If - and - are any vectors- then - - - - 0-

The correct option is A TrueLet α⃗=a_1î+a_2ĵ+a_3k̂ and β⃗=b_1î+b_2ĵ+b_3k̂ ⇒ β⃗.(α⃗×β⃗)=| b_1 amp; #160; b_2 amp; b_3 a_1 amp; a_2 ...

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Byju's Answer

Standard XII

Physics

Work Done as Dot Product

If α and ...

Question

If $α$ and $β$ are any vectors, then $\to β . (\to α \times \to β) = \to 0$ .

True

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False

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Solution

The correct option is A True
Let $\to α = a_{1}^i + a_{2}^j + a_{3}^k$
and $\to β = b_{1}^i + b_{2}^j + b_{3}^k$
$\Rightarrow \to β . (\to α \times \to β) = ∣ ∣ ∣ ∣ \begin{matrix} b_{1} & b_{2} & b_{3} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} \end{matrix} ∣ ∣ ∣ ∣$
We know the property that:
If any two of the vectors are same, two rows of the determinant will be same, then it will be zero by property
And here $R_{1}$ and $R_{3}$ are same
Hence the determinant is zero.
$∴ \to β . (\to α \times \to β) = 0$
Hence the given statement is true

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Similar questions

Q. If

\to b

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\to c

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\to a

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(\to a, \to b) \to b + (\to a . \to c) \to c + \frac{\to a . (\to b \times \to c)}{∣ ∣ ∣ \to b \times \to c ∣ ∣ ∣} (\to b \times \to c) =

Q. If

\to a, \to b, \to c

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(\to a + \to b) \cdot \to c = (\to a - \to b) \cdot ¯ ¯ c = 0,

then

(\to a \times \to b) \times \to c =

Q. If

\to b

and

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\to a

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(\to a . \to b) \to b + (\to a . \to c) \to c + \frac{\to a (\to b \times \to c)}{{∣ ∣ ∣ \to b \times \to c ∣ ∣ ∣}^{2}} (\to b \times \to c)

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Q. If

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Q. Vector

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If α and β are any vectors, then →β.(→α×→β)=→0.

If $α$ and $β$ are any vectors, then $\to β . (\to α \times \to β) = \to 0$ .