find range for dependent

Question

There are three vector $a$, $b$, $c$ in three-dimensional real vector space, and the inner product between them $a\cdot a=b\cdot b = a\cdot c= 1, a\cdot b= 0, c\cdot c= 4$. When setting $x = b\cdot c$, answer the following question: when $a, b, c$ are linearly dependent, find all possible values ​of $x$. (dot here means dot product)

For dependent condition

$$\begin{align} (a×b)·c &= 0\\ a·(b×c) &= 0\\ a(bc \sin θ)&=0 \end{align}$$

So $\theta = 0, \pi$.

Then $$\begin{align} x&=|b||c| \cos \theta \\ x&=2 \cos \theta \\ \implies x &= 2 \cos 0, x = 2 \cos \pi \\ x &= \mp 2 \end{align}$$

Am I right using triple cross product? or should i find $\theta$?

Answer 1

Linear dependence places the vectors in the same plane. This makes the problem easier because we can focus on the angle.

From $c\cdot c=4$ we get that $|c|=2$.

So $a\cdot c=2\cos\theta=1\implies\cos\theta=\frac{1}{2}\implies\theta=60^{\circ}$.

From $b\cdot b=1$ and $a\cdot b=0$ we get that $b$ is a unit vector perpendicular to $a$.

Since the angle between $a$ and $c$ is $60^{\circ}$, the angle between $b$ and $c$ is either $30^{\circ}$ or $150^{\circ}$.

Therefore, $x=2\cos{30^{\circ}}$ or $x=2\cos{150^{\circ}}$ so $x=\pm \sqrt{3}$

Answer 2

For the first identity, $a \times b$ is orthogonal to both $a,b$, hence $c$ must be in the plane spanned by the orthonormal $a,b$.

So we can write $c$ as $$ c = \alpha a + \beta b $$

plugging this in we see that $$ 1 = a \cdot c = \alpha a \cdot a = \alpha $$

and $$ x = b \cdot c = \beta b \cdot b = \beta $$

hence the norm squared of $c$ is $$ c\cdot c = \alpha^2 + \beta^2 = 1 + x^2 = 4 $$

this gives $$ x = \pm \sqrt{3} $$