Vector x satisfying the relation A - x -c- A - x - B is

The correct option is B c A ( A × B ) | A | ^ 2 A⃗×x⃗ =B⃗ ⇒( A⃗×x⃗ #160; ) A⃗ =B⃗×A⃗ ⇒ A⃗( x⃗ .A⃗ #160; ) +x⃗( A⃗×A⃗ #160; ) =B ...

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Standard X

Mathematics

Section Formula

Vector x sa...

Question

Vector $\to x$ satisfying the relation $\to A . \to x = c; \to A \times \to x = \to B$ is

\frac{c \to A - (\to A \times \to B)}{∣ ∣ ∣ \to A ∣ ∣ ∣}

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\frac{c \to A - (\to A \times \to B)}{{∣ ∣ ∣ \to A ∣ ∣ ∣}^{2}}

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\frac{c \to A + (\to A \times \to B)}{{∣ ∣ ∣ \to A ∣ ∣ ∣}^{2}}

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\frac{c \to A - 2 (\to A \times \to B)}{{∣ ∣ ∣ \to A ∣ ∣ ∣}^{2}}

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Solution

The correct option is B $\frac{c \to A - (\to A \times \to B)}{{∣ ∣ ∣ \to A ∣ ∣ ∣}^{2}}$
$\to A \times \to x = \to B$
$\Rightarrow (\to A \times \to x) \to A = \to B \times \to A$
$\Rightarrow - \to A (\to x . \to A) + \to x (\to A \times \to A) = \to B \times \to A$
$\Rightarrow= C \to A + \to x {| A |}^{2} = \to B \times \to A$
$\Rightarrow \to x {| A |}^{2} = C \to A + \to B \times \to A \Rightarrow \to x = \frac{C \to A - (\to A \times \to B)}{{∣ ∣ \to A ∣ ∣}^{2}}$

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Vector →x satisfying the relation →A.→x=c;→A×→x=→B is

The correct option is B c→A−(→A×→B)∣∣∣→A∣∣∣2→A×→x=→B⇒(→A×→x)→A=→B×→A⇒−→A(→x.→A)+→x(→A×→A)=→B×→A⇒=C→A+→x|A|2=→B×→A⇒→x|A|2=C→A+→B×→A⇒→x=C→A−(→A×→B)∣∣→A∣∣2

Vector $\to x$ satisfying the relation $\to A . \to x = c; \to A \times \to x = \to B$ is