Find the point on the line $\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}$ at a distance 3 $\sqrt{2}$ from the point
(1, 2, 3).

Solution

$Let \frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2} = λ x = - 2 + 3 λ, y = - 1 + 2 λ, z = 3 + 2 λ Therefore, a point on this line is : {(- 2 + 3 λ), (- 1 + 2 λ), (3 + 2 λ)} The distance of the point {(- 2 + 3 λ), (- 1 + 2 λ), (3 + 2 λ)} from point (1, 2, 3) = 3 \sqrt{2} ∴ \sqrt{{(- 2 + 3 λ - 1)}^{2} + {(- 1 + 2 λ - 2)}^{2} + {(3 + 2 λ - 3)}^{2}} = 3 \sqrt{2} \Rightarrow {(- 3 + 3 λ)}^{2} + {(- 3 + 2 λ)}^{2} + {(2 λ)}^{2} = 18 \Rightarrow 9 + 9 λ^{2} - 18 λ + 9 + 4 λ^{2} - 12 λ + 4 λ^{2} = 18 17 λ^{2} - 30 λ = 0 λ = 0, λ = \frac{30}{17} When λ = \frac{30}{17},$

$x = - 2 + 3 λ = - 2 + 3 (\frac{30}{17}) = - 2 + \frac{90}{17} = \frac{56}{17} y = - 1 + 2 λ = 1 + 2 (\frac{30}{17}) = 1 + \frac{60}{17} = \frac{43}{17} z = 3 + 2 λ = 3 + 2 (\frac{30}{17}) = \frac{51 + 60}{17} = \frac{111}{17} thus, when λ = \frac{30}{17}, the point is (\frac{56}{17}, \frac{43}{17}, \frac{111}{17}) and when λ = 0, the point is (- 2, - 1, 3) .$

For Daily Free Study Material Join wiredfaculty

Three Dimensional Geometry

Find the point on the line $\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}$ at a distance 3 $\sqrt{2}$ from the point
(1, 2, 3).

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

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Three Dimensional Geometry

Find the point on the line x + 23 = y + 12 = z - 32 at a distance 3 2 from the point(1, 2, 3).

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 point on the line $\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}$ at a distance 3 $\sqrt{2}$ from the point
(1, 2, 3).