The angle between the lines x-1-y-0-z-1 and x-3-y-4-z-5 is-Pb- CET 2002-A- cos -11-2B- cos -11-5C- cos -11-3D- cos -11-4

The correct option is B θ=cos^ 1(3+0 5/√(1+1)√(9+16+25) ) =cos^ 1( 2/± 10 )=cos^ 1(1/5 ). ...

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Slope Form of Tangent: Hyperbola

The angle bet...

Question

The angle between the lines $\frac{x}{1} = \frac{y}{0} = \frac{z}{- 1}$ and $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$ is
[Pb. CET 2002]

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Solution

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\frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z - 1}{2} and, \frac{x - 1}{- \sqrt{3} - 1} = \frac{y - 1}{\sqrt{3} - 1} = \frac{z - 1}{4}

\cos^{- 1} (\frac{1}{65})

\frac{π}{6}

\frac{π}{3}

\frac{π}{4}

x - 3 y - 4 = 0,

4 y - z + 5 = 0

x + 3 y - 11 = 0,

2 y - z + 6 = 0

2 x + 7 y + 11 z - 3 = 0

5 x + 3 y + 9 z + 1 = 0

x

2 x + y - 2 = 0

2 x + 4 y + 7 = 0

(2 cos α + 3 sin α) x + (3 cos α - 5 sin α) y = 5 cos α - 2 sin α

α

l + 3 m + 5 n = 0

5 l m - 2 m n + 6 n l = 0

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