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Question: If the angle between the line 2(x+1) = y = z+4 and the plane 2x – y + \(\sqrt \lambda \)z + 4 = 0 is...

If the angle between the line 2(x+1) = y = z+4 and the plane 2x – y + λ\sqrt \lambda z + 4 = 0 is π6\dfrac{\pi }{6}, then the value of λ\lambda is:
A. 1357\dfrac{{135}}{7}
B. 4511\dfrac{{45}}{{11}}
C. 457\dfrac{{45}}{7}
D. 13511\dfrac{{135}}{{11}}

Explanation

Solution

Hint: In this question, we will use the method of finding the angle between a line and a plane. The angle between θ\theta between the line x−x1l=y−y1m=z−z1n\dfrac{{x - {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z - {z_1}}}{n} and the plane ax+by+cz+d=0ax + by + cz + d = 0 is given by, sin⁡θ = al+bm+cna2+b2+c2l2+m2+n2\sin \theta {\text{ = }}\dfrac{{al + bm + cn}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{l^2} + {m^2} + {n^{{2^{}}}}} }} , where l, m, n are direction cosines and the line passes through a point (x1,y1,z1)({x_1},{y_1},{z_1}) is a point and a, b, c are the intercepts.

Complete step-by-step solution -
We have to know that the angle between the line and the plane is the complement of the angle between the line and the normal form to the plane.
Given that, the equation of the line is 2(x+1)=y=z+42(x+1) = y = z+4 . it can also be written as:
⇒\Rightarrow x−(−1)1/2=y−01=z−(−4)1\dfrac{{x - ( - 1)}}{{1/2}} = \dfrac{{y - 0}}{1} = \dfrac{{z - ( - 4)}}{1}
Equation of plane ,
⇒\Rightarrow 2x – y + λ\sqrt \lambda z + 4 = 0
And the angle between them is π6\dfrac{\pi }{6}.
We know that,
the general form of equation of line = x−x1l=y−y1m=z−z1n\dfrac{{x - {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z - {z_1}}}{n}
the general form of equation of plane = ax+by+cz+d=0ax + by + cz + d = 0
comparing these equations with the given equations of line and plane, we get
∴\therefore a = 2, b = (-1), c =λ\sqrt \lambda and d = 4. , l = 12\dfrac{1}{2}, m = 1, n = 1 and (x1,y1,z1)({x_1},{y_1},{z_1})= (-1,0,-4)
Now, put these values in the formulae of finding the angle between a line and the plane ,
⇒\Rightarrow sin⁡θ = al+bm+cna2+b2+c2l2+m2+n2\sin \theta {\text{ = }}\dfrac{{al + bm + cn}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{l^2} + {m^2} + {n^{{2^{}}}}} }}
⇒sin⁡π6 = 2×12+1×(−1)+λ×1(2)2+(−1)2+(λ)2(12)2+(1)2+(1)2 ⇒12=λ5+λ94 ⇒12=λ5+λ×23  ⇒35+λ=4λ  \Rightarrow \sin \dfrac{\pi }{6}{\text{ = }}\dfrac{{2 \times \dfrac{1}{2} + 1 \times ( - 1) + \sqrt \lambda \times 1}}{{\sqrt {{{(2)}^2} + {{( - 1)}^2} + {{(\sqrt \lambda )}^2}} \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{(1)}^2} + {{(1)}^2}} }} \\\ \Rightarrow \dfrac{1}{2} = \dfrac{{\sqrt \lambda }}{{\sqrt {5 + \lambda } \sqrt {\dfrac{9}{4}} }} \\\ \Rightarrow \dfrac{1}{2} = \dfrac{{\sqrt \lambda }}{{\sqrt {5 + \lambda } }} \times \dfrac{2}{3} \\\ \\\ \Rightarrow 3\sqrt {5 + \lambda } = 4\sqrt \lambda \\\
Squaring both sides, we get
⇒(35+λ)2=(4λ)2\Rightarrow {\left( {3\sqrt {5 + \lambda } } \right)^2} = {\left( {4\sqrt \lambda } \right)^2}
⇒9(5+λ)=16λ\Rightarrow 9(5 + \lambda ) = 16\lambda
⇒λ=457\Rightarrow \lambda = \dfrac{{45}}{7}.
Hence, the correct answer is option (C).

Note: In this type of question, We have to write all the given values like passing point and direction cosines of line and intercepts of the plane and then we will use the formula of finding the angle between the line and the plane. By using the given value of angle we will find the value of the required unknown term. And through this we will get our answer.