1
MHT CET 2019 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The vectors $(\mathbf{A}+\mathbf{B})$ and $(\mathbf{A}-\mathbf{B})$ are at right. angles to each other. This is possible under the condition

A
$|A|=|B|$
B
$A \cdot B=0$
C
$A \cdot B=1$
D
$A \times B=0$
2
MHT CET 2019 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A vector $P$ has $X$ and $Y$ components of magnitude 2 units and 4 units respectively. A vector $Q$ along negative $X$-axis has magnitude 6 units. The vector $(\mathbf{Q}-\mathbf{P})$ will be

A
$4(2 \hat{i}-\hat{j})$
B
$-4(2 \hat{\mathbf{i}}-\hat{\mathrm{j}})$
C
$4(2 \hat{i}+\hat{j})$
D
$-4(2 \hat{i}+\hat{j})$
3
MHT CET 2019 2nd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

$\mathbf{P}$ and $\mathbf{Q}$ are two non-zero vectors inclined to each other at an angle ' $\theta$ '. ' $p$ ' and ' $q$ ' are unit vectors along $\mathbf{P}$ and $\mathbf{Q}$ respectively. The component of $\mathbf{Q}$ in the direction of $\mathbf{Q}$ will be

A
$P \cdot Q$
B
$\frac{P \times Q}{P}$
C
$\frac{P \cdot Q}{Q}$
D
$p \cdot q$
4
MHT CET 2019 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The resultant $\mathbf{R}$ of $\mathbf{P}$ and $\mathbf{Q}$ is perpendicular to $\mathbf{P}$. Also $|\mathbf{P}|=|\mathbf{R}|$. The angle between $\mathbf{P}$ and $\mathbf{Q}$ is $\left[\tan 45^{\circ}=1\right]$

A
$\frac{5 \pi}{4}$
B
$\frac{7 \pi}{4}$
C
$\frac{\pi}{4}$
D
$\frac{3 \pi}{4}$
MHT CET Subjects
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