Scalar triple product practice · Mixed difficulty

If $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar vectors and $p, q$ are real numbers, then the equality $\left[ {\begin{array}{llllllllllllllllllll} {3\,\vec u}&{p\,\vec v}&{p\,\overrightarrow w } \end{array}} \right] - \left[ {\begin{array}{llllllllllllllllllll} {p\,\vec v}&{\vec w}&{q\,\vec u} \end{array}} \right] - \left[ {\begin{array}{llllllllllllllllllll} {2\vec w}&{q\vec v}&{q\vec u} \end{array}} \right] = 0$ holds for

If $a$ and ${b}$ are two non-parallel vectors satisfying $|{a}|=|{b}|$, then the vector $({a}+{b}) \times({a} \times {b})$ is parallel to

If the volume of parallelopiped formed by the vectors a, b, c as coterminous edges is 27 cu units, then the volume of the parallelopiped have

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit coplanar vectors, then the scalar triple product $\left[\begin{array}{ll}2 \vec{a}-\vec{b} & 2 \vec{b}-\vec{c} \quad 2 \vec{c}-\vec{a}\end{array}\right]=$

Let $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{k}}, \mathbf{b}=x \hat{\mathbf{i}}+\hat{\mathbf{j}}+(1-x) \hat{\mathbf{k}}$ and $\mathbf{c}=y \hat{\mathbf{i}}+x \hat{\mathbf{j}}+(1+x-y) \hat{\mathbf{k}}$. Then, $[\mathbf{a\,} \mathbf{b\,} \mathbf{c}]$ depends on

Let ${x_0}$ be the point of local maximum of $f(x) = a.(b \times c),$ where $a = x\widehat i - 2\widehat j + 3\widehat k,b = - 2\widehat i + x\widehat j - \widehat k\mathrm{ }\,\mathrm{and }\,\,c = 7\widehat i - 2\widehat j + x\widehat k$ Then the value $a.\,b + b.\,c + c.\,a$ at $x = {x_0}$ is

If the lines $\frac{x-1}{2}=\frac{y-2}{x_1}=\frac{z-3}{x_2}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ lie in the same plane, then for the equation $x_1 t^2+\left(x_2+2\right) t+a=0$, sum of roots of equation is

If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors such that $\mathbf{a}=\mathbf{b}+\mathbf{c}$ and the angle between $\mathbf{b}$ and $\mathbf{c}$ is $\pi / 2$, then

If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are any three vectors forming a linearly independent system, then $\forall \theta \in R$
$\left[\mathbf{a} \cos \theta+\mathbf{b} \sin \theta+\mathbf{c o s} 2 \theta, \mathbf{a} \cos \left(\frac{2 \pi}{3}+\theta\right)\right.$ $+ b\sin \left( {\frac{{2\pi }}{3} + \theta } \right) + c\cos 2\left( {\frac{{2\pi }}{3} + \theta } \right){\text{, }}$
$acos\left. {\left( {\theta - \frac{{2\pi }}{3}} \right) + b\sin \left( {\theta - \frac{{2\pi }}{3}} \right) + c\cos 2\left( {\theta - \frac{{2\pi }}{3}} \right)} \right]$ equals

Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors along the adjacent edges of a tetrahedron, if $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=2$ and $\mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=2$, then volume of tetrahedron is

$\left( {\vec a \times \vec b} \right) \times \left( {\vec a \times \vec c} \right)\cdot\vec d$ equals

If $\mathbf{a}$ and $\mathbf{b}$ are orthogonal unit vectors, then for any non-zero vector $\mathbf{r}$, the vector $(\mathbf{r} \times \mathbf{a})$ is equal to

The position vectors of three vertices A, B, C of a tetrahedron OABC with respect to its vertex O are $\hat i,\,6\hat j,\,\hat k,\,$ then its volume (in cu units) is

The three concurrent edges of a parallelopiped represent the vectors $\vec{a}, \vec{b}, \vec{c}$ such that $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=V$. Then, the volume of the parallelopiped whose three concurrent edges are the three diagonals of three faces of the given parallelopiped is

The three vectors $\mathbf{i}+\mathbf{j}, \mathbf{j}+\mathbf{k}, \mathbf{k}+\mathbf{i}$ taken two at a time form three planes. The three unit vectors drawn perpendicular to these three planes form a parallelopiped of volume

If $\mathbf{u}, \mathbf{v}$ and $\mathbf{w}$ are three non-coplanar vectors, then $(\mathbf{u}+\mathbf{v}-\mathbf{w}) \cdot[(\mathbf{u}-\mathbf{v}) \times(\mathbf{v}-\mathbf{w})]$ equal to

If $\mathbf{\vec a},\mathbf{\vec b},\mathbf{\vec c}$ are unit coplanar vectors, then $\left[ {2\mathbf{\vec a} - \mathbf{\vec b}\;2\;\mathbf{\vec b} - \mathbf{\vec c}\;2\;\mathbf{\vec c} - \mathbf{\vec a}} \right]$ is equal to

If $\mathbf{\vec u},\;\mathbf{\vec v}\;\mathrm{and}\;\mathbf{\vec w}\;$ are three non-coplanar vectors, then ($\mathbf{\vec u} + \mathbf{\vec v} - \mathbf{\vec w})\cdot\left[ {\left( {\mathbf{\vec u} - \mathbf{\vec v}} \right) \times \left( {\mathbf{\vec v} - \mathbf{\vec w}} \right)} \right]$
equals

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=x \hat{i}+(x-2) \hat{j}-\hat{k}$. If the vector $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$, then $x$ equals

If $\vec{d}=\lambda(\vec{a} \times \vec{b})+\mu(\vec{b} \times \vec{c})+\gamma(\vec{c} \times \vec{a})$ and $[\vec{a}, \vec{b}, \vec{c}]=1$ Then $\lambda+\mu+\gamma$ is equal to