1
AP EAPCET 2025 - 21st May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three unit vectors such that $|\mathbf{a}-\mathbf{b}|^2+|\mathbf{b}-\mathbf{c}|^2+|\mathbf{c}-\mathbf{a}|^2=15$, then $|\mathbf{a}-\mathbf{b}-\mathbf{c}|^2-4(\mathbf{b} \cdot \mathbf{c})=$

A

6

B

15

C

12

D

10

2
AP EAPCET 2025 - 21st May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\mathbf{a}=\hat{\mathbf{i}}+p \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \mathbf{b}=p \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are three vectors such that $|\mathbf{a} \times \mathbf{b}|=\mid \mathbf{a} \times \mathbf{c}$, then $p=$

A

-2

B

-1

C

1

D

2

3
AP EAPCET 2025 - 21st May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$, $\mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{c}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{d}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ are four vectors, then $(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=$

A

$17 \hat{\mathbf{i}}-15 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}$

B

$31 \hat{\mathbf{i}}-\mathbf{j}+23 \hat{\mathbf{k}}$

C

$17 \hat{\mathbf{i}}-\hat{\mathbf{j}}+23 \hat{\mathbf{k}}$

D

$31 \hat{\mathbf{i}}-15 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}$

4
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

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