The angle between the line r - - -3- -3k- -t 2- -3- -6k- and the plane r - - - -k- is

The correct option is A sin ^ 1 1 √( 3 ) #160; #160; Angle between the line and the plane is given by sinθ #160; =( 2î +3ĵ +6k̂ #160;) .( î ...

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Byju's Answer

Standard XII

Mathematics

Angle between a Plane and a Line

The angle bet...

Question

The angle between the line $\to r = (-^i + 3^j + 3^k) + t (2^i + 3^j + 6^k)$ and the plane $\to r . (-^i +^j +^k)$ is

{sin}^{- 1} \frac{1}{\sqrt{3}}

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{sin}^{- 1} \frac{1}{\sqrt{2}}

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{sin}^{- 1} \frac{2}{\sqrt{3}}

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{sin}^{- 1} \frac{3}{\sqrt{2}}

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Solution

The correct option is A ${sin}^{- 1} \frac{1}{\sqrt{3}}$
Angle between the line and the plane is given by
$sin θ = \frac{(2^i + 3^j + 6^k) . (-^i +^j +^k)}{\sqrt{4 + 9 + 36} \sqrt{1 + 1 + 1}}$
$= \frac{- 2 + 3 + 6}{7 \times \sqrt{3}} = \frac{7}{7 \sqrt{3}} = \frac{1}{\sqrt{3}}$
$\Rightarrow θ = {sin}^{- 1} (\frac{1}{\sqrt{3}})$

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The angle between the line →r=(−^i+3^j+3^k)+t(2^i+3^j+6^k) and the plane →r.(−^i+^j+^k) is

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The angle between the line $\to r = (-^i + 3^j + 3^k) + t (2^i + 3^j + 6^k)$ and the plane $\to r . (-^i +^j +^k)$ is