AP EAPCET 2025 - 21st May Morning Shift | Vector Algebra Question 46 | Mathematics

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 $(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=$

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

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

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

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

AP EAPCET 2024 - 23th May Morning Shift

MCQ (Single Correct Answer)

-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

-1

AP EAPCET 2024 - 23th May Morning Shift

MCQ (Single Correct Answer)

-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

$\sqrt{74}+14$

$\sqrt{74}+19$

$\sqrt{74}+3$

AP EAPCET 2024 - 23th May Morning Shift

MCQ (Single Correct Answer)

-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

$-\frac{1}{6}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$

$\frac{1}{6}(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$

$\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$

$-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$

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