Let a- b and c be non-zero vectors such that no two are collinear and a - b - c - 13 -b- -c- a- If - is the acute angle between the vectors b and c- then sin - equals

The correct option is B 2 √(2)3 Given, (a⃗×b⃗) ×c⃗ = 13 |b⃗| |c⃗| a⃗ Using the vector triple cross product we have (a⃗·c⃗) b⃗ #8722; (b⃗·c⃗) a⃗ ...

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Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

Let a, b and ...

Question

Let a, b and c be non-zero vectors such that no two are collinear and $(a \times b) \times c = \frac{1}{3} | b | | c | a .$ If $θ$ is the acute angle between the vectors b and c, then $s i n θ$ equals

\frac{2 \sqrt{2}}{3}

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\frac{\sqrt{2}}{3}

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\frac{2}{3}

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\frac{1}{3}

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Solution

The correct option is B $\frac{2 \sqrt{2}}{3}$
Given, $(\to a \times \to b) \times \to c = \frac{1}{3} | \to b | | \to c | \to a$

Using the vector triple cross product we have

$(\to a \cdot \to c) \to b - (\to b \cdot \to c) \to a = \frac{1}{3} | \to b | | \to c | \to a$

Equating the coefficient of $\to a$ , we get

$- | \to b | | \to c | cos θ = \frac{1}{3} | \to b | | \to c | ⟹ cos θ = - \frac{1}{3}$

$sin θ = \sqrt{1 - {cos}^{2} θ} = \sqrt{1 - \frac{1}{9}} = \frac{2 \sqrt{2}}{3}$

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Let a, b and c be non-zero vectors such that no two are collinear and (a×b)×c=13|b||c|a. If θ is the acute angle between the vectors b and c, then sinθ equals

Let a, b and c be non-zero vectors such that no two are collinear and $(a \times b) \times c = \frac{1}{3} | b | | c | a .$ If $θ$ is the acute angle between the vectors b and c, then $s i n θ$ equals