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

The equation of the line, through $\mathrm{A}(1,2,3)$ and perpendicular to the vector $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\hat{i}+3 \hat{j}+2 \hat{k}$, is

A
$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$
B
$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})$
C
$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$
D
$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$
2
MHT CET 2024 2nd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

A
$-4 y-x+7 z=0$
B
$x-4 y-7 z=0$
C
$x-4 y+7 z=0$
D
$x+4 y+7 z=0$
3
MHT CET 2024 2nd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The incentre of the triangle whose vertices are $P(0,3,0), Q(0,0,4)$ and $R(0,3,4)$ is

A
$(0,3,2)$
B
$(0,2,3)$
C
$(2,0,3)$
D
$(2,3,0)$
4
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The vector equation of a line whose Cartesian equations are $y=2,4 x-3 z+5=0$ is

A
$\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \mathrm{k})+\lambda\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)$
B
$\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \mathrm{k})+\lambda\left(2 \hat{\mathrm{j}}-\frac{5}{3} \hat{\mathrm{k}}\right)$
C
$\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \mathrm{k})$
D
$\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}-\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \mathrm{k})$
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