SATHEE: 3D Geometry 2 Question 5

3D Geometry 2 Question 5

####5. If the lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar, then $k$ can have

(a) any value

(b) exactly one value

(d) exactly three values

(2012 Main)

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Answer:

Correct Answer: 5. (c)

Solution:

Condition for two lines are coplanar.

$| \begin{array}{ccc} x_{1} - x_{2} & y_{1} - y_{2} & z_{1} - z_{2} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array} | = 0$

where, $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ are the points lie on lines (i) and (ii) respectively and $< l_{1}, m_{1}, n_{1} >$ and $< l_{2}, m_{2}, n_{2} >$ are the direction cosines of the lines (i) and (ii), respectively.

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

If 0 appears in the denominator, then the correct way of representing the equation of straight line is

$\frac{x - 2}{1} = \frac{y - 3}{1}; z = 4$

3D Geometry 2 Question 5

Answer:

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3D Geometry 2 Question 5

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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