1
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
$$\frac{3}{2}$$
2
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$[\vec{a}\ \vec{b}\ \vec{c}\ ] \neq 0$$, then $$\frac{[\vec{a}\ +\vec{b}\ \vec{b}\ +\vec{c}\ \vec{c}\ +\vec{a}\ ]}{[\vec{b}\ \vec{c}\ \vec{a}\ ]}=$$

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

If the scalar triple product of the vectors $-3 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ and $7 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ is 272 then $\lambda=\ldots \ldots$

A
9
B
11
C
8
D
10
4
MHT CET 2019 3rd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

$\mathbf{a}$ and $\mathbf{b}$ are non-collinear vectors. If $\mathbf{c}=(x-2) \mathbf{a}+\mathbf{b}$ and $\mathbf{d}=(2 x+1) \mathbf{a}-\mathbf{b}$ are collinear vectors, then the value of $x=\ldots \ldots$

A
$\frac{1}{2}$
B
$\frac{1}{4}$
C
$\frac{1}{5}$
D
$\frac{1}{3}$
MHT CET Subjects
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