Let a-i-k- b-xi-j-1-xk- and c-yi-xj-1-x-yk- then -a- b- c- depends on

The correct option is D Neither x nor y We have #160; [a⃗ b⃗ c⃗] = #160;a⃗.(b⃗× #160;c⃗) = #160;c⃗.(a⃗× #160;b⃗) = (y #160;i⃗ ...

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Standard VI

Mathematics

Idea of a Set

Let a⃗=i⃗-k...

Question

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

Only

x

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Only

y

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Neither

x

nor

y

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Both

x

and

y

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Solution

The correct option is D Neither $x$ nor $y$
We have
$[\to a \to b \to c] = \to a . (\to b \times \to c) = \to c . (\to a \times \to b)$

$= (y \to i + x \to j + (1 + x - y) \to k) . [(\to i - \to k) \times (x \to i + \to j + (1 - x) \to k)]$

$= (y \to i + x \to j + (1 + x - y) \to k) . [\to k + (x - 1) \to j - x \to j + \to i]$

$= y - x + 1 + x - y$

$= 1$

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Let →a=→i−→k, →b=x→i+→j+(1−x)→k and →c=y→i+x→j+(1+x−y)→k, then [→a→b→c] depends on

The correct option is D Neither x nor yWe have [→a→b→c]=→a.(→b×→c)=→c.(→a×→b)=(y→i+x→j+(1+x−y)→k).[(→i−→k)×(x→i+→j+(1−x)→k)]=(y→i+x→j+(1+x−y)→k).[→k+(x−1)→j−x→j+→i]=y−x+1+x−y=1

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