1
BITSAT 2021
+3
-1

Consider the two lines

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$ and $${L_2}:{{x - 2} \over 1} = {{y + 2} \over 2} = {{z - 3} \over 3}$$

The unit vector perpendicular to both the lines L1 and L2 is

A
$${{ - \widehat i + 7\widehat j + 7\widehat k} \over {\sqrt {99} }}$$
B
$${{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$$
C
$${{ - \widehat i + 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$$
D
$${{7\widehat i - 7\widehat j + \widehat k} \over {\sqrt {99} }}$$
2
BITSAT 2021
+3
-1

The distance between the line $$r = 2\widehat i - 2\widehat j + 3\widehat k + \lambda (\widehat i - \widehat j + 4\widehat k)$$ and the plane $$a\,.\,(\widehat i + 5\widehat j + \widehat k) = 5$$ is

A
$${10 \over {9}}$$
B
$${{10} \over {3\sqrt 3 }}$$
C
$${10 \over {3}}$$
D
None of these
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