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

Let $\mathrm{L}_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\mathrm{L}_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$ be two given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

A
$\frac{-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}}{\sqrt{99}}$
B
$\frac{-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{5 \sqrt{3}}$
C
$\frac{-\hat{i}+7 \hat{j}+5 \hat{k}}{5 \sqrt{3}}$
D
$\frac{7 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}}{\sqrt{99}}$
2
MHT CET 2024 4th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let $a, b \in R$. If the mirror image of the point $\mathrm{p}(\mathrm{a}, 6,9)$ w.r.t. line $\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9}$ is $(20, b,-a-9)$, then $|a+b|$ is equal to

A
88
B
86
C
90
D
84
3
MHT CET 2024 4th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

A plane which is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, passes through $(1,-2,1)$. The distance of the plane from the point $(1,2,2)$ is

A
0 units
B
1 units
C
$\sqrt{2}$ units
D
$2 \sqrt{2}$ units
4
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $\mathrm{L}_1$ $\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1}$ and $\mathrm{L}_2: \frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3}$ be the given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

A
$\frac{-5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{\sqrt{78}}$
B
$\frac{5 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{5 \sqrt{3}}$
C
$\frac{5 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{5 \sqrt{3}}$
D
$\frac{5 \hat{i}+7 \hat{j}-\hat{k}}{5 \sqrt{3}}$
MHT CET Subjects
EXAM MAP