Let - R and the three vectors a -3 k- b -2 -k- and c -2 -3 k- Then the set S -a - b and c are coplanar A- is emptyB- is singletonC- contains exactly two positive numbersD- contains exactly two numbers only one of which is positive

For these vectors to be coplanar, [a b c]=0 ⇒α amp;1 amp;3 2 amp;1 amp; α α amp; 2 amp;3=0 R_1→ R_1 R_3 0 amp;3 amp;0 2 amp;1 amp; α α a ...

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Standard XII

Mathematics

Test for Coplanarity

Let α∈ R and ...

Question

Let $α \in R$ and the three vectors $\to a = α^i +^j + 3^k, \to b = 2^i +^j - α^k$ and $\to c = α^i - 2^j + 3^k$ . Then the set $S = {α : \to a, \to b and \to c are coplanar}$

is empty

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is singleton

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contains exactly two positive numbers

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contains exactly two numbers only one of which is positive

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Solution

The correct option is A is empty
For these vectors to be coplanar,
$[\to a \to b \to c] = 0 \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} α & 1 & 3 2 & 1 & - α α & - 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$R_{1} \to R_{1} - R_{3} ∣ ∣ ∣ ∣ \begin{matrix} 0 & 3 & 0 2 & 1 & - α α & - 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow - 3 (6 + α^{2}) = 0$
$\Rightarrow α^{2} = - 6$
Not possible for real $α$
So, $S$ is empty.

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Test for Coplanarity

Standard XII Mathematics

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Let α∈R and the three vectors →a=α^i+^j+3^k, →b=2^i+^j−α^k and →c=α^i−2^j+3^k. Then the set S={α:→a,→b and →c are coplanar }

The correct option is A is emptyFor these vectors to be coplanar, [→a →b →c]=0⇒∣∣ ∣∣α1321−αα−23∣∣ ∣∣=0 R1→R1−R3∣∣ ∣∣03021−αα−23∣∣ ∣∣=0 ⇒−3(6+α2)=0 ⇒α2=−6 Not possible for real α So, S is empty.

Let $α \in R$ and the three vectors $\to a = α^i +^j + 3^k, \to b = 2^i +^j - α^k$ and $\to c = α^i - 2^j + 3^k$ . Then the set $S = {α : \to a, \to b and \to c are coplanar}$