Consider the lines {\L}_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{ {\z}+1}{2} {\L}_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{ {\z}-3}{3}, then the unit vector perpendicular to both {\L}_1 and {\L}_2 is

Correct Answer is : −^i−7^j+5^k5√3

\stackrel{\^}{-{i}+7 \stackrel{\^}{j}+5 \stackrel{\^}{k}}/{5 \sqrt{3}}

\stackrel{\^}{-{i}-7 \stackrel{\^}{j}+5 \stackrel{\^}{k}}/{5 \sqrt{3}}

\stackrel{\^}{+{i}-7 \stackrel{\^}{j}+5 \stackrel{\^}{k}}/{5 \sqrt{3}}

\stackrel{\^}{{i}+7 \stackrel{\^}{j}+5 \stackrel{\^}{k}}/{5 \sqrt{3}}

MHTCET 14 May 2023 Shift 1 Solved Paper

Section: Mathematics

Question : 99 of 150

Marks: +1, -0

Consider the lines L1:‌

x+1

=‌

y+2

=‌

z+1

L2:‌

x−2

=‌

y+2

=‌

z−3

, then the unit vector perpendicular to both L1 and L2 is

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