Question

A line OP where O =(0,0,0) makes equal angles with OX,OY,OZ. The point on OP, which is at a distance of 6 from O is
1)(12/√3,12/√3,12/√3)
2)(2√3,-2√3,2√3)
3)(2/√3,2/√3,2/√3)
4)(6√3,6√3,6√3)​

Answer 1

Answer:

69

Step-by-step explanation:

1 girl one boy obvio

Answer 2

Answer:

Required point is

$( \frac{6}{ \sqrt{3} } . \frac{6}{ \sqrt{3} } . \frac{6}{ \sqrt{3} } )$

Step-by-step explanation:

Let l,m,n be the direction cosines.

We have

${l}^{2} + {m}^{2} + {n}^{2} = \: 1$

Given,

$l = m = n \\ so \: 3 {l}^{2} = 1$

Solving we get

$l = \frac{1}{ \sqrt{3} }$

Any point on the line is a multiple of direction cosines.

Let the point P is given by

$( \frac{k}{ \sqrt{3} } . \frac{k}{ \sqrt{3} } . \frac{k}{ \sqrt{3} } )$

Using the distance formula we get k = 6

So option d is the correct answer.