If 2,3,4 and 1,2,3 are direction ratios of OA→andOB→, then the direction cosines of a line perpendicular to both the lines OA→ andOB→ are

# If $\left(2,3,4\right)$ and $\left(1,2,3\right)$ are direction ratios of $\stackrel{\to }{OA}$and$\stackrel{\to }{OB}$, then the direction cosines of a line perpendicular to both the lines $\stackrel{\to }{OA}$ and$\stackrel{\to }{OB}$ are

1. A

$\left(\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}}\right)$

2. B

$\left(\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{-1}{\sqrt{6}}\right)$

3. C

$\left(\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}}\right)$

4. D

None

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### Solution:

The direction ratio of any line perpendicular to both lines $OA,OB$ is along the vector$\overline{OA}×\overline{OB}$

It implies,

$\begin{array}{c}\overline{OA}×\overline{OB}=\left|\begin{array}{ccc}i& j& k\\ 2& 3& 4\\ 1& 2& 3\end{array}\right|\\ =i\left(9-8\right)-j\left(6-4\right)+k\left(4-3\right)\\ =i-2j+k\end{array}$

Hence, the direction ratios of the line perpendicular to the lines are$〈1,-2,1〉$

Therefore, the required direction cosines are  $\overline{)〈\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}}〉}$

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