The volume of a tetrahedron whose vertices are $4 \hat{i}+5 \hat{j}+\hat{k}$, $-\hat{j}+\hat{k}$, $3 \hat{i}+9 \hat{j}+4 \hat{k}$ and $-2 \hat{i}+4 \hat{j}+4 \hat{k}$ is (in cubic units)

  • A
    $\frac{14}{3}$
  • B
    $5$
  • C
    $6$
  • D
    $30$

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If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors,then $\frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{\vec{c} \cdot (\vec{a} \times \vec{b})} + \frac{\vec{b} \cdot (\vec{a} \times \vec{c})}{\vec{c} \cdot (\vec{a} \times \vec{b})} = \dots$

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If $a, b,$ and $c$ are coplanar unit vectors,find the value of the scalar triple product $[2a - b, 2b - c, 2c - a]$.

If $\overline{a}, \overline{b}, \overline{c}$ are mutually perpendicular vectors having magnitudes $1, 2, 3$ respectively,then the value of $[\overline{a}+\overline{b}+\overline{c} \quad \overline{b}-\overline{a} \quad \overline{c}]$ is

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

Let $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{B} = \hat{i}$,and $\overrightarrow{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$. If $C_2 = -1$ and $C_3 = 1$,then to make the three vectors coplanar:

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