1
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let P be a plane passing through the points $(2,1,0),(4,1,1)$ and $(5,0,1)$ and $R$ be the point $(2,1,6)$. Then image of $R$ in the plane $P$ is

A
$(6,5,2)$
B
$(4,3,2)$
C
$(6,5,-2)$
D
$(3,4,-2)$
2
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The equation of the plane, passing through the point $(-1,2,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$, is

A
$8 x-14 y-13 z-3=0$
B
$8 x-14 y+13 z+75=0$
C
$8 x+14 y+13 z+19=0$
D
$8 x+14 y-13 z-59=0$
3
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The co-ordinates of the point where the line through $\mathrm{A}(3,4,1)$ and $\mathrm{B}(5,1,6)$ crosses the $x y$-plane are

A
$\left(\frac{13}{5}, \frac{23}{5}, 0\right)$
B
$\left(-\frac{13}{5}, \frac{23}{5}, 0\right)$
C
$\left(\frac{13}{5},-\frac{23}{5}, 0\right)$
D
$\left(-\frac{13}{5},-\frac{23}{5}, 0\right)$
4
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The Cartesian equation of a line is $2 x-2=3 y+1=6 z-2$, then the vector equation of the line is

A
$\overline{\mathrm{r}}=\left(\hat{\mathrm{i}}-\frac{\hat{\mathrm{j}}}{3}+\frac{\hat{\mathrm{k}}}{3}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$
B
$\overline{\mathrm{r}}=\left(-\hat{\mathrm{i}}+\frac{\hat{\mathrm{j}}}{3}-\frac{\hat{\mathrm{k}}}{3}\right)+\lambda\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{6} \hat{\mathrm{k}}\right)$
C
$\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$
D
$\overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{6} \hat{\mathrm{k}}\right)$
MHT CET Subjects
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