question

The distance of the point $A(3,-4,5)$ from the plane $2x+5y-6z=16$ measured along the line $\frac{x}{2}=\frac{y}{1}=\frac{z}{-2}$ is}

• A. $60/7$ units
• B. $7/60$ units
• C. $5/7$ units
• D. $9$ units

Hint:

Distance measured along a line $\vec{d}$ is $|r| \cdot |\vec{d}|$.

Answer 1

The Correct Option is D

Step 1: Concept
Find a point $P$ on the plane such that the line $AP$ is parallel to the given line.

Step 2: Analysis
Line through $A(3, -4, 5)$ parallel to $\vec{d}(2, 1, -2)$ is:
$x = 3 + 2r, y = -4 + r, z = 5 - 2r$.

Step 3: Calculation
Substitute into plane $2x + 5y - 6z = 16$:
$2(3+2r) + 5(-4+r) - 6(5-2r) = 16$
$6 + 4r - 20 + 5r - 30 + 12r = 16 \implies 21r - 44 = 16 \implies 21r = 60 \implies r = 60/21 = 20/7$.
Distance $= |r| \sqrt{2^2 + 1^2 + (-2)^2} = \frac{20}{7} \times 3 = 60/7$ units.

Step 4: Conclusion
Hence, the distance is $60/7$ units. Final Answer: (A)