Engineering
Mathematics
Equation of Straight Line in 3D and Distance Of a Point From Line
Question

If the angle between the line x=y12=z3λ and the plane x + 2y + 3z = 4 is cos–1 (514), then l equals :

5/2

2/3

2/5

3/2

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Solution

\(\frac{{x - 0}}{1} = \frac{{y - 1}}{2} = \frac{{z - 3}}{\lambda }\)                      ......(1)

x + 2y + 3z = 4                           ….(2)

Angle between the line and plane is

\(\cos (90 - \theta ) = \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \sqrt {a_2^2 + b_2^2 + c_2^2} }}\)

\( \Rightarrow \sin \theta = \frac{{1 + 4 + 3\lambda }}{{\sqrt {14} \times \sqrt {5 + {\lambda ^2}} }} = \frac{{5 + 3\lambda }}{{\sqrt {14} \times \sqrt {5 + {\lambda ^2}} }}\)         .....(3)

But given that angle between line and plane is

\(\theta = {\cos ^{ - 1}}\left( {\sqrt {\frac{5}{{14}}} } \right) = {\sin ^{ - 1}}\left( {\frac{3}{{\sqrt {14} }}} \right)\)

\( \Rightarrow \sin \theta = \frac{3}{{\sqrt {14} }}\)

\ from (3)

                           \(\frac{3}{{\sqrt {14} }} = \frac{{5 + 3\lambda }}{{\sqrt {14} \times \sqrt {5 + {\lambda ^2}} }}\)

Þ 9(5 + l2) = 25 + 9l2 + 30l

Þ 30l = 20

                            \(\lambda = \frac{2}{3}\)