If the 2 lines \(\rm {x -1\over 2}={y+1\over3}={z\over-1}\) and \(\rm {x+2\over 1}={y-1\overλ}={z-2\over2}\) are coplanar, then λ = ?

Question

Question

Answer 1

Concept:

If the 2 lines \(\rm {x -x_1\over a}={y-y_1\over b}={z-z_1\over c}\) and \(\rm {x -x_2\over p}={y-y_2\over q}={z-z_2\over r}\) are coplanar, then

\(\begin{vmatrix} x_2-x_1& y_2-y_1 & z_2-z_1\\ a& b &c \\ p& q &r \end{vmatrix} = 0\)

Calculation:

Given equations of the coplanar lines are

\(\rm {x -1\over 2}={y+1\over3}={z\over-1}\) and \(\rm {x+2\over 1}={y-1\overλ}={z-2\over2}\)

x₁= 1, y₁ = - 1, z₁ = 0 and x₂ = - 2, y₂ = 1, z₂ = 2

∵ The lines are coplanar so,

\(\begin{vmatrix} x_2-x_1&y_2-y_1& z_2-z_1\\ 2& 3 &-1 \\ 1&λ &2 \end{vmatrix} = 0\)

\(\begin{vmatrix} -2-1& 1-(-1)& 2-0\\ 2& 3 &-1 \\ 1&λ &2 \end{vmatrix} = 0\)

\(\begin{vmatrix} -2-1& 1+1& 2-0\\ 2& 3 &-1 \\ 1&λ &2 \end{vmatrix} = 0\)

\(\begin{vmatrix} -3& 2& 2\\ 2& 3 &-1 \\ 1&λ &2 \end{vmatrix} = 0\)

-3(6 + λ) - 2(4 + 1) + 2(2λ - 3) = 0

λ - 34 = 0

λ = 34

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