SATHEE: 3D Geometry 2 Question 6

3D Geometry 2 Question 6

####6. If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then the value of $k$ is

(a) $\frac{3}{2}$

(b) $\frac{9}{2}$

(d) $- \frac{3}{2}$

(2004, 1M)

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Answer:

Correct Answer: 6. (b)

Solution:

Since, the lines intersect, therefore they must have a point in common, i.e.

and

$\begin{aligned} \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = λ \\ \frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1} = μ \\ x = 2 λ + 1, y = 3 λ - 1 \\ z = 4 λ + 1 \end{aligned}$

$\Rightarrow x = 2 λ + 1, y = 3 λ - 1$

and $x = μ + 3, y = 2 μ + k, z = μ$ are same.

$\Rightarrow \begin{aligned} 2 λ + 1 & = μ + 3 \\ 3 λ - 1 & = 2 μ + k \\ 4 λ + 1 & = μ \end{aligned}$

On solving Ist and IIIrd terms, we get,

$\begin{aligned} λ & = - \frac{3}{2} and μ = - 5 \\ ∴ & k & = 3 λ - 2 μ - 1 \end{aligned}$

$\begin{array}{lll} \Rightarrow & k = 3 - \frac{3}{2} - 2 (- 5) - 1 = \frac{9}{2} \\ ∴ & k & = \frac{9}{2} \end{array}$

3D Geometry 2 Question 6

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3D Geometry 2 Question 6

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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