1
AP EAPCET 2024 - 23th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
In $\triangle A B C,\left(r_2+r_3\right) \operatorname{cosec}^2 \frac{A}{2}$ is equal to
A
$4 R$
B
$4 R \cot ^2 \frac{\mathrm{~A}}{2}$
C
$4 R \tan ^2 \frac{A}{2}$
D
$R \tan ^2 \frac{A}{2}$
2
AP EAPCET 2024 - 23th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
If the vectors $a \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+b \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{j}}+c \hat{\mathbf{k}}$ $(a \neq b \neq c \neq 1)$ are coplanar, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is equal to
A
0
B
2
C
1
D
-1
3
AP EAPCET 2024 - 23th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
If $\mathbf{A B}=2 \mathbf{i}+3 \mathbf{j}-6 \mathbf{k}, \mathbf{B C}=6 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k}$ are the vectors along two sides of a $\triangle A B C$. Then, perimeter of $\triangle A B C$ is
A
21
B
$\sqrt{74}+14$
C
$\sqrt{74}+19$
D
$\sqrt{74}+3$
4
AP EAPCET 2024 - 23th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
The orthogonal projection vector of $a=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ on $\mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ is
A
$-\frac{1}{6}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$
B
$\frac{1}{6}(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$
C
$\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
D
$-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
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