If $\vec{x} \cdot \vec{y} = 0$ then,$(\vec{y} \times \vec{x}) \times \vec{x} = $ . . . . . . . where,$|\vec{x}| = 1$.

  • A
    $\vec{y}$
  • B
    $\vec{x}$
  • C
    $-\vec{y}$
  • D
    $\vec{x} \times \vec{y}$

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Let $\vec{x}, \vec{y}$ and $\vec{z}$ be three vectors each of magnitude $\sqrt{2}$ and the angle between each pair of them is $\frac{\pi}{3}$. If $\vec{a}$ is a nonzero vector perpendicular to $\vec{x}$ and $\vec{y} \times \vec{z}$ and $\vec{b}$ is a nonzero vector perpendicular to $\vec{y}$ and $\vec{z} \times \vec{x}$,then
$(A)$ $\vec{b}=(\vec{b} \cdot \vec{z})(\vec{z}-\vec{x})$
$(B)$ $\vec{a}=(\vec{a} \cdot \vec{y})(\vec{y}-\vec{z})$
$(C)$ $\vec{a} \cdot \vec{b}=-(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})$
$(D)$ $\vec{a}=(\vec{a} \cdot \vec{y})(\vec{z}-\vec{y})$

Let $\vec{a}=-\hat{i}+\hat{j}+2\hat{k}$, $\vec{b}=\hat{i}-\hat{j}-3\hat{k}$, $\vec{c}=\vec{a}\times\vec{b}$ and $\vec{d}=\vec{c}\times\vec{a}$. Then $(\vec{a}-\vec{b}) \cdot \vec{d}$ is equal to :

If $\vec{a}$ and $\vec{b}$ are vectors in space given by $\vec{a}=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}$ and $\vec{b}=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}$,then the value of $(2 \vec{a}+\vec{b}) \cdot[(\vec{a} \times \vec{b}) \times(\vec{a}-2 \vec{b})]$ is

If $\overline{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\overline{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$,then the value of $(\overline{a}-2 \overline{b}) \cdot \{(\overline{a} \times \overline{b}) \times (2 \overline{a}+\overline{b})\}$ is

For any vector $r$, the expression $i \times(r \times i) + j \times(r \times j) + k \times(r \times k)$ is equal to:

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