The vector form of the sphere 2(x² + y² + z²) - 4x + 6y + 8z - 5 = 0 is

• $\overrightarrow{\mathrm{r}} . \left[\overrightarrow{\mathrm{r}} - \left(2\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}\right)\right] = \frac{2}{5}$

• $\overrightarrow{\mathrm{r}} . \left[\overrightarrow{\mathrm{r}} - \left(2\hat{\mathrm{i}} - 3\hat{\mathrm{j}} - 4\hat{\mathrm{k}}\right)\right] = \frac{1}{2}$

• $\overrightarrow{\mathrm{r}} . \left[\overrightarrow{\mathrm{r}} - \left(2\hat{\mathrm{i}} + 3\hat{\mathrm{j}} + 4\hat{\mathrm{k}}\right)\right] = \frac{5}{2}$

• $\overrightarrow{\mathrm{r}} . \left[\overrightarrow{\mathrm{r}} - \left(2\hat{\mathrm{i}} - 3\hat{\mathrm{j}} - 4\hat{\mathrm{k}}\right)\right] = \frac{5}{2}$

If $\left|\overrightarrow{\mathrm{a}}\right| = 5, \left|\overrightarrow{\mathrm{b}}\right| = 6 \mathrm{and} \overrightarrow{\mathrm{a}} . \overrightarrow{\mathrm{b}} = - 25, \mathrm{then} \left|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right| \mathrm{is} \mathrm{equal} \mathrm{to}$

• 25

• $6\sqrt{11}$

• 11$\sqrt{5}$

• $5\sqrt{11}$

If $\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{q}} \mathrm{and} \overrightarrow{\mathrm{r}}$ are perpendicular to $\overrightarrow{\mathrm{q}} + \overrightarrow{\mathrm{r}},$ $\overrightarrow{\mathrm{r}} + \overrightarrow{\mathrm{p}},$ $\overrightarrow{\mathrm{p}} + \overrightarrow{\mathrm{q}}$ respectively and if $\left|\overrightarrow{\mathrm{p}} + \overrightarrow{\mathrm{q}}\right| = 6,$ $\left|\overrightarrow{\mathrm{q}} + \overrightarrow{\mathrm{r}}\right|= 4\sqrt{3}$ and $\left|\overrightarrow{\mathrm{r}} + \overrightarrow{\mathrm{p}}\right| = 4,$  then $\left|\overrightarrow{\mathrm{p}} + \overrightarrow{\mathrm{q}} + \overrightarrow{\mathrm{r}}\right|$ is

• 5$\sqrt{2}$

• 10

• 15

• 5

434.

The vectors of magnitude a, 2a, 3a meet at a point and their directions are along the diagonals of three adjacent faces of a cube. Then, the magnitude of their resultant is

• 5a

• 6a

• 10a

• 9a

Which one of the following vectors is of magnitude 6 and perpendicular to both $\overrightarrow{\mathrm{a}} = 2\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + \hat{\mathrm{k}}$ $\mathrm{and} \overrightarrow{\mathrm{b}} = \hat{\mathrm{i}} - 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$ ?

• $2\hat{\mathrm{i}} - \hat{\mathrm{j}} - 2\hat{\mathrm{k}}$

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

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

• $2\left(2\hat{\mathrm{i}} - \hat{\mathrm{j}} - 2\hat{\mathrm{k}}\right)$

If the vectors $\overrightarrow{\mathrm{a}} = 2\hat{\mathrm{i}} + \hat{\mathrm{j}} + 4\hat{\mathrm{k}},$ $\overrightarrow{\mathrm{b}} = 4\hat{\mathrm{i}} - 2\hat{\mathrm{j}} + 3\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}} = 2\hat{\mathrm{i}} - 3\hat{\mathrm{j}} - \mathrm{\lambda }\hat{\mathrm{k}}$ are coplanar, then the value of $\mathrm{\lambda }$ is equal to

• 2

• 1

• 3

• - 1

Let A(1, - 1, 2) and B (2, 3, - 1) be two points. If a point P divides AB internally in the ratio 2 : 3, then the position vector of P is

• $\frac{1}{\sqrt{5}}\left(\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}\right)$

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

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

• $\frac{1}{5}\left(7\hat{\mathrm{i}} + 3\hat{\mathrm{j}} + 4\hat{\mathrm{k}}\right)$

If the scalar product of the vector $\hat{\mathrm{i}} + \hat{\mathrm{j}} + 2\hat{\mathrm{k}}$ with the unit vector along $\mathrm{m}\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 3\hat{\mathrm{k}}$ is equal to 2, then one of the values of m is

• 3

• 4

• 5

• 6

The vector equation of the straight line $\frac{1 - \mathrm{x}}{3} = \frac{\mathrm{y} + 1}{- 2} = \frac{3 - \mathrm{z}}{- 1}$ is

• $\overrightarrow{\mathrm{r}} = \left(\hat{\mathrm{i}} - \hat{\mathrm{j}} + 3\hat{\mathrm{k}}\right) + \mathrm{\lambda }\left(3\hat{\mathrm{i}} + 2\hat{\mathrm{j}} - \hat{\mathrm{k}}\right)$

• $\overrightarrow{\mathrm{r}} = \left(\hat{\mathrm{i}} - \hat{\mathrm{j}} + 3\hat{\mathrm{k}}\right) + \mathrm{\lambda }\left(3\hat{\mathrm{i}} - 2\hat{\mathrm{j}} - \hat{\mathrm{k}}\right)$

• $\overrightarrow{\mathrm{r}} = \left(3\hat{\mathrm{i}} - 2\hat{\mathrm{j}} - \hat{\mathrm{k}}\right) + \mathrm{\lambda }\left(\hat{\mathrm{i}} - \hat{\mathrm{j}} + 3\hat{\mathrm{k}}\right)$

• $\overrightarrow{\mathrm{r}} = \left(3\hat{\mathrm{i}} + 2\hat{\mathrm{j}} - \hat{\mathrm{k}}\right) + \mathrm{\lambda }\left(\hat{\mathrm{i}} - \hat{\mathrm{j}} + 3\hat{\mathrm{k}}\right)$

If a is perpendicular to b, then the vector $\mathrm{a} \times \left\{\mathrm{a} \times \left\{\mathrm{a} \times \left(\mathrm{a} \times \mathrm{b}\right)\right\}\right\}$ is equal to

• ${\left|\mathrm{a}\right|}^2\mathrm{b}$

• $\left|\mathrm{a}\right|\mathrm{b}$

• ${\left|\mathrm{a}\right|}^3\mathrm{b}$

• ${\left|\mathrm{a}\right|}^4\mathrm{b}$