1
MHT CET 2021 21th September Morning Shift
+2
-0

The distance between parallel lines

\begin{aligned} & \bar{r}=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(2 \hat{i}+\hat{j}-2 \hat{k}) \text { and } \\ & \bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+\hat{j}-2 \hat{k}) \text { is } \end{aligned}

A
$$\sqrt{2}$$
B
$$\frac{1}{3}$$ units
C
$$\frac{1}{\sqrt{3}}$$ units
D
$$\frac{\sqrt{2}}{3}$$ units
2
MHT CET 2021 21th September Morning Shift
+2
-0

The vertices of triangle $$\mathrm{ABC}$$ are $$\mathrm{A} \equiv(3,0,0) ; \mathrm{B} \equiv(0,0,4) ; \mathrm{C} \equiv(0,5,4)$$. Find the position vector of the point in which the bisector of angle A meets B C is

A
$$5 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}$$
B
$$\frac{5 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{3}$$
C
$$\frac{5 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}}{13}$$
D
$$\frac{5 \hat{\mathrm{i}}-12 \hat{\mathrm{j}}}{3}$$
3
MHT CET 2021 21th September Morning Shift
+2
-0

In a quadrilateral PQRS, M and N are mid-points of the sides PQ and RS respectively. If $$\overline {PS} + \overline {QR} = t\overline {MN}$$, then t =

A
$$\frac{1}{2}$$
B
4
C
$$\frac{3}{2}$$
D
2
4
MHT CET 2021 21th September Morning Shift
+2
-0

If vectors $$\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$$ and $$\bar{c}=3 \hat{i}+\hat{j}+2 \hat{k}$$ are such that, $$\bar{a}+\lambda \bar{b}$$ is perpendicular to $$\bar{c}$$, then $$\lambda=$$

A
$$-14$$
B
$$14$$
C
$$2$$
D
$$-2$$
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