The angle between two diagonals of a cube is:

  • A
    $\cos ^{-1}\left(\frac{1}{3}\right)$
  • B
    $\sin ^{-1}\left(\frac{1}{3}\right)$
  • C
    $\frac{\pi}{2}-\cos ^{-1}\left(\frac{1}{3}\right)$
  • D
    $\frac{\pi}{2}-\sin ^{-1}\left(\frac{1}{3}\right)$

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If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}-\bar{b}|^2+|\bar{b}-\bar{c}|^2+|\bar{c}-\bar{a}|^2=15$,then $|\bar{a}-\bar{b}-\bar{c}|^2-4(\bar{b} \cdot \bar{c})=$

The value of $\lambda$ for which the vectors $2\lambda \hat{i} + \hat{j} - \hat{k}$ and $2\hat{j} + \hat{k}$ are perpendicular is:

Consider the following Assertion $(A)$ and Reason $(R)$:
Assertion $(A)$: The two lines $\bar{r}=\bar{a}+t(\bar{b})$ and $\bar{r}=\bar{b}+s(\bar{a})$ intersect each other.
Reason $(R)$: The shortest distance between the lines $\bar{r}=\bar{p}+t(\bar{q})$ and $\bar{r}=\bar{c}+s(\bar{d})$ is equal to the length of the projection of the vector $(\bar{p}-\bar{c})$ on $(\bar{q} \times \bar{d})$.
The correct answer is:

Find the sine of the angle between the vectors $\vec{a}=3 \hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}$.

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View Solution

If $\vec{a}$ and $\vec{b}$ are perpendicular unit vectors and vector $\vec{c}$ is such that $\vec{c} = \vec{a} + \vec{b}$,then $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b})$ is

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