Consider the following 3 lines in space L1-r - 3i - - j - - 2 k - - - 2 i - - 4 j - - k - L2-r - i - - j - - 3 k - - - 4 i - - 2 j - - 4 k - L3-r - 3i - - 2j - - 2 k - - t 2 i - - j - - 2k -Then which one of the following pairs are in the same plane-

The correct option is A Only L_1L_2 Given lines: L_1:r #160;= 3i^ ∧ #160; j^ ∧ #160; + 2 k^ ∧ #160; + λ( 2 i^ ∧ #160; + 4 j^ ∧ #160; ...

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Byju's Answer

Standard XII

Mathematics

Angle between Two Planes

Consider the ...

Question

Consider the following 3 lines in space
$L_{1} : \to r = \land 3 i - \land j + 2 \land k + λ (2 \land i + 4 \land j - \land k)$
$L_{2} : \to r = \land i - \land j + 3 \land k + μ (4 \land i + 2 \land j + 4 \land k)$
$L_{3} : \to r = \land 3 i + \land 2 j - 2 \land k + t (2 \land i + \land j + \land 2 k)$
Then which one of the following pair(s) are in the same plane.

Only

L_{1} L_{2}

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Only

L_{2} L_{3}

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Only

L_{3} L_{1}

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L_{1} L_{2}

and

L_{2} L_{1}

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Solution

The correct option is A Only $L_{1} L_{2}$
Given lines:
$L_{1} : \to r = \land 3 i - \land j + 2 \land k + λ (2 \land i + 4 \land j - \land k)$
$L_{2} : \to r = \land i - \land j + 3 \land k + μ (4 \land i + 2 \land j + 4 \land k)$
$L_{3} : \to r = \land 3 i + \land 2 j - 2 \land k + t (2 \land i + \land j + \land 2 k)$
The condition for lines to be on same plane is that their determinant = 0
Lets do it for $L_{1}, L_{2}$
$L_{1} : \to r = \land 3 i - \land j + 2 \land k + λ (2 \land i + 4 \land j - \land k)$
$L_{2} : \to r = \land i - \land j + 3 \land k + μ (4 \land i + 2 \land j + 4 \land k)$
In the 3 X 3 determinant, we take direction vectors of two lines and difference of position vector in 3rd .
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 2 & 4 & - 1 4 & 2 & 4 2 & 0 & - 1 \end{matrix} ∣ ∣ ∣ ∣ = 2 [- 2 - 0] - 4 [- 4 - 8] - 1 [0 - 4] = - 4 + 48 + 4 = 48 \neq 0$
Here, lets check $L_{2} and L_{3}$
$L_{2} : \to r = \land i - \land j + 3 \land k + μ (4 \land i + 2 \land j + 4 \land k)$
$L_{3} : \to r = \land 3 i + \land 2 j - 2 \land k + t (2 \land i + \land j + \land 2 k)$
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 4 & 2 & 4 2 & 1 & 2 - 2 & - 3 & 5 \end{matrix} ∣ ∣ ∣ ∣ = 4 [5 - (- 6)] - 2 [10 - (- 4)] + 4 [- 6 - (- 2)] = 44 - 28 - 16 = 0$
As $Δ = 0$
$L_{2} and L_{3}$ are in same plane.
Similarly for
$L_{3} : \to r = \land 3 i + \land 2 j - 2 \land k + t (2 \land i + \land j + \land 2 k)$
$L_{1} : \to r = \land 3 i - \land j + 2 \land k + λ (2 \land i + 4 \land j - \land k)$
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 2 & 1 & 2 2 & 4 & - 1 0 & 3 & - 4 \end{matrix} ∣ ∣ ∣ ∣ = 2 [- 16 - (- 3)] - 1 [- 8 - (0)] + 2 [6 - (0)] = 26 + 8 + 12 = - 6$
As $Δ \neq 0$
$L_{1} and L_{3}$ are also not in same plane.

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Similar questions

Q. The point of intersection of the following lines is:

l_{1} : r (t) = (i - 6 j + 2 k) + t (i + 2 j + k)

l_{2} : R (u) = (4 j + k) + u (2 i + j + 2 k)

Q. Consider the following statements relating to

3

lines

L_{1}, L_{2}

and

L_{3}

in the same plane

1.

L_{2}

and

L_{3}

are both parallel to

L_{1}

, then they are parallel to each other

2.

L_{2}

and

L_{3}

are both perpendicular to

L_{1}

, then they are parallel to each other

3.

If the acute angle between

L_{1}

and

L_{2}

is equal to the acute angle between

L_{1}

and

L_{3}

then

L_{2}

is parallel to

L_{3}

.
Of these statements which one is correct?

Q. AB, AC are tangents to a parabola

y^{2} = 4 a x

l_{1}, l_{2} . l_{3}

are the lengths of perpendiculars from A, B, C on any tangent to the parabola, then

The representation of $l_{1} parallel to l_{2}$ and $l_{3} perpendicular to l_{2}$ is respectively,

Q. If line

l_{1} ⊥ l_{2}

and

l_{2} ⊥ l_{3}

then what is relationship between

l_{1}

and

l_{3}

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Consider the following 3 lines in space L1:→r=∧3i−∧j+2∧k+λ(2∧i+4∧j−∧k)L2:→r=∧i−∧j+3∧k+μ(4∧i+2∧j+4∧k)L3:→r=∧3i+∧2j−2∧k+t(2∧i+∧j+∧2k)Then which one of the following pair(s) are in the same plane.