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JEE MainMathematicsToughThree dimension

A line L is perpendicular to the lines having direction ratios
(3, 2, 1) and (1, 2, 3).The sum of direction cosines of line L is

Show answer and solution
a vector perpendicular to both (3, 2, 1) and (1, 2, 3)
will be
$\overset{\rightarrow}{c} = \overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = \left| \begin{matrix}
\overset{\hat{}}{i}\ \ \ \ \overset{\hat{}}{j}\ \ \ \ \overset{\hat{}}{k} \\
3\ \ \ \ 2\ \ \ \ 1 \\
1\ \ \ \ 2\ \ \ \ 3
\end{matrix} \right|$
c→=4i^+8j^+4k^\overset{\rightarrow}{c} = 4\overset{\hat{}}{i} + 8\overset{\hat{}}{j} + 4\overset{\hat{}}{k}
directions cosines ofc→\overset{\rightarrow}{c},sa
=±(442+82+42,−842+82+42,442+82+42)= \pm \left( \frac{4}{\sqrt{4^{2} + 8^{2} + 4^{2}}},\frac{- 8}{\sqrt{4^{2} + 8^{2} + 4^{2}}},\frac{4}{\sqrt{4^{2} + 8^{2} + 4^{2}}} \right)
sum of d.c.s
=±(4−8+442+82+42)=0= \pm \left( \frac{4 - 8 + 4}{\sqrt{4^{2} + 8^{2} + 4^{2}}} \right) = 0

Source: Grass-G / M-21 Mathematics Paper 19 Dec Hindi Eng.docx

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