1
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+2
-0

The direction cosines of a line which is perpendicular to lines whose direction ratios are $$3,-2,4$$ and $$1,3,-2$$ are

A
$$\frac{4}{\sqrt{297}}, \frac{5}{\sqrt{297}}, \frac{16}{\sqrt{297}}$$
B
$$\frac{8}{\sqrt{285}}, \frac{10}{\sqrt{285}}, \frac{11}{\sqrt{285}}$$
C
$$\frac{-8}{\sqrt{285}}, \frac{10}{\sqrt{285}}, \frac{11}{\sqrt{285}}$$
D
$$\frac{-8}{\sqrt{285}}, \frac{-10}{\sqrt{285}}, \frac{11}{\sqrt{285}}$$
2
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the lines given by $$\frac{x-1}{2 \lambda}=\frac{y-1}{-5}=\frac{z-1}{2}$$ and $$\frac{x+2}{\lambda}=\frac{y+3}{\lambda}=\frac{z+5}{1}$$ are parallel, then the value of $$\lambda$$ is

A
$$\frac{2}{5}$$
B
$$-\frac{5}{2}$$
C
$$-\frac{2}{5}$$
D
$$\frac{5}{2}$$
3
MHT CET 2019 3rd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The vector equation of the plane $\mathbf{r}=(2 \hat{\mathbf{i}}+\hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}})+\mu(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})$ in scalar product form is $\mathbf{r} \cdot(3 \hat{\mathbf{i}}+2 \hat{\mathbf{k}})=\alpha$, then $\alpha=\ldots$

A
2
B
3
C
1
D
0
4
MHT CET 2019 3rd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes $x+2 y+3 z=4$ and $4 x+3 y+2 z=1$ are $\ldots \ldots$

A
$2,3,1$
B
$1,2,3$
C
$3,1,2$
D
$3,2,1$
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