Question

Find the angle between the line
(x + 1)/2 = y/3 = (z - 3 )/6 and the plane 10x + 2y - 11z = 3 ​

Answer 1

EXPLANATION.

$\sf \: \implies \: angle \: between \: the \: line \: = \dfrac{x + 1}{2} = \dfrac{y}{3} = \dfrac{z - 3}{6}$

And the plane = 10x + 2y - 11z = 3.

As we know that,

$\sf \: angle \: between \: the \: line \: = \dfrac{x - x_{1} }{a} = \dfrac{y - y_{1} }{b} = \dfrac{z - z_{1} }{c}$

normal to the plane = Ax + By + Cz = 0.

As we know that,

Formula of angle between line and plane.

$\sf \: \sin( \phi) = \bigg| \dfrac{Aa + Bb \: + Cc}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2} } \: \sqrt{ {A}^{2} + {B}^{2} + {C}^{2} } } \bigg|$

= a = 2 , b = 3 , c = 6.

= A = 10 , B = C = -11.

Put the value in equation, we get.

$\sf \: \sin( \phi) = \bigg| \dfrac{(10 \times 2) + (2 \times 3) + ( - 11 \times 6)}{ \sqrt{ {2}^{2} + {3}^{2} + {6}^{2} } \sqrt{10 {}^{2} + {2}^{2} + ( - 11) {}^{2} } } \bigg|$

$\sf \: \sin( \phi) = \bigg| \dfrac{20 + 6 - 66}{ \sqrt{4 + 9 + 36} \sqrt{100 + 4 + 121} } \bigg|$

$\sf \: \sin( \phi) = \bigg| \dfrac{ - 40}{7 \times 15} \bigg|$

$\sf \: \sin( \phi) = \bigg| \dfrac{ - 8}{21} \bigg|$

$\sf \: \sin( \phi) = \dfrac{8}{21}$

$\sf \: \phi \: = \sin {}^{ - 1} ( \dfrac{8}{21} )$

Answer 2

To find the angle between line and plane

$\large\underline{\bold{❥︎Step :- 1 }}$

Equation of line is

$\sf \:  ⟼ \: \dfrac{x + 1}{2} = \dfrac{y}{3} = \dfrac{z - 3}{6} \: \sf \:  ⟼ \: (1)$

So, it implies line (1) passing through the point (-1, 0, 3) and having direction ratios (2, 3, 6).

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$\large\underline{\bold{❥︎Step :- 2 }}$

☆ Equation of plane is

$\sf \:  ⟼ \: 10x + 2y - 11z = 3 \: \sf \:  ⟼ \: (2)$

☆ So, it implies, direction ratios of the plane is (10, 2, - 11).

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$\large\underline{\bold{❥︎Step :- 3 }}$

$\begin{gathered}\begin{gathered}\bf So, \: we \: get \: \begin{cases} &\sf{direction \: ratios \: of \: line} \\ &\sf{a_1 = 2}\\ &\sf{b_1 = 3}\\ &\sf{c_1 = 6}\\ &\sf{direction \: of \: plane}\\ &\sf{a_2 = 10}\\ &\sf{b_2 = 2}\\ &\sf{c_2 = - 11} \end{cases}\end{gathered}\end{gathered}$

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$\large\underline{\bold{❥︎Step :- 4 }}$

☆ Let angle between line and plane be α.

Then angle between line and plane is given by

$\bf \:  ⟼ sin \alpha = \dfrac{ |a_1a_2 + b_1b_2 + c_1c_2| }{ \sqrt{ {a_1}^{2} + {b_1}^{2} + {c_1}^{2} } \sqrt{ {a_2}^{2} + {b_2}^{2} + {c_2}^{2} } }$

☆ On substituting the values, we get

$\sf \:  ⟼sin \alpha = \dfrac{ |2 \times 10 + 3 \times 2 - 6 \times 11| }{ \sqrt{4 + 9 + 36} \sqrt{100 + 4 + 121} }$

$\sf \:  ⟼sin \alpha = \dfrac{ |20 + 6 - 66| }{ \sqrt{49} \times \sqrt{225} }$

$\sf \:  ⟼sin \alpha = \dfrac{ | - 40| }{7 \times 15}$

$\sf \:  ⟼sin \alpha \: = \dfrac{40}{105}$

$\sf \:  ⟼sin \alpha \: = \dfrac{8}{21}$

$\bf\implies \: \alpha = {sin}^{ - 1} \bigg( \dfrac{8}{21} \bigg)$

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