Let a - - b - - k- and c be a vector such that a-c-b-0 and a-c-4- then -c-2 is equal to-

Given : nbsp; nbsp; a×c= b ⇒ (a× nbsp; nbsp; nbsp; c)×a= b×a ⇒ (a×c)×a= a×b ⇒ (a·a)c (c·a) a= a× nbsp; b ⇒ 2c 4a=a× nbsp; b nbsp; We calcula ...

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Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

Let a = î-ĵ, ...

Question

Let $\to a =^i -^j, \to b =^i +^j +^k$ and $\to c$ be a vector such that $\to a \times \to c + \to b = \to 0$ and $\to a \cdot \to c = 4$ , then $| \to c |^{2}$ is equal to:

\frac{19}{2}

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\frac{17}{2}

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Solution

The correct option is A $\frac{19}{2}$
Given :
$\to a \times \to c = - \to b$
$\Rightarrow (\to a \times \to c) \times \to a = - \to b \times \to a$
$\Rightarrow (\to a \times \to c) \times \to a = \to a \times \to b$
$\Rightarrow (\to a \cdot \to a) \to c - (\to c \cdot \to a) \to a = \to a \times \to b$
$\Rightarrow 2 \to c - 4 \to a = \to a \times \to b$
We calculate $\to a \times \to b$ .
$\to a \times \to b = ∣ ∣ ∣ ∣ ∣ \begin{matrix} i & ^j & ^k 1 & - 1 & 0 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = -^i -^j +^2 k$
So,
$2 \to c - 4 \to a = -^i -^j +^2 k$
$\Rightarrow 2 \to c = 3^i - 5^j + 2^k$
$\Rightarrow \to c = \frac{3}{2}^i - \frac{5}{2}^j +^k$

$\Rightarrow | \to c |^{2} = {(\frac{3}{2})}^{2} + {(\frac{5}{2})}^{2} + 1^{2} = \frac{19}{2}$

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Let →a=^i−^j,→b=^i+^j+^k and →c be a vector such that →a×→c+→b=→0 and →a⋅→c=4, then |→c|2 is equal to:

Let $\to a =^i -^j, \to b =^i +^j +^k$ and $\to c$ be a vector such that $\to a \times \to c + \to b = \to 0$ and $\to a \cdot \to c = 4$ , then $| \to c |^{2}$ is equal to: