JEE-Mains 2015 (23/30)

Geometry Level 3

The distance of the point $(1,0,2)$ from the point of intersection of the line $\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}$ and the plane $x-y+z=16$ is :

3\sqrt{21}

2\sqrt{14}

1 solution

Uros Stojkovic
Jan 10, 2017

The coordinates of the point of intersection of mentioned line and plane are solution of the following system of equations: $\frac { x-2 }{ 3 } =\frac { y+1 }{ 4 } \\ \frac { x-2 }{ 3 } =\frac { z-2 }{ 12 } \\ x-y+z=16$

After doing the math, we get:

$x=5\\ y=3\\ z=14$

Now we have a point $T=(5,3,14)$

At last, the distance between these two points is:

$\sqrt { { (5-1) }^{ 2 }+{ (3-0) }^{ 2 }+{ (14-2) }^{ 2 } } =\sqrt { { 16+ }{ 9 }+{ 144 } } =\sqrt { 169 } =13$

JEE-Mains 2015 (23/30)

1 solution

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