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

For any non-zero vectors $$\mathbf{a}$$ and $$\mathbf{b}$$,

A
$$a \times b$$
B
$$|a \times b|^2$$
C
0
D
$$|\mathbf{a} \times \mathbf{b}|$$
2
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the vectors $$\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$$ and $$2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+m \hat{\mathbf{k}}$$ are coplanar, then $$m=$$

A
3
B
2
C
$$-$$3
D
$$-$$2
3
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+2
-0

The angles between the lines $$\mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \text { and } \mathbf{r}=(3 \hat{\mathbf{i}}+\hat{\mathbf{k}})+\lambda^{\prime}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}), \lambda, \lambda^{\prime} \in \mathbf{R}$$ is

A
$$\cos ^{-1}\left(\frac{1}{5}\right)$$
B
$$\cos ^{-1}\left(\frac{2}{3}\right)$$
C
$$\cos ^{-1}\left(\frac{1}{3}\right)$$
D
$$\cos ^{-1}\left(\frac{1}{6}\right)$$
4
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+2
-0

In a quadrilateral $$ABCD, M$$ and $$N$$ are the mid-points of the sides $$A B$$ and $$C D$$ respectively. If $$\mathbf{A D}+\mathbf{B C}=t \mathbf{M N}$$, then $$t=$$

A
2
B
$$\frac{1}{2}$$
C
4
D
$$\frac{3}{2}$$
MHT CET Subjects
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