Given value of vectors a,b, b.c and a+(b×c), Find (c.a)

[Aurelius120]

Homework Statement

$a=\hat i-2\hat j+3\hat k$ and $b=\hat i+\hat j+\hat k$
$c$ is a vector such that $a+(b\times c)=0$ and $b.c=5$
Find $3(c.a)$

Relevant Equations

$$(p\times q)\times r=(p.r)q-(q.r)p$$

I thought this was too easy
$$a+(b\times c)=0\implies a=-(b\times c)=(c\times b)$$
Then
$$3(c.a)=3(c.(c\times b))=0$$
Since cross product of vectors is perpendicular to both vectors and dot product of perpendicular vectors is zero.

Now here's the problem, correct answer given is 10. But how do I get that?Why is my answer wrong?
It seems to hint at using vector triple product

 

[Delta2]

I crack my head to find why your method is wrong, but I find indeed 10 IF I use another method where I take the triple cross product $a\times(b\times c)=0$ (you can deduce this equation by taking the cross product with $a$ of the given equation.

 

[Delta2]

Well the problem statement has a contradiction, it gives as $a$ clearly different than the zero vector, but it also gives that $a+(b\times c)=0$ from which you can deduce that either a is zero or that a is perpendicular to b (and perpendicular to c), neither can hold according to what is given for a and b.

Anyway that's what I concluded with my internal thinking but lets wait what @Orodruin or @pasmith have to say for this, two of the advisors that almost always find a "jet" and rigorous explanation.

@PeroK too plz have a look at this.

 

[Orodruin]

There is an obvious self-contraction in the problem statement. From $\vec a + \vec b \times \vec c = 0$ it is necessary that $\vec a \perp \vec b$ but the given $\vec a$ and $\vec b$ are not perpendicular by virtue of $\vec a\cdot \vec b = 1-2+3 = 2 \neq 0$.

 

[Aurelius120]

There is an obvious self-contraction in the problem statement. From $\vec a + \vec b \times \vec c = 0$ it is necessary that $\vec a \perp \vec b$ but the given $\vec a$ and $\vec b$ are not perpendicular by virtue of $\vec a\cdot \vec b = 1-2+3 = 2 \neq 0$.

Well the problem statement has a contradiction, it gives as $a$ clearly different than the zero vector, but it also gives that $a+(b\times c)=0$ from which you can deduce that either a is zero or that a is perpendicular to b (and perpendicular to c), neither can hold according to what is given for a and b.

Anyway that's what I concluded with my internal thinking but lets wait what @Orodruin or @pasmith have to say for this, two of the advisors that almost always find a "jet" and rigorous explanation.

@PeroK too plz have a look at this.

And this was a 'Previous Year Question' in an exam. They tried too hard to make it difficult.
[Attachment missing. Link removed by moderator]
This solution that gives 10.

 

[Orodruin]

It is not the first and not the last time someone makes a mistake when formulating an exam.

My favourite was when our classical mechanics teacher asked us to find the natural frequency of oscillations around a stable equilibrium that was, in fact, unstable. The rest of the class got imaginary frequencies. I showed that the equilibrium was unstable, found the actual stable equilibrium of the system (which was significantly more complex), and found the frequency for that equilibrium. I got 6 out of 3 points for that problem.