The plane passing through the line L: l x-y+3(1-l) z=1, x+2 y-z=2 and perpendicular to the plane 3 x+2 y+z=6 is 3 x-8 y+7 z=4. If is the acute angle between the line L and the y-axis, then 415 ^2 is equal to _____________.

L:lx - y + 3(1 - l)z = 1, x + 2y - z = 2 and plane containing the line p:3x - 8y + 7z = 4 Let n be the vector parallel to L. then n = | i & j & k l & - 1 & 3(1 - l) 1 & 2 & - 1 | = (6l - 5) i + (3 - 2l) j + (2l + 1) k R containing L 3(6l - 5) - 8(3 - 2l) + 7(2l + 1) = 0 18l + 16l + 14l - 15 - 24 + 7 =

The plane passing through the line L l xy31l z1 x2 yz2 and p: null (Mathematics)

L:lx - y + 3(1 - l)z = 1

x + 2y - z = 2

and plane containing the line

p:3x - 8y + 7z = 4

Let

\overrightarrow n

be the vector parallel to L.

then

\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr l & { - 1} & {3(1 - l)} \cr 1 & 2 & { - 1} \cr } } \right|

= (6l - 5)\widehat i + (3 - 2l)\widehat j + (2l + 1)\widehat k

\because

R containing L

3(6l - 5) - 8(3 - 2l) + 7(2l + 1) = 0

18l + 16l + 14l - 15 - 24 + 7 = 0

\therefore

l = {{32} \over {48}} = {2 \over 3}

Let

\theta

be the acute angle between L and y-axis

\therefore

\cos \theta = {{{5 \over 3}} \over {\sqrt {1 + {{25} \over 9} + {{49} \over 9}} }} = {5 \over {\sqrt {83} }}

\therefore

415{\cos ^2}\theta = 125

Explanation

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