The distance of the point A ( 3 , − 4 , 5 ) from the plane 2 x + 5 y − 6 z = 16 measured along the line 2 x = 1 y = − 2 z is

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The distance of the point A ( 3 , − 4 , 5 ) from the plane 2 x + 5 y − 6 z = 16 measured along the line 2 x ​ = 1 y ​ = − 2 z ​ is

TestHub · Accepted Answer

The equation of the line passing through $A (3, - 4, 5)$ and parallel to the given line $\frac{x}{2} = \frac{y}{1} = \frac{z}{- 2}$ is:

$\frac{x - 3}{2} = \frac{y + 4}{1} = \frac{z - 5}{- 2} = r$

Any point $P$ on this line can be written as $(2 r + 3, r - 4, - 2 r + 5)$ .

Since $P$ lies on the plane $2 x + 5 y - 6 z = 16$ , we substitute its coordinates into the equation of the plane:

$2 (2 r + 3) + 5 (r - 4) - 6 (- 2 r + 5) = 16$

$4 r + 6 + 5 r - 20 + 12 r - 30 = 16$

$21 r - 44 = 16$

$21 r = 60 \Rightarrow r = \frac{20}{7}$

The distance of point $A$ from the plane measured along the line is the distance $A P$ .

$A P = (2 r + 3 - 3)^{2} + (r - 4 + 4)^{2} + (- 2 r + 5 - 5)^{2}$

$A P = (2 r)^{2} + (r)^{2} + (- 2 r)^{2}$

$A P = 4 r^{2} + r^{2} + 4 r^{2} = 9 r^{2} = 3∣ r ∣$

Substituting $r = \frac{20}{7}$ :

$A P = 3 \times \frac{20}{7} = \frac{60}{7}$ units

Answer: $\frac{60}{7}$ units

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