1
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

A line $$\mathrm{L}_1$$ passes through the point, whose p. v. (position vector) $$3 \hat{i}$$, is parallel to the vector $$-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$$. Another line $$\mathrm{L}_2$$ passes through the point having p.v. $$\hat{i}+\hat{j}$$ is parallel to vector $$\hat{i}+\hat{k}$$, then the point of intersection of lines $$L_1$$ and $$L_2$$ has p.v.

A
$$2 \hat{i}+2 \hat{j}+\hat{k}$$
B
$$2 \hat{i}+\hat{j}+\hat{k}$$
C
$$2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$$
D
$$2 \hat{i}-2 \hat{j}+\hat{k}$$
2
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The equation of the line passing through the point $$(-1,3,-2)$$ and perpendicular to each of the lines $$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$$ and $$\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$$ is

A
$$\frac{x+1}{2}=\frac{y-3}{7}=\frac{z+2}{4}$$
B
$$\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}$$
C
$$\frac{x-1}{2}=\frac{y+3}{-7}=\frac{z+2}{4}$$
D
$$\frac{x-1}{2}=\frac{y+3}{7}=\frac{z-2}{4}$$
3
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$A(1,4,2)$$ and $$C(5,-7,1)$$ are two vertices of triangle $$A B C$$ and $$G\left(\frac{4}{3}, 0, \frac{-2}{3}\right)$$ is centroid of the triangle $$A B C$$, then the mid point of side $$B C$$ is

A
$$\left(-2,-2, \frac{3}{2}\right)$$
B
$$\left(2,2, \frac{3}{2}\right)$$
C
$$\left(\frac{3}{2}, 2,-2\right)$$
D
$$\left(\frac{3}{2},-2,-2\right)$$
4
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The distance of the point $$(-1,-5,-10)$$ from the point of intersection of the line $$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$$ and the plane $$x-y+z=5$$ is

A
13 units.
B
12 units.
C
5 units.
D
16 units.
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