Ques. -13

Geometry Level 4

Let $\vec{a},\vec{b},\vec{c}$ be non-zero vectors such that no two are collinear and $(\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3} \cdot |\vec{b}| \cdot |\vec{c}| \cdot \vec{a}$ . If $\theta$ is the acute angle between the vectors $\vec{b} \ and \ \vec{c}$ , then $sin\theta$ is equal to :

If you're looking to promote your Rank in JEE-MAINS-2015, then go for solving this set : Expected JEE-MAINS-2015 .

\frac{1}{3}

\frac{2}{3}

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

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

2 solutions

Cody Martin
Feb 25, 2015

apply vector triple product on LHS $(c.b)a-(c.a)b=(\frac{ -b.c }{ 3\cos \theta })a$ $a[(b.c)+\frac{ b.c }{ 3\cos \theta }]=(c.a)b$ $since,a \neq \lambda b;\lambda \in R$ $c.a \neq0$ as c and a vectors are non collinear $b.c=\frac{ -b.c }{ 3\cos \theta }=>\cos \theta=\frac{ -1 }{ 3 }$ $\sin \theta=\frac{ 2\sqrt{2} }{ 3 }$

cant understand

prajwal kavad - 6 years, 2 months ago

oh! i input cosine value :(

Rohith M.Athreya - 4 years, 5 months ago

Aman Gautam
Apr 4, 2015

this question appeared in jee mains..!!

Ques. -13

If you're looking to promote your Rank in JEE-MAINS-2015, then go for solving this set : Expected JEE-MAINS-2015 .

2 solutions

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