Let L1- x-12 - y-2-1 - z-33- L2 - x-1-1 - y-23 - 3z-35 and L3 - x-1-32 - y-2-19 - z-315 be three lines- A plane is intersecting these lines at A- B and C respectively such that PA - 2- PB - 3 and PC - 6- where P - 1- 2- 3- If V is the volume of the tetrahedron PABC and d is the perpendicular distance of the plane from the point P- then

Direction ratios of lines L_1, L_2, L_3 are n_1(2, 1,3), n_2 ( 1,3, 53), n_3( 32, 19,15) respectively. Since n_1· n_2 = 0, n_2 · n_3 = 0, n_1 · n_3 =0 nbsp; ...

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Byju's Answer

Standard XII

Physics

A Triangular Plate

Let L1: x-12 ...

Question

Let $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{- 1} = \frac{z - 3}{3},$ $L_{2} : \frac{x - 1}{- 1} = \frac{y - 2}{3} = \frac{3 (z - 3)}{5}$ and $L_{3} : \frac{x - 1}{- 32} = \frac{y - 2}{- 19} = \frac{z - 3}{15}$ be three lines. A plane is intersecting these lines at $A,$ $B$ and $C$ respectively such that $P A = 2,$ $P B = 3$ and $P C = 6,$ where $P \equiv (1, 2, 3)$ . If $V$ is the volume of the tetrahedron $P A B C$ and $d$ is the perpendicular distance of the plane from the point $P,$ then

V = 18

cubic units

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V = 6

cubic units

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d = \frac{6}{\sqrt{14}}

units

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d = 7

units

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Solution

The correct option is C $d = \frac{6}{\sqrt{14}}$ units
Direction ratios of lines $L_{1}, L_{2}, L_{3}$ are $n_{1} (2, - 1, 3), n_{2} (- 1, 3, \frac{5}{3}), n_{3} (- 32, - 19, 15)$ respectively.
Since $n_{1} \cdot n_{2} = 0, n_{2} \cdot n_{3} = 0, n_{1} \cdot n_{3} = 0$
$∴ L_{1}, L_{2}$ and $L_{3}$ are mutually perpendicular lines and all these lines are passing through $P (1, 2, 3) .$
So, volume of tetrahedron $V = \frac{1}{6} [- - \to P A - - \to P B - - \to P C] = \frac{1}{6} ∣ ∣ ∣ - - \to P A ∣ ∣ ∣ ∣ ∣ ∣ - - \to P B ∣ ∣ ∣ ∣ ∣ ∣ - - \to P C ∣ ∣ ∣ = \frac{1}{6} \times 2 \times 3 \times 6 = 6$

In tetrahedron $P A B C,$
$V = \frac{1}{3} (Base area) (Height)$
$\Rightarrow 6 = \frac{1}{3} \times \sqrt{{(\frac{1}{2} \cdot 2 \cdot 3)}^{2} + {(\frac{1}{2} \cdot 3 \cdot 6)}^{2} + {(\frac{1}{2} \cdot 2 \cdot 6)}^{2}} \times d$
$∴ d = \frac{18}{\sqrt{9 + 81 + 36}} = \frac{18}{\sqrt{126}} = \frac{6}{\sqrt{14}}$

Alternative :
Shifting origin at $(1, 2, 3)$ and considering lines $L_{1}, L_{2}, L_{3}$ as $x -$ axis, $y -$ axis, $z -$ axis
Equation of plane $A B C$ is
$\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1 \Rightarrow 3 x + 2 y + z - 6 = 0$
Perpendicular distance of the plane from the origin,
$d = \frac{6}{\sqrt{9 + 4 + 1}} = \frac{6}{\sqrt{14}}$

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Similar questions

Q. Let

L_{1} : \frac{x - 1}{2} = \frac{y - 2}{- 1} = \frac{z - 3}{3},

L_{2} : \frac{x - 1}{- 1} = \frac{y - 2}{3} = \frac{3 (z - 3)}{5}

and

L_{3} : \frac{x - 1}{- 32} = \frac{y - 2}{- 19} = \frac{z - 3}{15}

be three lines.
A plane is intersecting these lines at

A, B

and

C

respectively such that

P A = 2,

P B = 3

and

P C = 6

where

P \equiv (1, 2, 3) .

V

is the volume of the tetrahedron

P A B C

and

d

is the perpendicular distance of the plane from the point

P,

then

Q. Assertion :

Consider three planes:

P_{1} : x - y + z = 1 P_{2} : x + y - z = 1 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.

At least two of the lines

L_{1}, L_{2}

and

L_{3}

are non-parallel. Reason: The three planes do not have a common point.

Q. Consider three planes

P_{1} : x - y + z = 1

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

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}

, and

P_{1}

and

P_{2}

, respectively.
STATEMENT-

1

: At least two of the lines

L_{1}, L_{2}

and

L_{3}

are non-parallel.
and
STATEMENT -

2

: The three planes do not have a common point.

Q. Two lines whose equation are

L_{1} : \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{λ}

and

L_{2} : \frac{x - 2}{3} = \frac{y - 3}{2} = \frac{z - 2}{3}

lie in the same plane. If

L_{1}

intersects a plane

x + y + z = 15

at P, then distance of P from (3, 4, 3) is

Q. Let

L_{1}

and

L_{2}

be the lines whose equation are

\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}

and

\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}

respectively A and B are two points on

L_{1}

and

L_{2}

respectively such that AB is perpendicular both the lines

L_{1}

and

L_{2}

.Find points A, B and hence find shortest distance between lines

L_{1}

and

L_{2}

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