Let $\dpi{100} L$ be the line of intersection of the planes $\dpi{100} 2x+3y+z=1\; \; and\; \; x+3y+2z=2.$ If $\dpi{100} L$ makes an angle $\dpi{100} \alpha$ with the positive $\dpi{100} x$-axis,then $\dpi{100} \cos \alpha$ equals Option 1) 1 Option 2) $\frac{1}{\sqrt{2}}$ Option 3) $\frac{1}{\sqrt{3}}$ Option 4) $\frac{1}{2}$

As we learnt in

Direction Cosines -

i)    $\sin^{2} \alpha+ \sin^{2} \beta+\sin^{2} \gamma= 2$

ii)    If OP =r then the co-ordinates of P will be (lr,mr,nr)

iii)    Direction cosines of X-axis are (1,0,0)

iv)    Direction cosines of Y-axis are (0,1,0)

v)    Direction cosines of Z-axis are (0,0,1)

-

Let DCs of L be (l,m,n) then 2l+3m+2n=0 and l + 3m + 2n = 0

On solving

$\frac{l}{3}=\frac{m}{-3}=\frac{n}{3}$

$So,\: \: \: \: l:m:n= \frac{l}{\sqrt{}3},\frac{-1}{\sqrt3},\frac{1}{\sqrt3}$

$cos\alpha =\frac{1}{\sqrt{3}}$

Option 1)

1

Incorrect Option

Option 2)

$\frac{1}{\sqrt{2}}$

Incorrect Option

Option 3)

$\frac{1}{\sqrt{3}}$

Correct Option

Option 4)

$\frac{1}{2}$

Incorrect Option

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