If the lines 1 x - my - n -0- mx - ny - l -0 and nx - ly - m -0 are concurrent then l - m - p -0A- 1-m-n-0B- 1- m - n -0C- m - n - l -0D- l- m - n -0

The correct option is A l + m + n = 0 Since the lines are concurrent, so l amp;m amp;n m amp;n amp;l n amp;l amp;n = 0 ⇒ 3lmn l^3 ...

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Byju's Answer

Standard XII

Mathematics

Consistency of Linear System of Equations

If the lines ...

Question

If the lines $l x + m y + n = 0, m x + n y + l = 0$ and $n x + l y + m = 0$ are concurrent then ( $l, m, n \neq 0$ )

$l + m + n = 0$

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$l - m - n = 0$

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$l + m - n = 0$

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$m + n - l = 0$

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Solution

The correct option is A
$l + m + n = 0$

Since the lines are concurrent, so

$∣ ∣ ∣ ∣ \begin{matrix} l & m & n m & n & l n & l & n \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow 3 l m n - l^{3} - m^{3} - n^{3} = 0$

$\Rightarrow (l + m + n) (l^{2} + m^{2} + n^{2} - l m - m n - n l) = 0$

$\Rightarrow (l + m + n) \frac{1}{2} ((l - m)^{2} + (m - n)^{2} + (n - l)^{2}) = 0$

$\Rightarrow l + m + n = 0$

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Similar questions

Q. Find the angle between the lines whose direction cosines are given by the equations
(i) l + m + n = 0 and l² + m² − n² = 0
(ii) 2l − m + 2n = 0 and mn + nl + lm = 0
(iii) l + 2m + 3n = 0 and 3lm − 4ln + mn = 0
(iv) 2l + 2m − n = 0, mn + ln + lm = 0

Q. In a multi-electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic and electric fields?
(A)

n = 1, l = 0, m = 0

(B)

n = 2, l = 0, m = 0

(C)

n = 2, l = 1, m = 1

(D)

n = 3, l = 2, m = 1

(E)

n = 3, l = 2, m = 0

Q. If the direction cosines of a line are l, m and n then which of the following is true?

Q. In a multi electron atom, which of the following orbitals described by the three quantum numbers will lose degeneracy in the presence of magnetic and electric field?
A. n = 1, l = 0, m = 0
B. n = 2, l = 0, m = 0
C. n = 2, l = 1, m = 1
D. n = 3, l = 2, m = 1
E. n = 3, l = 2, m = 0

Explain, giving reasons, which of the following sets of quantum numbers are not possible.
a) n = 0 l = 0 ml = 0 $m, = + \frac{1}{2}$
b) n = 1 l = 0 ml = 0 $m, = - \frac{1}{2}$
c) n = 1 l = 1 ml = 0 $m, = + \frac{1}{2}$
d) n = 2 l = 1 ml = 0 $m, = - \frac{1}{2}$
e) n = 3 l = 3 ml = – 3 $m, = + \frac{1}{2}$
f) n = 3 l = 1 ml = 0 $m, = + \frac{1}{2}$

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If the lines lx+my+n=0, mx+ny+l=0 and nx+ly+m=0 are concurrent then (l,m,n≠0)

The correct option is A l+m+n=0 Since the lines are concurrent, so ∣∣ ∣∣lmnmnlnln∣∣ ∣∣=0 ⇒3lmn−l3−m3−n3=0 ⇒(l+m+n)(l2+m2+n2−lm−mn−nl)=0 ⇒(l+m+n)12((l−m)2+(m−n)2+(n−l)2)=0 ⇒l+m+n=0

If the lines $l x + m y + n = 0, m x + n y + l = 0$ and $n x + l y + m = 0$ are concurrent then ( $l, m, n \neq 0$ )