SATHEE: 3D Geometry 3 Question 54

3D Geometry 3 Question 54

####54. If the straight lines $\frac{x - 1}{2} = \frac{y + 1}{K} = \frac{z}{2}$ and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is/are

(2012)

(a) $y + 2 z = - 1$

(b) $y + z = - 1$

(d) $y - 2 z = - 1$

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Answer:

Correct Answer: 54. (b, c)

Solution:

PLAN If the straight lines are coplanar. They the should lie in same plane.

Description of Situation If straight lines are coplanar.

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

Since, $\frac{x - 1}{2} = \frac{y + 1}{K} = \frac{z}{2}$

and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{k}$ are coplanar.

$\begin{array}{lll} \Rightarrow & | \begin{array}{ccc} 2 & 0 & 0 \\ 2 & K & 2 \\ 5 & 2 & K \end{array} | & = 0 \Rightarrow K^{2} = 4 \Rightarrow K = \pm 2 \\ ∴ & n_{1} = b_{1} \times d_{1} = 6 j - 6 k, for k = 2 \\ ∴ & n_{2} = b_{2} \times d_{2} = 14 j + 14 k, for k = - 2 \end{array}$

So, equation of planes are $(r - a) \cdot n_{1} = 0$

$\begin{array}{ll} \Rightarrow & y - z = - 1 and (r - a) \cdot n_{2} = 0 \\ \Rightarrow & y + z = - 1 \end{array}$

3D Geometry 3 Question 54

Numerical Value

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3D Geometry 3 Question 54

Numerical Value

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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