SATHEE: 3D Geometry 3 Question 59

3D Geometry 3 Question 59

####59. Consider three planes $P_{1} : x - y + z = 1$

	$P_{2} : x + y - z = - 1$
and	$P_{3} : x - 3 y + 3 z = 2$

Let $L_{1}, L_{2}, L_{3}$ be the lines of intersection of the planes $P_{2}$ and $P_{3}, P_{3}$ and $P_{1}, P_{1}$ and $P_{2}$ , respectively.

Statement I Atleast two of the lines $L_{1}, L_{2}$ and $L_{3}$ are non-parallel.

Statement II The three planes do not have a common point.

(2008, 3M)

Show Answer

Answer:

Correct Answer: 59. (d)

Solution:

Given three planes are

$\begin{aligned} P_{1} : x - y + z & = 1 \dots . . (1) \\ P_{2} : x + y - z & = - 1 \dots . . (2) \\ and P_{3} : x - 3 y + 3 z & = 2 \dots . . (3) \end{aligned}$

On solving Eqs. (i) and (ii), we get

$x = 0, z = 1 + y$

which does not satisfy Eq. (iii).

As $x - 3 y + 3 z = 0 - 3 y + 3 (1 + y) = 3 (\neq 2)$

So, Statement II is true.

Next, since we know that direction ratios of line of intersection of planes $a_{1} x + b_{1} y + c_{1} z + d_{1} = 0$

$and a_{2} x + b_{2} y + c_{2} z + d_{2} = 0 is$

$b_{1} c_{2} - b_{2} c_{1}, c_{1} a_{2} - a_{1} c_{2}, a_{1} b_{2} - a_{2} b_{1}$

Using above result,

Direction ratios of lines $L_{1}, L_{2}$ and $L_{3}$ are

$0, 2, 2; 0, - 4, - 4; 0, - 2, - 2$

Since, all the three lines $L_{1}, L_{2}$ and $L_{3}$ are parallel pairwise.

Hence, Statement I is false.

3D Geometry 3 Question 59

Answer:

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3D Geometry 3 Question 59

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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