If OP=102and direction cosines of OP→ are3210,4210,5210, then

# If $OP=10\sqrt{2}$and direction cosines of $\stackrel{\to }{OP}$ are$\left(\frac{3\sqrt{2}}{10},\frac{4\sqrt{2}}{10},\frac{5\sqrt{2}}{10}\right)$, then

1. A

$\left(3,4,5\right)$

2. B

$\left(-3,-4,-5\right)$

3. C

$\left(-3,-4,5\right)$

4. D

$\left(6,8,10\right)$

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### Solution:

If the length of the line segment $OP=r$ and direction cosines of  $\overline{OP}$are $〈l,m,n〉$ then the coordinates of the point $P\left(lr,mr,nr\right)$

Here, $\left(l,m,n\right)=\left(\frac{3\sqrt{2}}{10},\frac{4\sqrt{2}}{10},\frac{5\sqrt{2}}{10}\right)$ and $r=10\sqrt{2}$

Therefore, the coordinates of the point $P$ are as below.

Here ,$P\left(\frac{3\sqrt{2}}{10}\cdot 10\sqrt{2},\frac{4\sqrt{2}}{10}\cdot 10\sqrt{2},\frac{5\sqrt{2}}{10}\cdot 10\sqrt{2}\right)=\overline{)\left(6,8,10\right)}$

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