Let a- b and c be non-zero vectors such that no two are collinear and a- b - c- 1 3 - b - - c - a- If - is the acute angle between the vectors b and c- then sin- is equal to

The correct option is A 2√( 2 ) 3 We have, ( a× b ) × c= 1 3 | b | | c | a ⇒( a· c ) · b ( b· c ) · a= 1 3 | b | | c | a ⇒( a· c ) · b {( ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

Let a, b ...

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 $sin θ$ is equal to

\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 A $\frac{2 \sqrt{2}}{3}$
We have, $(a \times b) \times c = \frac{1}{3} | b | | c | a$
$\Rightarrow (a \cdot c) \cdot b - (b \cdot c) \cdot a = \frac{1}{3} | b | | c | a$
$\Rightarrow (a \cdot c) \cdot b - {(b \cdot c) + \frac{1}{3} | b | | c |} a = 0$
As $a$ and $b$ are not parallel, $(a \cdot c) = 0$ and $b \cdot c + \frac{1}{3} | b | | c | = 0$
$\Rightarrow | b | | c | cos θ + \frac{1}{3} | b | | c | = 0$
$\Rightarrow cos θ = - \frac{1}{3}$
$∴ sin θ = \sqrt{1 - {cos}^{2} θ} = \sqrt{1 - \frac{1}{9}}$
$= \sqrt{\frac{8}{9}} = \frac{2 \sqrt{2}}{3}$

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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 is equal to

The correct option is A 2√23We have, (a×b)×c=13|b||c|a⇒(a⋅c)⋅b−(b⋅c)⋅a=13|b||c|a⇒(a⋅c)⋅b−{(b⋅c)+13|b||c|}a=0As a and b are not parallel, (a⋅c)=0 and b⋅c+13|b||c|=0⇒|b||c|cosθ+13|b||c|=0⇒cosθ=−13∴sinθ=√1−cos2θ=√1−19 =√89=2√23

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 $sin θ$ is equal to