If $\vec{a}$, $\vec{b}$, $\vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=1$ & $\vec{a} \cdot \vec{c}=1 / 2$ then

(P) $\vec{a}$, $\vec{b}$ and $\vec{c}$ are non-coplanar

(Q) $\vec{a}$, $\vec{b}$ and $\vec{d}$ are non-coplanar

(R) $\vec{b}$ and $\vec{d}$ are non-parallel

(S) $\vec{a}$ and $\vec{d}$ are parallel and $\vec{b}$ and $\vec{c}$ are parallel.

$|\vec{a}|=|\vec{b}|=|\vec{c}|=|\vec{d}|=1$

Let $\theta$ be the angle bet $(\vec{a} \times \vec{b})$ and $(\vec{c} \times \vec{d})$

$\theta_1$ be the angle between $\vec{a}$ and $\vec{b}$

$\theta_2$ be the angle between $\vec{c}$ and $\vec{d}$

$I=(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=|\vec{a} \times \vec{b}||\vec{c} \times \vec{d}| \cos \theta=|\vec{a}||\vec{b}| \sin \theta_1|\vec{c}||\vec{d}| \sin \theta_2 \cos \theta$

$=\sin \theta_1 \sin \theta_2 \cos \theta$

$\theta_1=\theta_2=\frac{\pi}{2}$

$\theta=0$ $\Rightarrow \vec{a} \perp \vec{b}, \vec{c} \perp \vec{d},(\vec{a} \times \vec{b}) |(\vec{c} \times \vec{d})$

$(\vec{a} \times \vec{b})=k(\vec{c} \times \vec{d})-\cdots *$

$(\vec{a} \times \vec{b}) \cdot \vec{c}=k(\vec{c} \times \vec{d}) \cdot \vec{c}=0$

$\Rightarrow[\vec{a} \vec{b} \vec{c}]=0 \Rightarrow$ $\vec{a}$, $\vec{b}$ and $\vec{c}$ one couple

$(\vec{a} \times \vec{b}) \cdot \vec{d}=k(\vec{c} \times \vec{d}) \cdot \vec{d}=0$

$\Rightarrow[\vec{a} \vec{b} \vec{d}]=0 \Rightarrow$ $\vec{a}$, $\vec{b}$ and $\vec{d}$ one

Suppose

$\vec{b} || \vec{d}$

$\Rightarrow \vec{b} =k \vec{d}$

$= \pm \vec{d}$ $(\vec{d}$ and $\vec{b}$ are unit vectors $)$

$1=(\vec{a} \times \vec{b}) \cdot\left(\vec{c} \times d^2\right) \Rightarrow \pm 1=(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{b})$

$\Rightarrow[\vec{a} \times \vec{b} \vec{c}, \vec{b}]= \pm 1=[\vec{c} \vec{b} \vec{a} \times \vec{b}]$

$\Rightarrow \vec{c} \cdot(\vec{b} \times(\vec{a} \times \vec{b}))= \pm 1 \Rightarrow \vec{c}(\vec{a}-(\vec{b} \cdot \vec{a}) \vec{b})= \pm 1$

$\Rightarrow \vec{c} \cdot \vec{a}= \pm 1 \Rightarrow \leftarrow$

$\vec{a}|\vec{a}$ and $\vec{b}| \vec{c}$

$\vec{a} \cdot \vec{d}=1=\vec{b} \cdot \vec{c}$

$\vec{a}= \pm \vec{d}$ and $\vec{b}= \pm \vec{c}$

$(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=(\vec{a} \times \vec{b})(\vec{b} \times \vec{a})= \pm 1$

$(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=(\vec{a} \times \vec{b})(\vec{b} \times \vec{a})= \pm 1$

$(R)$ is correct

Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ are three non-coplanar vectors.

$(\vec{a}+\vec{b}+\vec{c}) \cdot[(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})]$ equals

(P) 0

(Q) $[\vec{c} \quad \vec{b} \vec{c}]$

(R) $2[\vec{c} \vec{b} \vec{c}]$

(S) $-\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$

$(\vec{a}+\vec{b}+\vec{c}) \cdot[(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})]$

$=(\vec{a}+\vec{b}+\vec{c}) \cdot[\vec{a} \times \vec{a}+\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}]$

$=\vec{a} [ \vec{a} \times \vec{a}+\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c} ] +\vec{b}[ \vec{a} \times \vec{a}+\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c} ] +\vec{c}[ \vec{a} \times \vec{a}+\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c} ]$

$=\left[\begin{array}{lll}\vec{a} & \vec{a} & \vec{a}\end{array}\right]+\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{a}\end{array}\right]+\left[\begin{array}{lll}\vec{a} & \vec{a} & \vec{c}\end{array}\right]$

$+\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{a}\end{array}\right]+\left[\begin{array}{lll}\vec{b} & \vec{b} & \vec{a}\end{array}\right]$

$+\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{b} & \vec{b} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{c} & \vec{a} & \vec{a}\end{array}\right]$

$+\left[\begin{array}{lll}\vec{c} & \vec{b} & \vec{a}\end{array}\right]+\left[\begin{array}{lll}\vec{c} & \vec{a} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{c} & \vec{b} & \vec{c}\end{array}\right]$

$=\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{c} & \vec{b} & \vec{a}\end{array}\right]$

$=\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]-\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]-\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$

(S) is correct.

Let $\vec{a}, \vec{b} \& \vec{c}$ be three non-zero vectors such that no two of them are collinear & $(\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}$. If $\theta$ is the angle between ${ }^n$ $\vec{b} \& \vec{c}$ then find the value of $\theta$.

Solution:

$(\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}$

$-\vec{c} \times(\vec{a} \times \vec{b})=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}$

$-(\vec{c} \cdot \vec{b}) \vec{a}+(\vec{c} \cdot \vec{a}) \vec{b}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a} .$

$(\vec{c} \cdot \vec{a}) \vec{b}=\left(\frac{1}{3}|\vec{b}||\vec{c}|+\vec{c} \cdot \vec{b}\right) \vec{a}$

$\vec{c} \cdot \vec{a}=0$

and $\left(\frac{1}{3}|\vec{b}||\vec{c}|+|\vec{b}||\vec{c}| \cos \theta\right)=0$

$\Rightarrow \quad \cos \theta+\frac{1}{3}=0$

$\rightarrow \cos \theta=\frac{-1}{3}$

$\quad \sin \theta-\frac{\sqrt{8}}{3}=\frac{2 \sqrt{2}}{3}$

$\vec{a}, \vec{b}$ , $\vec{c}$ are vector such that $|\vec{b}|=|\vec{c}|$

Show that $[[(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})] \times(\vec{b} \times \vec{c})] \cdot(\vec{b}+\vec{c})=0$

Solution:

$(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})= \vec{a} \times \vec{a}+\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}$

${[(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})] \times(\vec{b} \times \vec{c})=[\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}] \times(\vec{b} \times \vec{c}) }$

$=(\vec{b} \times \vec{a}) \times(\vec{b} \times \vec{c})+(\vec{a} \times \vec{c}) \times(\vec{b} \times \vec{c})$

$= {[(\vec{b} \times \vec{a}) \cdot \vec{c}] \vec{b}-[(\vec{b} \times \vec{a}) \cdot \vec{b}] \vec{c} } +[(\vec{a} \times \vec{c}) \cdot \vec{c}] \vec{b}-[(\vec{a} \times \vec{c}) \cdot \vec{b}] \vec{c}$

$= {[\vec{b} \vec{a} \vec{c}] \vec{b}-[\vec{b}\vec{a} \vec{b}] \vec{c} } +[\vec{a} \vec{c} \vec{c}] \vec{b}-[\vec{a} \vec{c} \vec{b}] \vec{c}$

$=[\vec{b} \vec{a} \vec{c} ](\vec{b}-\vec{c})$

${[[(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})] \times(\vec{b} \times \vec{c})] \cdot(\vec{b}+\vec{c})}$

$=\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{c}\end{array}\right] (\vec{b}-\vec{c}) \cdot(\vec{b}+\vec{c})$

$=\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{c}\end{array}\right](\vec{b} \cdot \vec{b}+\vec{c} \vec{b}-\vec{c} \vec{b}-\vec{c} \cdot \vec{c})$

$=\left[\begin{array}{lll}\vec{b} & \vec{a} &\vec{c}\end{array}\right]\left(|\vec{b}|^2-|\vec{c}|^2\right)=\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{c}\end{array}\right] 0=0$

Solution:

$\vec{d}=x i+y j+z k$

$\vec{a} \cdot \vec{d}=0$

$1 \cdot x-y. 1=0 \Rightarrow x=y$

$\left[\begin{array}{lll}\vec{b} & \vec{c} & \vec{d}\end{array}\right]=0$

$\left|\begin{array}{ccc}0 & 1 & -1 \\ -1 & 0 & 1 \\ x & y & z\end{array}\right|=0$

$\Rightarrow 2 x+z=0 \Rightarrow z=-2 x$

$|\alpha| =1 =\sqrt{x^2+y^2+z^2}$

$x^2+y^2+z^2=1 \Rightarrow x^2+x^2+4x^2=1$

$6x^2=1 \Rightarrow 1/6$

$x\pm \frac{1}{\sqrt{6}} \Rightarrow 1/6$

$\Rightarrow xi +yj +zk = xi +xj + (-2x)k$ $=\pm\frac{1}{\sqrt{6}}(i + j -2k)$

Which of the following statements are meaningful?

(P) ${\vec{u}} \cdot(\vec{v} \times \vec{w})$

(Q) $(\vec{u} \cdot \vec{v}) \cdot \vec{\omega}$

(R) $(\vec{u} \cdot \vec{v}) \vec{\omega}$

(s) $\vec{u} \times(\vec{v} \cdot \vec{w})$

Answer:

$(P)$ and $(R)$ are meaningful.