If the angle between the plane x-3y+2z=1 and the line (x-1)/2=(y-1)/(-1)=(z-1)/(-3) is θ, then sec⁡2θ is equal to

Correct answer: (B) 98/97. A normal vector to the plane n→= A vector parallel to the line b→= sinθ=(b → . n →)/(b → ⋅ n →) =(2+3−6)/(√14 √14) =(−1)/14 sec⁡2θ=1/(cos⁡2θ)=1/(1-2sin^2θ)=1/(1-2196)=196/194=98/97

Q40 / 105 · NTA Abhyas (JEE Main) 2020 · Mathematics · Three Dimensional Geometry

If the angle between the plane $x - 3 y + 2 z = 1$ and the line $\frac{x - 1}{2} = \frac{y - 1}{- 1} = \frac{z - 1}{- 3}$ is $θ$ , then $\sec 2 θ$ is equal to

A
$\frac{107}{11}$
Correct answer
B
$\frac{98}{97}$
C
$\frac{100}{9}$
D
$\frac{87}{79}$

Solution

A normal vector to the plane

(\vec{n}) = < 1, - 3, 2 >

A vector parallel to the line

(\vec{b}) = < 2, - 1, - 3 >

\sin θ = \frac{\vec{b} . \vec{n}}{|\vec{b}| \cdot |\vec{n}|} = \frac{2 + 3 - 6}{\sqrt{14} \sqrt{14}} = \frac{- 1}{14}

\sec 2 θ = \frac{1}{\cos 2 θ} = \frac{1}{1 - 2 s i n^{2} θ} = \frac{1}{1 - \frac{2}{196}} = \frac{196}{194} = \frac{98}{97}

Hence, the correct answer is option: 'B'.

If the angle between the plane x-3y+2z=1 and the line x-12=y-1-1=z-1-3 is θ, then sec⁡2θ is equal to

Solution

More Three Dimensional Geometry Questions

If the angle between the plane $x - 3 y + 2 z = 1$ and the line $\frac{x - 1}{2} = \frac{y - 1}{- 1} = \frac{z - 1}{- 3}$ is $θ$ , then $\sec 2 θ$ is equal to