1
MHT CET 2024 3rd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{A} \equiv(1,-1,0), \mathrm{B} \equiv(0,1,-1)$ and $\mathrm{C} \equiv(-1,0,1)$, then the unit vector $\overline{\mathrm{d}}$ such that $\overline{\mathrm{a}}$ and $\overline{\mathrm{d}}$ are perpendiculars and $\overline{\mathrm{b}}, \overline{\mathrm{c}}, \overline{\mathrm{d}}$ are coplanar is

A
$+\frac{1}{\sqrt{3}}(1,1,1)$
B
$+\frac{1}{\sqrt{3}}(-1,-1,1)$
C
$+\frac{1}{\sqrt{6}}(1,1,-2)$
D
$+\frac{1}{\sqrt{2}}(1,1,0)$
2
MHT CET 2024 3rd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let the vectors $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ be such that $|\overline{\mathrm{a}}|=2,|\overline{\mathrm{~b}}|=4$ and $|\bar{c}|=4$. If the projection of $\bar{b}$ on $\bar{a}$ is equal to the projection of $\overline{\mathrm{c}}$ on $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ is perpendicular to $\overline{\mathrm{c}}$, then the value of $|\overline{\mathrm{a}}+\overline{\mathrm{b}}-\overline{\mathrm{c}}|$ is equal to

A
$2\sqrt5$
B
6
C
4
D
$4\sqrt2$
3
MHT CET 2024 3rd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$. If $\overline{\mathrm{c}}$ is a vector such that $\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=|\overline{\mathrm{c}}|,|\overline{\mathrm{c}}-\overline{\mathrm{a}}|=2 \sqrt{2}$ and the angle between $(\overline{\mathrm{a}} \times \overline{\mathrm{b}})$ and $\overline{\mathrm{c}}$ is $30^{\circ}$, then the value of $|(\bar{a} \times \bar{b}) \times \bar{c}|$ is equal to

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

Let $\bar{a}, \bar{b}, \bar{c}$ be three non-coplanar vectors and $\overline{\mathrm{p}}, \overline{\mathrm{q}}, \overline{\mathrm{r}}$ defined by the relations

$$\overline{\mathrm{p}}=\frac{\overline{\mathrm{b}} \times \overline{\mathrm{c}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}, \overline{\mathrm{q}}=\frac{\overline{\mathrm{c}} \times \overline{\mathrm{a}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}, \overline{\mathrm{r}}=\frac{\overline{\mathrm{a}} \times \overline{\mathrm{b}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}$$

then the value of the expression $(\overline{\mathrm{a}}+\overline{\mathrm{b}}) \cdot \overline{\mathrm{p}}+(\overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot \overline{\mathrm{q}}+(\overline{\mathrm{c}}+\overline{\mathrm{a}}) \cdot \overline{\mathrm{r}}$ is equal to

A
0
B
1
C
2
D
3
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