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

Let two non-collinear vectors $$\hat{a}$$ and $$\hat{b}$$ form an acute angle. A point $$\mathrm{P}$$ moves, so that at any time $$t$$ the position vector $$\overline{\mathrm{OP}}$$, where $$\mathrm{O}$$ is origin, is given by $$\hat{a} \sin t+\hat{b} \cos t$$, when $$P$$ is farthest from origin $$O$$, let $$M$$ be the length of $$\overline{\mathrm{OP}}$$ and $$\hat{\mathrm{u}}$$ be the unit vector along $$\overline{\mathrm{OP}}$$, then

A
$$\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|\hat{\mathrm{a}}+\hat{\mathrm{b}}|}$$ and $$\mathrm{M}=(1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}$$
B
$$\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}-\hat{\mathrm{b}}}{|\hat{\mathrm{a}}-\hat{\mathrm{b}}|}$$ and $$\mathrm{M}=(1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}$$
C
$$\hat{u}=\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}$$ and $$M=(1+2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}$$
D
$$\hat{u}=\frac{\hat{a}-\hat{b}}{|\hat{a}-\hat{b}|}$$ and $$M=(1-2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}$$
2
MHT CET 2023 9th May Morning Shift
+2
-0

The distance of the point having position vector $$\hat{i}-2 \hat{j}-6 \hat{k}$$, from the straight line passing through the point $$(2,-3,-4)$$ and parallel to the vector $$6 \hat{i}+3 \hat{j}-4 \hat{k}$$ is units.

A
$$\sqrt{\frac{340}{61}}$$
B
$$\frac{341}{61}$$
C
$$\frac{\sqrt{341}}{61}$$
D
$$\sqrt{\frac{341}{61}}$$
3
MHT CET 2023 9th May Morning Shift
+2
-0

The scalar product of the vector $$\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$$ with a unit vector along the sum of the vectors $$2 \hat{i}+4 \hat{j}-5 \hat{k}$$ and $$\lambda \hat{i}+2 \hat{j}+3 \hat{k}$$ is equal to 1 , then value of $$\lambda$$ is

A
1
B
2
C
3
D
4
4
MHT CET 2023 9th May Morning Shift
+2
-0

If $$[(\bar{a}+2 \bar{b}+3 \bar{c}) \times(\bar{b}+2 \bar{c}+3 \bar{a})] \cdot(\bar{c}+2 \bar{a}+3 \bar{b})=54$$ then the value of $$\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$$ is

A
0
B
1
C
3
D
2
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