JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

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Application +4 | -1

The shortest distance of the point

(1,2,3)

from

X, Y

and

Z

-axes, respectively are

View answer and explanation

Correct option: D

\sqrt{13}, \sqrt{10}, \sqrt{5}

Option D (

\sqrt{13}, \sqrt{10}, \sqrt{5}

) is correct because the shortest distance from the X, Y, and Z axes are calculated using the formulas

\sqrt{y^2+z^2}

\sqrt{x^2+z^2}

, and

\sqrt{x^2+y^2}

respectively.

More Questions from 3D GEOMETRY SYSTEM

Practice All Chapter MCQs →

\left( { - 10,\;8} \right)

\left( { - 6,\;12} \right)

\left( {4,\; - 2} \right)

\left( { - 2,\;4} \right)

y z

(1,-2,4)

(3,-4

5)

p

2 x+y-3=0

3 x-y-2=0

x-p y+5=0

O\left( {0,0} \right),A\left( {a,0} \right)

B\left( {0,a} \right).

A(1,0,3)

B(4,7,1)

C(3,5,3)

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}, \frac{x}{2}=\frac{y}{1}=\frac{z}{3}

\frac{x-1}{1}=\frac{2-y}{2}, z=3