Find the value of λ so that the lines, $\frac{1 - x}{3} = \frac{y - 2}{2 λ} = \frac{z - 3}{2} and \frac{x - 1}{3 λ} = \frac{y - 1}{1} = \frac{6 - z}{7}$ are perpendicular to each other.

Solution

Given lines are

$\frac{1 - x}{3} = \frac{y - 2}{2 λ} = \frac{z - 3}{2} and \frac{x - 1}{3 λ} = \frac{y - 1}{1} = \frac{6 - z}{7}$

let us rewrite the equations of the given lines as follows:

$\frac{- (x - 1)}{3} = \frac{y - 2}{2 λ} = \frac{z - 3}{2} and \frac{x - 1}{3 λ} = \frac{y - 1}{1} = \frac{- (z - 6)}{7}$

That is we have,

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

And

$\frac{x - 1}{3 λ} = \frac{y - 1}{1} = \frac{z - 6}{- 7}$

The lines are perpendicular so angle between them is $90 °$

So, cos $θ$ = 0

Here ( a1, b1, c1 ) = ( -3, $2 λ$ , 2 )

and

( a2, b2, c2 ) = ( $3 λ$ , 1, -7 )

For perpendicular lines

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \Rightarrow - 9 λ + 2 λ - 14 = 0 \Rightarrow - 7 λ - 14 = 0 \Rightarrow λ = \frac{14}{- 7} \Rightarrow λ = - 2$

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Three Dimensional Geometry

Find the value of λ so that the lines, $\frac{1 - x}{3} = \frac{y - 2}{2 λ} = \frac{z - 3}{2} and \frac{x - 1}{3 λ} = \frac{y - 1}{1} = \frac{6 - z}{7}$ are perpendicular to each other.

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.

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Three Dimensional Geometry

Find the value of λ so that the lines, 1 - x3 = y - 22λ = z - 32 and x - 13λ = y - 11 = 6 - z7 are perpendicular to each other.

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the value of λ so that the lines, $\frac{1 - x}{3} = \frac{y - 2}{2 λ} = \frac{z - 3}{2} and \frac{x - 1}{3 λ} = \frac{y - 1}{1} = \frac{6 - z}{7}$ are perpendicular to each other.