A unit vector parallel to the straight line $\frac{\mathrm{x} - 2}{3} = \frac{3 + \mathrm{y}}{- 1} = \frac{\mathrm{z} - 2}{- 4}$ is

• $\frac{1}{\sqrt{26}}\left(3\hat{\mathrm{i}} - \hat{\mathrm{j}} + 4\hat{\mathrm{k}}\right)$

• $\frac{1}{\sqrt{26}}\left(\hat{\mathrm{i}} + 3\hat{\mathrm{j}} - \hat{\mathrm{k}}\right)$

• $\frac{1}{\sqrt{26}}\left(3\hat{\mathrm{i}} - \hat{\mathrm{j}} - 4\hat{\mathrm{k}}\right)$

• $\frac{1}{\sqrt{26}}\left(3\hat{\mathrm{i}} + \hat{\mathrm{j}} + 4\hat{\mathrm{k}}\right)$

The angle between the two vectors $\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}$ and $2\hat{\mathrm{i}} - 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$ is equal to

• ${\cos }^{-1}\left(\frac{2}{3}\right)$

• ${\cos }^{-1}\left(\frac{1}{6}\right)$

• ${\cos }^{-1}\left(\frac{5}{6}\right)$

• ${\cos }^{-1}\left(\frac{1}{3}\right)$

If a = $\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}$, b = $4\hat{\mathrm{i}} + 3\hat{\mathrm{j}} + 4\hat{\mathrm{k}}$ and $\mathrm{c} = \hat{\mathrm{i}} + \mathrm{\alpha }\hat{\mathrm{j}} + \mathrm{\beta }\hat{\mathrm{k}}$ are copalnar and $\left|\mathrm{c}\right| = \sqrt{3}$, then

• $\mathrm{\alpha } = \sqrt{2}, \mathrm{\beta } = 1$

• $\mathrm{\alpha } = 1, \mathrm{\beta } = \pm 1$

• $\mathrm{\alpha } = \pm 1, \mathrm{\beta } = 1$

• $\mathrm{\alpha } = \pm 1, \mathrm{\beta } = - 1$

Let P (1, 2, 3) and Q (- 1, - 2, - 3) be the two points and let O be the origin. Then, $\left|\mathrm{PQ} + \mathrm{OP}\right|$ is equal to

• $\sqrt{13}$

• $\sqrt{14}$

• $\sqrt{24}$

• $\sqrt{12}$

Let ABCD be a parallelogram. If $\mathrm{AB} = \hat{\mathrm{i}} + 3\hat{\mathrm{j}} + 7\hat{\mathrm{k}}$, AD = $2\hat{\mathrm{i}} + 3\hat{\mathrm{j}} - 5\hat{\mathrm{k}}$ and p is a unit vector parallel to AC, then p is equal to

• $\frac{1}{3}\left(2\hat{\mathrm{i}} + \hat{\mathrm{j}} + 2\hat{\mathrm{k}}\right)$

• $\frac{1}{3}\left(2\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}\right)$

• $\frac{1}{7}\left(3\hat{\mathrm{i}} + 6\hat{\mathrm{j}} + 2\hat{\mathrm{k}}\right)$

• $\frac{1}{7}\left(6 + 2\hat{\mathrm{j}} + 3\hat{\mathrm{k}}\right)$

Let OB = $\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$ and OA = $4\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$. The distance of the point B from the straight line passing through A and parallel to the vector $2\hat{\mathrm{i}} + 3\hat{\mathrm{j}} + 6\hat{\mathrm{k}}$ is

• $\frac{7\sqrt{5}}{9}$

• $\frac{5\sqrt{7}}{9}$

• $\frac{3\sqrt{5}}{7}$

• $\frac{9\sqrt{5}}{7}$

127.

If a = $\mathrm{\lambda }\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$ and b = $2\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + \mathrm{\lambda }\hat{\mathrm{k}}$  are at right angle, then the value of $\left|\mathrm{a} + \mathrm{b}\right| - \left|\mathrm{a} - \mathrm{b}\right|$ is

• 2

• 1

• 0

• - 1

Let the position vectors of the points A, B and C be a, b and c, respectively. Let Q be the point of intersection of the medians of the $∆\mathrm{ABC}$. Then, QA + QB + QC is equal to

• $\frac{\mathrm{a} + \mathrm{b} + \mathrm{c}}{2}$

• 2a + b + c

• a + b + c

• 0

If $\mathrm{a} = 2\hat{\mathrm{i}} - \hat{\mathrm{j}} - \mathrm{m}\hat{\mathrm{k}}$ and $\mathrm{b} = \frac{4}{7}\hat{\mathrm{i}} - \frac{2}{7}\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$  are collinear, then the value of m is equal to

• - 7

• - 1

• 2

• 7

Let $\mathrm{a} = 2\hat{\mathrm{i}} + 5\hat{\mathrm{j}} - 7\hat{\mathrm{k}}$ and $\mathrm{b} = \hat{\mathrm{i}} + 3\hat{\mathrm{j}} - 5\hat{\mathrm{k}}$.  Then, (3a - 5b) . (4a $\times$ 5b) is equal to

• - 7

• 0

• - 13

• 1