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Byju's Answer

Standard XII

Mathematics

Condition for Two Lines to be on the Same Plane

If a1, a2, ...

Question

If $(a_{1}, a_{2}, a_{3})$ , $(b_{1}, b_{2}, b_{3})$ and $(c_{1}, c_{2}, c_{3})$ are linearly independent and
$x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0$ then

x = y = z

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x = y = z = 0

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x + y + z = 0

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x + y + z \neq 0

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Solution

The correct option is B $x = y = z = 0$
As $∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ \neq 0$
$a_{1} x + b_{1} y + c_{1} z = 0 a_{2} x + b_{2} y + c_{2} z = 0 a_{3} x + b_{3} y + c_{3} z = 0$
For $x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0 x = y = z = 0$

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Similar questions

A set of vectors

{(a_{1}, a_{2}, a_{3}), (b_{1}, b_{2}, b_{3}), (c_{1}, c_{2}, c_{3})}

is said to be linearly independent if and only if
$\begin{vmatrix}
a_{1} & a_{2} & a_{3}\\
b_{1} & b_{2} & b_{3}\\
c_{1} & c_{2} & c_{3}
\end{vmatrix}\neq 0$
otherwise the set is said to be linearly dependent. A similar result holds for

{(a_{1}, a_{2}), (b_{1}, b_{2})}

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

and

(c_{1}, c_{2}, c_{3})

are linearly independent and

x, y, z ϵ R

, then

Three lines given by $L_{1} : a_{1} x + b_{1} y + c_{1} = 0, L_{2} : a_{2} x + b_{2} y + c_{2} = 0 and L_{3} : a_{3} x + b_{3} y + c_{3} = 0$ are always concurrent given that

$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

Q. If

| a_{1} | > | a_{2} | + | a_{3} |, | b_{2} | > | b_{1} | + | b_{3} |

and

| c_{3} | > | c_{1} | + | c_{2} |

, then show

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0

Q. If each pair of equations

a_{1} x^{2} + b_{1} x + c_{1} = 0, a_{2} x^{2} + b_{2} x + c_{2} = 0

and

a_{3} x^{2} + b_{3} x + c_{3} = 0

has a common root, then show that
(i)

\frac{c_{1} a_{2} + c_{2} a_{1}}{c_{1} a_{2} - c_{2} a_{1}} + \frac{a_{1} a_{2} b_{3}}{a_{3} (a_{1} b_{2} - a_{2} b_{1})} = 0

(ii)

{(\frac{a_{1} b_{2} - a_{2} b_{1}}{a_{1} c_{2} - a_{2} c_{1}})}^{2} = \frac{a_{1} a_{2} c_{3}}{c_{1} c_{2} a_{3}}

Q. Let

a_{1}, b_{1}, c_{1}

be natural numbers. We define

a_{2} = g c d (b_{1}, c_{1}), b_{2} = g c d (c_{1}, a_{1}), c_{2} = g c d (a_{1}, b_{1}),

and

a_{3} = l c m (b_{2}, c_{2}), b_{3} = l c m (c_{2}, a_{2}), c_{3} = l c m (a_{2}, b_{2}) .

Show that

g c d (b_{3}, c_{3}) = a_{2}

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If (a1,a2,a3), (b1,b2,b3) and (c1,c2,c3) are linearly independent andx(a1,a2,a3)+y(b1,b2,b3)+z(c1,c2,c3)=0 then

The correct option is B x=y=z=0As ∣∣ ∣∣a1a2a3b1b2b3c1c2c3∣∣ ∣∣≠0a1x+b1y+c1z=0a2x+b2y+c2z=0a3x+b3y+c3z=0For x(a1,a2,a3)+y(b1,b2,b3)+z(c1,c2,c3)=0x=y=z=0

If $(a_{1}, a_{2}, a_{3})$ , $(b_{1}, b_{2}, b_{3})$ and $(c_{1}, c_{2}, c_{3})$ are linearly independent and
$x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0$ then