1
MHT CET 2024 2nd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If C is a given non-zero scalar and $\overline{\mathrm{A}}$ and $\overline{\mathrm{B}}$ are given non-zero vectors such that $\overline{\mathrm{A}}$ is perpendicular to $\overline{\mathrm{B}}$. If vector $\overline{\mathrm{X}}$ is such that $\overline{\mathrm{A}} \cdot \overline{\mathrm{X}}=\mathrm{C}$ and $\overline{\mathrm{A}} \times \overline{\mathrm{X}}=\overline{\mathrm{B}}$ then $\overline{\mathrm{X}}$ is given by

A
$\mathrm{\frac{C \bar{A}+\overline{\mathrm{A}} \times \overline{\mathrm{B}}}{|\overline{\mathrm{A}}|^2}}$
B
$\mathrm{\frac{C \bar{A} \times \bar{B}}{|\overline{\mathrm{~A}}|^2}}$
C
$\mathrm{\frac{C \overline{\mathrm{~A}}-\overline{\mathrm{A}} \times \overline{\mathrm{B}}}{|\overline{\mathrm{A}}|^2}}$
D
$\mathrm{\frac{C \bar{A}+\bar{B}}{|\overline{\mathrm{~A}}|^2}}$
2
MHT CET 2024 2nd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\overline{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\bar{b}=\hat{i} \times(\bar{a} \times \hat{i})+\hat{j} \times(\bar{a} \times \hat{j})+\hat{k} \times(\bar{a} \times \hat{k})$ then $|\bar{b}|$ is

A
$\sqrt{12}$
B
$2 \sqrt{12}$
C
$3 \sqrt{14}$
D
$2 \sqrt{14}$
3
MHT CET 2024 2nd 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 $\overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$. Let $\overline{\mathrm{c}}$ be a vector such that $|\bar{c}-\bar{a}|=3$ and $|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|=3$ and the angle between $\overline{\mathrm{c}}$ and $\overline{\mathrm{a}} \times \overline{\mathrm{b}}$ is $30^{\circ}$, then $\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}$ is equal to

A
$2$
B
$-\frac{1}{8}$
C
$\frac{25}{8}$
D
$5$
4
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The scalar $\overline{\mathrm{a}} \cdot[(\overline{\mathrm{b}}+\overline{\mathrm{c}}) \times(\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}})]$ equals

A
$0$
B
$[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]+[\overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{a}}]$
C
$[\bar{a} \bar{b} \bar{c}]$
D
$1$
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