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

If $$|\bar{a} \times \bar{b}|^2+(\bar{a} \cdot \bar{b})^2=144$$ and $$|\bar{a}|=4$$, then $$|\bar{b}|=$$

A
8
B
12
C
3
D
16
2
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
3
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
$$\frac{5 \hat{\mathrm{i}}-12 \hat{\mathrm{j}}}{3}$$
4
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
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