JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

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Application +4 | -1

The point on Y-axis, which is at a distance of

\sqrt{10}

units from the point

(1,2,3)

View answer and explanation

Correct option: B

(0,2,0)

Option B ((0,2,0)) is correct because the distance between (1,2,3) and (0,y,0) is

\sqrt{{(1-0)^2 + (2-y)^2 + (3-0)^2}} = \sqrt{{10}}

, which gives

y=2

More Questions from 3D GEOMETRY SYSTEM

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\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