Find the angle between the line x/6 = y+32/2 = z-2/3 and the plane 3x + 4y - 12z - 7 = 0.

Correct Answer is : sin⁡−1(1091)^{-1}≤ft(10/91)sin−1(9110)

Correct Answer is : sin⁡−1(1091)^{-1}≤ft(10/91)sin−1(9110)

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UPSC NDA Math Model Paper 2

Question : 80 of 120

Marks: +1, -0

\frac{x}{6} = \frac{y+32}{2} = \frac{z-2}{3}

3x + 4y - 12z - 7 = 0.

Solution:

Given: Equation of line is

\frac{x}{6} = \frac{y+32}{2} = \frac{z-2}{3}

and equation of plane is

3x + 4y - 12z - 7 = 0.

As we know that the angle between the line

\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z_{1}^{-z}}{c_1}

and the plane

a_2 x + b_2 y + c_2 z + d = 0

is given by:

\sin\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{(\sqrt{a_1^2 + b_1^2 - c_1^2})(\sqrt{a_2^2 + b_2^2 + c_2^2})}

Here,

a_1 = 6, b_1 = 2, c_1 = 3, a_2 = 3, b_2 = 4

and

c_2 = -12

\Rightarrow a_1 \cdot a_2 + b_1 \cdot b_2 + c_1 \cdot c_2

= 18 + 8 - 36 = -10

\Rightarrow \sqrt{a_1^2 + b_1^2 + c_1^2} = 7

and

\sqrt{a_2^2 + b_2^2 + c_2^2} = 13

\Rightarrow \sin\theta = \frac{10}{7 \times 13} = \frac{10}{91}

\Rightarrow \theta = \sin^{-1}\left(\frac{10}{91}\right)

Hence, option

B

is the correct answer

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