Find the angle between the line 3x−2=−1y+1=−2z−3 and the plane 3x+4y+z+5=0

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Find the angle between the line {x - 2}{3} = {y + 1}{-1} = {z

Question

Find the angle between the line $\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 3}{- 2}$ and the plane $3 x + 4 y + z + 5 = 0$

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Noah Discussed

Find the angle between the line

\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 3}{- 2}

and the plane

3 x + 4 y + z + 5 = 0

5 mins ago

Discuss this question LIVE

5 mins ago

The given line is

\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{2 - 3}{- 2}

\Rightarrow r = (2 i - j + k) + t (3 i - j - 2 k)

∴

the vector parallel to the given line is

r = 3 i - j - 2 k

The given plane is

3 x + 4 y + z = - 5

\Rightarrow r = (3 i + 4 j + k) = 5

∴

the normal to the given plane is

n = 3 i + 4 j + k

Let

θ

be the angle between the line and the plane, then

sin θ = \frac{b . n}{∣ b ∣∣ n ∣}

= \frac{( 3 i - j - 2 k ) . ( 3 i + 4 j + k )}{∣3 i - j - 2 k ∣∣3 i + 4 j + k ∣}

= \frac{( 3 ) ( 3 ) + ( - 1 ) ( 4 ) + ( - 2 ) ( 1 )}{3 ^{2} + ( - 1 ) ^{2} + ( - 2 ) ^{2} 3 ^{2} + 4 ^{2} + 1 ^{2}}

= \frac{9 - 4 - 2}{9 + 1 + 4 9 + 16 + 1} = \frac{3}{14 26} = \frac{3}{2 91}

θ = sin^{- 1} (\frac{3}{2 91})

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Question Text	Find the angle between the line $\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 3}{- 2}$ and the plane $3 x + 4 y + z + 5 = 0$
Answer Type	Text solution:1
Upvotes	150