751.

If three points A,B,and C have position vectors $\left(1, \mathrm{x}, 3\right), \left(3, 4, 7\right) \mathrm{and} \left(\mathrm{y}, - 2, - 5\right)$ respectively and if they are collinear, then (x, y) is

• $(2, - 3)$

• ( - 2, 3)

• ( - 2, - 3)

• $(2, 3)$

The orthogonal projection of a on b is

• $\frac{\left(\mathrm{a} . \mathrm{b}\right)\mathrm{b}}{{\left|\mathrm{a}\right|}^2}$

• $\frac{\left(\mathrm{a} . \mathrm{b}\right)\mathrm{b}}{{\left|\mathrm{b}\right|}^2}$

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

• $\frac{\mathrm{b}}{\left|\mathrm{b}\right|}$

If the position vectors of the vertices of atriangle are $2\hat{\mathrm{i}} - \hat{\mathrm{j}} + \hat{\mathrm{k}}$, $\hat{\mathrm{i}} - 3\hat{\mathrm{j}} - 5\hat{\mathrm{k}}$ and $3\hat{\mathrm{i}} - 4\hat{\mathrm{j}} - 4\hat{\mathrm{k}}$, then the probability of triangle is

• equilateral

• isosceless

• right angled isosceless

• right angled

If [a b c] = 3, then the volume (in cubic units) of e parallelepiped with 2a + b, 2b + c and 2c + a as edges, is

• 15

• 22

• 25

• 27

$\left(\mathrm{a} +\hspace{0.17em}\mathrm{b}\right) . \left(\mathrm{b} +\hspace{0.17em}\mathrm{c}\right) \times \left(\mathrm{a} + \mathrm{b} +\hspace{0.17em}\mathrm{c}\right)$ is equal to

• 0

• - [a b c]

• 2[a b c]

• [a b c]

If D, E and F are respectively the mid-points of AB, AC and BC in $∆\mathrm{ABC}$, then $\overrightarrow{\mathrm{BE}} + \overrightarrow{\mathrm{AF}}$ is equal to :

• $\overrightarrow{\mathrm{DC}}$

• $\frac{1}{2}\overrightarrow{\mathrm{BF}}$

• $2\overrightarrow{\mathrm{BF}}$

• $\frac{3}{2}\overrightarrow{\mathrm{BF}}$

Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ be the position vectors of the vertices A, B, C respectively of $∆$ABC. Thevector area of $∆$ABC is :

• $\frac{1}{2}\left(\overrightarrow{\mathrm{a}}\left(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}\right) + \overrightarrow{\mathrm{b}}\left(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}\right) + \overrightarrow{\mathrm{c}}\left(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right)\right)$

• $\frac{1}{2}\left(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} + \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} + \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}\right)$

• $\frac{1}{2}\left(\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}} + \overrightarrow{\mathrm{c}}\right)$

• $\frac{1}{2}\left(\overrightarrow{\mathrm{a}}\left(\overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}\right) + \overrightarrow{\mathrm{b}}\left(\overrightarrow{\mathrm{c}} \overrightarrow{\mathrm{a}}\right) + \overrightarrow{\mathrm{c}}\left(\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}}\right)\right)$

If $\overrightarrow{\mathrm{a}} = \hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}$, $\overrightarrow{\mathrm{b}} = \hat{\mathrm{i}} + \hat{\mathrm{j}}$, $\overrightarrow{\mathrm{c}} = \hat{\mathrm{i}}$ and $\left(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right) \times \overrightarrow{\mathrm{c}} = \mathrm{\lambda }\overrightarrow{\mathrm{a}} + \mathrm{\mu }\overrightarrow{\mathrm{b}}$, then $\mathrm{\lambda } + \mathrm{\mu }$ is equal to :

• 0

• 1

• 2

• 3

$$\begin{gathered} \mathrm{If} \hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 3\hat{\mathrm{k}}, 3\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + \hat{\mathrm{k}} \mathrm{are} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram}, \mathrm{then} \mathrm{a} \\ \mathrm{unit} \mathrm{vector} \mathrm{is} \mathrm{parallel} \mathrm{to} \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{diagonals} \mathrm{of} \mathrm{the} \mathrm{parallelogram} \mathrm{is} \end{gathered}$$

• $\frac{\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}}{\sqrt{3}}$

• $\frac{\hat{\mathrm{i}} + \hat{\mathrm{j}} - \hat{\mathrm{k}}}{\sqrt{3}}$

• $\frac{\hat{\mathrm{i}} - \hat{\mathrm{j}} + \hat{\mathrm{k}}}{\sqrt{3}}$

• $\frac{- \hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}}{\sqrt{3}}$

If G is the centroid of the $∆$ABC, then GA + BG + GC is equal

• 2GB

• 2GA

• 0

• 2BG