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

Correct Answer is : sin⁡−1(752)^{-1}≤ft({7/52})sin−1(527)

Correct Answer is : sin⁡−1(752)^{-1}≤ft({7/52})sin−1(527)

Test Index

UPSC NDA Math Model Paper 10

Question : 104 of 120

Marks: +1, -0

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

3x+4y+z+5=0

Solution:

Given: Equation of line is

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

and equation of plane is

3x+4y+z+5=0

As we know that the angle between the line

\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{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}{\left(\sqrt{a_1^2+b_1^2+c_1^2}\right)\left(\sqrt{a_2^2+b_2^2+c_2^2}\right)}

Here,

a_1=3, b_1=-1, c_1=2, a_2=3, b_2=4

and

c_2=1

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

=9-4+2=7

\Rightarrow\sqrt{a_1^2+b_1^2+c_1^2}=\sqrt{14}

and

\sqrt{a_2^2+b_2^2+c_2^2}=\sqrt{26}

\Rightarrow \sin\theta = \frac{7}{\sqrt{14} \cdot \sqrt{26}} = \sqrt{\frac{7}{52}}

\Rightarrow \theta = \sin^{-1}\left(\sqrt{\frac{7}{52}}\right)

Go to Question:

Prev Question

Next Question