Let a-k-b-x -1-xk- and c-y-x -1-x-yk- Then -a-b-c- depends on

The correct option is C both x and y a⃗= k̂ b⃗=x î++(1 x) k̂ c⃗=y +x +(1+x y) k̂ [a⃗b⃗c⃗] amp;=|[ 0 amp; 1 amp; 1; x ...

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Question

Let $\to a =^j -^k, \to b = x^i +^j + (1 - x)^k$ and $\to c = y^i + x^j + (1 + x - y)^k$ . Then $[\to a, \to b, \to c]$ depends on

only y

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only x

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both x and y

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neither x nor y

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Similar questions

\to a =^i -^k, \to b = x^i +^j + (1 - x)^k

\to c = y^i + x^j + (1 + x - y)^k,

[\to a \to b \to c]

¯ a =^i -^k, ¯ b = x^i +^j + (1 - x)^k

¯ c = y^i + x^j + (1 + x - y)^k

[¯ a ¯ b ¯ c]

\to a =^i +^j +^k

\to b = - 3^i +^k

\to c =^i + 2^j + 3^k

A)

[\to a \times \to b

\to b \times \to c

\to c \times \to a]

B)

[\to a + \to b

\to b + \to c

\to c + \to a]

C)

[\to a

\to b

\to c]

[\to a

\to b

\to c]^{- 1}

D)

[\to a - \to b

\to b - \to c

\to c - \to a]

\to a =^j -^k

\to c =^i -^j -^k

\to b

\to a \times \to b + \to c = \to 0

\to a \cdot \to b = 3

\to a =^i +^j +^k, \to c =^j -^k

\to b

\to a \times \to b = \to c

\to a \cdot \to b = 3.

| \to b |

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