If $\theta$ is the angle between the vectors $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,then:

  • A
    $\cos \theta = \frac{4}{21}$
  • B
    $\cos \theta = \frac{3}{19}$
  • C
    $\cos \theta = \frac{2}{19}$
  • D
    $\cos \theta = \frac{5}{21}$

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If $l\vec{a} + m\vec{b} + n\vec{c} = \vec{0},$ where $l, m, n$ are scalars and $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular non-zero vectors,then

Let $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$. Let $\hat{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{c}$ 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 $(\bar{q} \times \bar{d})$.
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