The angle between the pair of lines given by r 1-6 -4 k-2 -3 k- r -2-4 -2 -k-2 -k- isa sin -13-2 -21b cos -13-2 -21c -sin -13-2 -21d -cos -13-2 -21

Lines are in the form of L_1=a_1+λ b_1 L_2=a_2+λ b_2 Then cosθ=|b_1.b_2|b_1||b_2|| =|( 2î ĵ+3k̂).( î+2ĵ k̂)(√(2^2+( 1)^2+3^2))(√(1^2+2^2+( 1)^2))| ...

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Byju's Answer

Standard XII

Mathematics

Determinant Form of a Line

The angle bet...

Question

The angle between the pair of lines given by
${\to r}_{1} =^i + 6^j + 4^k + λ (2^i -^j + 3^k)$ ${\to r}_{2} = 4^i + 2^j -^k + λ (^i + 2^j -^k)$ is

$(a) {sin}^{- 1} (\frac{3}{2 \sqrt{21}}) (b) {cos}^{- 1} (\frac{3}{2 \sqrt{21}})$
$(c) - {sin}^{- 1} (\frac{3}{2 \sqrt{21}}) (d) - {cos}^{- 1} (\frac{3}{2 \sqrt{21}})$

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Solution

Lines are in the form of
$L_{1} = a_{1} + λ b_{1} L_{2} = a_{2} + λ b_{2}$
Then
$cos θ = ∣ ∣ ∣ ∣ \frac{{\to b}_{1} . {\to b}_{2}}{| {\to b}_{1} | | {\to b}_{2} |} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ \frac{(2^i -^j + 3^k) . (^i + 2^j -^k)}{(\sqrt{2^{2} + (- 1)^{2} + 3^{2}}) (\sqrt{1^{2} + 2^{2} + (- 1)^{2}})} ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ \frac{- 3}{2 \sqrt{21}} ∣ ∣ ∣ \Rightarrow θ = {cos}^{- 1} (\frac{3}{2 \sqrt{21}})$

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Similar questions

The acute angle between two lines $\to r = 2^i - 5^j +^k + α (3^i + 2^j + 6^k) a n d \to r = 7^i - 6^j +^k + β (^i + 2^j + 2^k)$ is

Q. The angle between the planes

r . (2^i -^j + 2^k) = 3

and

r . (3^i - 6^j + 2^k) = 4

Q. Find the angle between the pair of lines

\to r =^i +^j -^k + λ (^i + 2^j + 3^k)

and

\to r = 3^i -^j +^k + μ (^i +^j +^k)

Find the shortest distance between the following pair of lines.

¯ r = (^i + 2^j +^k) + λ (2^i -^j + 3^k)

¯ r = (^i - 2^j +^k) + μ (3^i + 2^j - 5^k)

Q. The line through

^i + 3^j + 2^k

and

⊥

to the line

^r = (^i + 2^j -^k) + λ (2^i +^j +^k)

and

\to r = (2^i + 6^j +^k) + μ (^i + 2^j + 3^k)

is:

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The angle between the pair of lines given by →r1=^i+6^j+4^k+λ(2^i−^j+3^k) →r2=4^i+2^j−^k+λ(^i+2^j−^k) is (a) sin−1(32√21) (b) cos−1(32√21) (c) −sin−1(32√21) (d) −cos−1(32√21)

Lines are in the form of L1=a1+λb1L2=a2+λb2 Then cosθ=∣∣ ∣∣→b1.→b2|→b1||→b2|∣∣ ∣∣ =∣∣ ∣ ∣ ∣∣( 2^i−^j+3^k).( ^i+2^j−^k)(√22+(−1)2+32)(√12+22+(−1)2)∣∣ ∣ ∣ ∣∣ =∣∣∣−32√21∣∣∣⇒θ=cos−1(32√21)