Given a- - x - - y - - 2 k- - b- - - - - - k- c- - - - 2 - - a-b- - -2 - a- - c- - 4 then

The correct option is D [a⃗b⃗c⃗] = |a⃗|^2 Given: a⃗= xî+yĵ+2k̂ b⃗= î ĵ+k̂ c⃗= î+2ĵ Now given that (a⃗∧b⃗)= π2 So, a⃗.b⃗=0⇒ x y+2= ...

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Conic Section

Given a⃗ = ...

Question

Given $\to a = x^i + y^j + 2^k, \to b =^i -^j +^k, \to c =^i + 2^j; (\to a \land \to b) = \frac{π}{2}, \to a . \to c = 4$ then

[\to a \to b \to c]^{2} = | \to a |

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[\to a \to b \to c] = | \to a |

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[\to a \to b \to c] = 0

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[\to a \to b \to c] = | \to a |^{2}

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If \to a = 2^i + 3^j +^k, \to b =^i - 2^j +^k and \to c = - 3^i +^j + 2^k,

[\to a \to b \to c] .

If $\to a, \to b$ and $\to c$ are three non-coplanar vectors, then $(\to a + \to b + \to c) \cdot [(\to a + \to b) \times (\to a + \to c)]$ equals

\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]

\to a, \to b, \to c

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}

\to q = \frac{\to c \times \to a}{[\to a \to b \to c]}

\to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

(2 \to a + 3 \to b + 4 \to c) . \to p + (2 \to b + 3 \to c + 4 \to a) . \to q +

(2 \to c + 3 \to a + 4 \to b) . \to r =

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

(\to a . \to b) \neq 0

[\to a \to b \to c] =

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