Statement 1 - The distance of the point of intersection of the line 1 x − 3 = 2 y − 4 = 2 z − 5 and the plane x + y + z = 17 from the point ( 3 , 4 , 5 ) is 3. Statement 2 - The distance from th

Question

Statement 1 - The distance of the point of intersection of the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ and the plane $x + y + z = 17$ from the point $(3, 4, 5)$ is 3.

Statement 2 - The distance from the point $(x_{1}, y_{1}, z_{1})$ to the plane $a x + b y + cz + d = 0$ measured parallel to the line $\frac{x - x _{1}}{c o s α} = \frac{y - y _{1}}{c o s β} =$ $\frac{z - z _{1}}{c o s γ}$ is $\frac{∣ a x _{1} + b y _{1} + c z _{1} + d ∣}{a c o s α + b c o s β + c c o s γ}$ .

TestHub · Accepted Answer

A point on the line is ( 3 + t , 4 + 2 t , 5 + 2 t ) . On the plane: 12 + 5 t = 17 , so t = 1 and the point is ( 4 , 6 , 7 ) . Its distance from ( 3 , 4 , 5 ) is 1 + 4 + 4 ​ = 3 , so Statement 1 is true. The correct distance measured along the given direction is ∣ a c o s α + b c o s β + c c o s γ ∣ ∣ a x 1 ​ + b y 1 ​ + c z 1 ​ + d ∣ ​ with no square root in the denominator, so Statement 2 as written is false.

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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