Let the direction cosines of two lines satisfy the equations and Then the cosine of the acute angle between these lines is: [2026]

Question

Let the direction cosines of two lines satisfy the equations $4 l + m - n = 0$ and $2 m n + 10 n l + 3 l m = 0 .$ Then the cosine of the acute angle between these lines is: [2026]

Smart Achievers · Accepted Answer

(4)

$Direction cosines of two lines satisfy the equation$

$\Rightarrow 4 ℓ + m - n = 0 . . . (1)$

$2 m n + 10 n ℓ + 3 ℓ m = 0 . . . (2)$

$And we know$

$\Rightarrow ℓ^{2} + m^{2} - n^{2} = 1 . . . (3)$

$\Rightarrow n = 4 ℓ + m putting in eqn. (1)$

$\Rightarrow n (2 m + 10 ℓ) + 3 ℓ m = 0$

$\Rightarrow (4 ℓ + m) (2 m + 10 ℓ) + 3 ℓ m = 0$

$\Rightarrow 8 ℓ m + 40 ℓ^{2} + 2 m^{2} + 10 ℓ m + 3 ℓ m = 0$

$\Rightarrow 40 ℓ^{2} + 21 ℓ m + 2 m^{2} = 0$

$\Rightarrow (8 ℓ + m) (5 ℓ + 2 m) = 0$

$Case 1: 8 ℓ + m = 0 \Rightarrow m = - 8 ℓ$

$Case 2: 5 ℓ + 2 m = 0 \Rightarrow m = - \frac{5}{2} ℓ$

$So direction ratio of L_{1} is ℓ, - 8 ℓ, - 4 ℓ$

$and direction ratio of L_{2} is ℓ, \frac{- 5 ℓ}{2}, \frac{3 ℓ}{2}$

$\cos θ =$ $| \frac{ℓ^{2} + 20 ℓ^{2} - 6 ℓ^{2}}{\sqrt{ℓ^{2} + 64 ℓ^{2} + 16 ℓ^{2}} \sqrt{ℓ^{2} + \frac{25 ℓ^{2}}{4} + \frac{9 ℓ^{2}}{4}}} |$

$= \frac{15 ℓ^{2}}{(9 ℓ) \frac{\sqrt{38} ℓ}{2}}$ $= \frac{10}{3 \sqrt{38}}$

$Ans. = \frac{10}{3 \sqrt{38}}$

Solution

Q.
Let the direction cosines of two lines satisfy the equations $4 l + m - n = 0$ and $2 m n + 10 n l + 3 l m = 0 .$ Then the cosine of the acute angle between these lines is: [2026]

1 $\frac{10}{\sqrt{38}}$

2 $\frac{20}{3 \sqrt{38}}$

3 $\frac{10}{7 \sqrt{38}}$

4 $\frac{10}{3 \sqrt{38}}$

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Solution

Q. Let the direction cosines of two lines satisfy the equations 4l+m-n=0 and 2mn+10nl+3lm=0. Then the cosine of the acute angle between these lines is: [2026]

1 1038

2 20338

3 10738