SATHEE: 3D Geometry 2 Question 8

3D Geometry 2 Question 8

####8. Two lines $L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$ and $L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}$ are coplanar. Then, $α$ can take value(s)

(2013 Adv.)

(a) 1

(b) 2

(d) 4

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Answer:

Correct Answer: 8. (a, d)

Solution:

Key Idea If two straight lines are coplanar,

$\begin{aligned} i.e. \frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}} \\ and \frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}} are coplanar \end{aligned}$

Then, $(x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1}), (a_{1}, b_{1}, c_{1})$ and $(a_{2}, b_{2}, c_{2})$ are coplanar,

i.e. $| \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0$

Here, $x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$

$\Rightarrow \frac{x - 5}{0} = \frac{y - 0}{- (α - 3)} = \frac{z - 0}{- 2}$

and

$x = α, \frac{y}{- 1} = \frac{z}{2 - α}$

$\Rightarrow \frac{x - α}{0} = \frac{y - 0}{- 1} = \frac{z - 0}{2 - α}$

$\begin{aligned} \Rightarrow | \begin{array}{ccc} 5 - α & 0 & 0 \\ 0 & 3 - α & - 2 \\ 0 & - 1 & 2 - α \end{array} | = 0 \\ \Rightarrow (5 - α) [(3 - α) (2 - α) - 2] = 0 \\ \Rightarrow (5 - α) [α^{2} - 5 α + 4] = 0 \\ \Rightarrow (5 - α) (α - 1) (α - 4) = 0 \\ ∴ α = 1, 4, 5 \end{aligned}$

3D Geometry 2 Question 8

Answer:

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3D Geometry 2 Question 8

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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