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

The unit vector which is orthogonal to the vector $5 \hat{i}+2 \hat{j}+6 \hat{k}$ and is coplanar with the vectors $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is

A
$\frac{2 \hat{i}-6 \hat{j}+\hat{k}}{\sqrt{41}}$
B
$\frac{2 \hat{i}-5 \hat{j}}{\sqrt{29}}$
C
$\frac{-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{10}}$
D
$\frac{2 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{69}$
3
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
$-\hat{\mathrm{i}}-8 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
4
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
$\mathrm{\frac{C \bar{A}+\bar{B}}{|\overline{\mathrm{~A}}|^2}}$
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