1.

Find the distance of point (3, 4 ,-5) form y-axis Please i need ans please please please ​

Answer» ONG>Answer:

To find : The perpendicular distance of point (3,4,5) from y axis ?

SOLUTION :

The point on y-axis is given by (0,4,0).

The distance between two points (x_1,y_1,z_1),\ (x_2,y_2,z_2)(x

1

,y

1

,z

1

), (x

2

,y

2

,z

2

) is given by,

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}d=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

+(z

2

−z

1

)

2

Here, (x_1,y_1,z_1)=(3,4,5),\ (x_2,y_2,z_2)=(0,4,0)(x

1

,y

1

,z

1

)=(3,4,5), (x

2

,y

2

,z

2

)=(0,4,0)

Substitute the value,

d=\sqrt{(0-3)^2+(4-4)^2+(0-5)^2}d=

(0−3)

2

+(4−4)

2

+(0−5)

2

d=\sqrt{9+0+25}d=

9+0+25

d=\sqrt{34}d=

34

THEREFORE, the perpendicular distance of point (3,4,5) from y axis is \sqrt{34}

34

unit.


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