If a- - i - j - k- b- - 4i - 3j - 4k and c- - i - -j - -k are linearly dependent vectors and -c- - -3- then

The correct option is A α = ± 1, β = 1 Given that a⃗ = #160;i + #160;j + #160;k, #160;b⃗ = 4i + 3j + 4k and c⃗ = #160;i + αj + βk ar ...

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Linear Dependence and Independence of Vectors

If a⃗ = i +...

Question

If $\to a = ˆ i + ˆ j + ˆ k, \to b = 4 ˆ i + 3 ˆ j + 4 ˆ k$ and $\to c = ˆ i + α ˆ j + β ˆ k$ are linearly dependent vectors and $| \to c | = \sqrt{3}$ , then

a = 1, b = - 1

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a = 1, b = \pm 1

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α = - 1, β = \pm 1

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α = \pm 1, β = 1

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Solution

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Similar questions

\to a

\to b

\to c

α

β

| c |

\sqrt{3}

a = i + j + k, b = 4 i + 3 j + 4 k

c = i + α j + β k

| c | = \sqrt{3}

α

β

A

\to a =^i +^i +^k, \to b = 4^i + 3^i + 4^k

\to c =^i + α^j + β^k

| \to c | = \sqrt{3}

α = \pm 1, β = 1

R

\to a =^i +^j +^k,^b = 4^i + 3^j + 4^k

\to c =^i + α^j + β^k

| \to c | = \sqrt{3},

\to a, \to b, \to c

\to b, \to c

(\to a, \to b) = α, (\to a, \to c) = β

\to a \times (\to b \times \to c) = \frac{\to b + \to c}{2}

c o s (α + β) =

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