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AP EAPCET 20-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 30 of 160

Marks: +1, -0

If the lines

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

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

are coplanar, then

\sin^{-1}(\sin \lambda)+\cos^{-1}(\cos \lambda)=

$8-2\pi$
$6-\pi$
$3\pi-8$
$4\pi-8$

Solution:

Since we know that two lines are coplanar if

\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}=0

...(i)

Given lines are

L_{1}: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{6}

\therefore x_{1}=3, y_{1}=2, z_{1}=1

a_{1}=2, b_{1}=3, c_{1}=\lambda

L_{2}: \frac{x-z}{3}=\frac{y-3}{2}=\frac{z-2}{3}

a_{2}=3, b_{2}=2, c_{2}=3

From Eq. (i)

\begin{vmatrix} 1 & -1 & -1 \\ 2 & 3 & \lambda \\ 3 & 2 & 3 \end{vmatrix}=0

\Rightarrow \begin{vmatrix} 1 & 0 & 0 \\ 2 & 5 & \lambda+2 \\ 3 & 5 & 6 \end{vmatrix}=0

\Rightarrow 30-5\lambda-10=0

\Rightarrow \lambda=4

\Rightarrow \sin^{-1}(\sin 4)+\cos^{-1}(\cos 4)

\Rightarrow (\pi-4)+(2\pi-4)=(3\pi-8)

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