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

Let $\overline{\mathrm{A}}=2 \hat{\mathrm{i}}+\hat{\mathrm{k}}, \overline{\mathrm{B}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{C}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$. If a vector $\bar{R}$ satisfies $\bar{R} \times \bar{B}=\bar{C} \times \bar{B}$ and $\bar{R} \cdot \overline{\mathrm{~A}}=0$, then $\overline{\mathrm{R}}$ is given by

A
$\hat{\mathrm{i}}-8 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
B
$\hat{i}+8 \hat{j}+2 \hat{k}$
C
$-\hat{i}-8 \hat{j}+2 \hat{k}$
D
$-\hat{\mathrm{i}}-8 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
2
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
$\mathrm{\frac{C \bar{A}+\bar{B}}{|\overline{\mathrm{~A}}|^2}}$
3
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
$2 \sqrt{14}$
4
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
$5$
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