1
MHT CET 2023 9th May Evening Shift
+2
-0

If $$\bar{a}, \bar{b}$$ and $$\bar{c}$$ are any three non-zero vectors, then $$(\bar{a}+2 \bar{b}+\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})]=$$

A
$$\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$$
B
$$2\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right]$$
C
$$3\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$$
D
$$4\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right]$$
2
MHT CET 2023 9th May Evening Shift
+2
-0

Vectors $$\overline{\mathrm{a}}$$ and $$\overline{\mathrm{b}}$$ are such that $$|\overline{\mathrm{a}}|=1 ;|\overline{\mathrm{b}}|=4$$ and $$\bar{a} \cdot \bar{b}=2$$. If $$\bar{c}=2 \bar{a} \times \bar{b}-3 \bar{b}$$, then the angle between $$\bar{b}$$ and $$\bar{c}$$ is

A
$$\frac{\pi}{6}$$
B
$$\frac{5 \pi}{6}$$
C
$$\frac{\pi}{3}$$
D
$$\frac{2 \pi}{3}$$
3
MHT CET 2023 9th May Evening Shift
+2
-0

Two adjacent sides of a parallelogram are $$2 \hat{i}-4 \hat{j}+5 \hat{k}$$ and $$\hat{i}-2 \hat{j}-3 \hat{k}$$, then the unit vector parallel to its diagonal is

A
$$\frac{3}{7} \hat{\mathrm{i}}-\frac{6}{7} \hat{\mathrm{j}}+\frac{2}{7} \hat{\mathrm{k}}$$
B
$$\frac{2}{7} \hat{\mathrm{i}}-\frac{6}{7} \hat{\mathrm{j}}+\frac{3}{7} \hat{\mathrm{k}}$$
C
$$\frac{6}{7} \hat{\mathrm{i}}-\frac{2}{7} \hat{\mathrm{j}}+\frac{3}{7} \hat{\mathrm{k}}$$
D
$$\frac{1}{7} \hat{\mathrm{i}}+\frac{1}{7} \hat{\mathrm{j}}-\frac{3}{7} \hat{\mathrm{k}}$$
4
MHT CET 2023 9th May Evening Shift
+2
-0

If $$\mathrm{D}, \mathrm{E}$$ and $$\mathrm{F}$$ are the mid-points of the sides $$\mathrm{BC}$$, $$\mathrm{CA}$$ and $$\mathrm{AB}$$ of triangle $$\mathrm{ABC}$$ respectively, then $$\overline{\mathrm{AD}}+\frac{2}{3} \overline{\mathrm{BE}}+\frac{1}{3} \overline{\mathrm{CF}}=$$

A
$$\frac{1}{2} \overline{\mathrm{AB}}$$
B
$$\frac{1}{2} \overline{\mathrm{AC}}$$
C
$$\frac{1}{2} \overline{\mathrm{BC}}$$
D
$$\frac{2}{3} \overline{\mathrm{AC}}$$
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