
Let a, b and c be non-zero vectors such that no two are collinear and $(a\times b)\times c=\dfrac{1}{3}|b||c|a$ . If $\theta $ is the acute angle between the vectors b and c, then $\sin \theta $ equals
(a). $\dfrac{2\sqrt{2}}{3}$
(b). $\dfrac{\sqrt{2}}{3}$
(c). $\dfrac{2}{3}$
(d). $\dfrac{1}{3}$
Answer
598.2k+ views
Hint: Vector triple product is given by $a\times (b\times c)=b(a\cdot c)-c(a\cdot b)$ . We can expand $(a\times b)\times c$ and compare it with $\dfrac{1}{3}|b||c|a$ and proceed.
Complete step-by-step solution -
We can see that the expression in the right is only in terms of $a$ only and the identity for vector triple products is as follows:
$(a\times b)\times c=b(a.c)-a(b.c)$
On comparing this identity with the given equation we get
$b(a.c)-a(b.c)=\dfrac{1}{3}|b||c|a$
This equation is true only if $b(a.c)=0$. Thus, we have to conditions- either $b=0$or $a.c=0$. The former cannot be true because then $(a\times b)\times c=0$.
On equating two sides we get
$\begin{align}
& b.c=-\dfrac{1}{3}|b||c|\ \\
& or|b||c|\cos \theta =-\dfrac{1}{3}|b||c| \\
& or\ \cos \theta =-\dfrac{1}{3} \\
\end{align}$
But, in question it has been given that $\theta $is the acute angle between the two vectors and cosine value of an acute angle can never be less than 0.
Therefore, $(\pi -\theta )$ must be the angle between the tails or heads of vectors $a$ and $b$ which is an obtuse angle. So, we can write
$\begin{align}
& \cos (\pi -\theta )=-\dfrac{1}{3} \\
& -\cos \theta =-\dfrac{1}{3} \\
& \cos \theta =\dfrac{1}{3} \\
& \therefore \sin \theta =\sqrt{1-{{\cos }^{2}}\theta }=\sqrt{1-\dfrac{1}{9}}=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt{2}}{3} \\
\end{align}$
Hence, the answer is (a)
Note: One might be misled into thinking that since the value of cosine of the angle $\theta $ is a negative number, while $\theta $ itself is acute and conclude that the question has some discrepancy. The subtlety in the semantics of the question is to blame for that. Please note that the question mentions “acute angle between the vectors $b$ and $c$” and not the actual angle between $b$ and $c$ and we conclude later that the actual angle is obtuse. One must also be careful in comparing equations as one might think that $b=0$ but that would render other information provided in the question redundant.
Complete step-by-step solution -
We can see that the expression in the right is only in terms of $a$ only and the identity for vector triple products is as follows:
$(a\times b)\times c=b(a.c)-a(b.c)$
On comparing this identity with the given equation we get
$b(a.c)-a(b.c)=\dfrac{1}{3}|b||c|a$
This equation is true only if $b(a.c)=0$. Thus, we have to conditions- either $b=0$or $a.c=0$. The former cannot be true because then $(a\times b)\times c=0$.
On equating two sides we get
$\begin{align}
& b.c=-\dfrac{1}{3}|b||c|\ \\
& or|b||c|\cos \theta =-\dfrac{1}{3}|b||c| \\
& or\ \cos \theta =-\dfrac{1}{3} \\
\end{align}$
But, in question it has been given that $\theta $is the acute angle between the two vectors and cosine value of an acute angle can never be less than 0.
Therefore, $(\pi -\theta )$ must be the angle between the tails or heads of vectors $a$ and $b$ which is an obtuse angle. So, we can write
$\begin{align}
& \cos (\pi -\theta )=-\dfrac{1}{3} \\
& -\cos \theta =-\dfrac{1}{3} \\
& \cos \theta =\dfrac{1}{3} \\
& \therefore \sin \theta =\sqrt{1-{{\cos }^{2}}\theta }=\sqrt{1-\dfrac{1}{9}}=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt{2}}{3} \\
\end{align}$
Hence, the answer is (a)
Note: One might be misled into thinking that since the value of cosine of the angle $\theta $ is a negative number, while $\theta $ itself is acute and conclude that the question has some discrepancy. The subtlety in the semantics of the question is to blame for that. Please note that the question mentions “acute angle between the vectors $b$ and $c$” and not the actual angle between $b$ and $c$ and we conclude later that the actual angle is obtuse. One must also be careful in comparing equations as one might think that $b=0$ but that would render other information provided in the question redundant.
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