1
AP EAPCET 2021 - 20th August Evening Shift
MCQ (Single Correct Answer)
+1
-0

Let $$\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}}$$ and $$\mathbf{c}=\hat{\mathbf{k}}-\hat{\mathbf{i}}$$ if $$\mathbf{d}$$ is a unit vector such $$\mathbf{a} \cdot \mathbf{b}=0=[\mathbf{b} \mathbf{c} \mathbf{d}]$$, then $$\mathbf{d}$$ is

A
$$\pm \frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}$$
B
$$\pm \frac{\hat{i}+\hat{j}-2 \hat{k}}{\sqrt{6}}$$
C
$$\pm \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$$
D
$$\pm \frac{\hat{i}+\hat{j}+2 \hat{k}}{\sqrt{6}}$$
2
AP EAPCET 2021 - 20th August Evening Shift
MCQ (Single Correct Answer)
+1
-0

Let $$u$$ and $$v$$ be two non-zero vectors in $$R^3$$ with the intermediate angle $$45^{\circ}$$. Then $$|\mathbf{u} \times \mathbf{v}|$$ is equal to

A
$$|u||v|$$
B
$$2|u||v|$$
C
$$u \cdot v$$
D
$$|u|+|v|$$
3
AP EAPCET 2021 - 20th August Evening Shift
MCQ (Single Correct Answer)
+1
-0

Given, $$\mathbf{a}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}$$ and $$\mathbf{b}=\mathbf{b}_1+\mathbf{b}_2$$ where $$\mathbf{b}_1$$ is parallel to $$\mathbf{a}$$ and $$\mathbf{b}_2$$ is perpendicular to $$\mathbf{a}$$. Then, $$\mathbf{b}_2$$ is equal to

A
$$\frac{1}{2} \hat{\mathbf{i}}+\frac{3}{2} \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$$
B
$$\frac{1}{2} \hat{\mathbf{i}}-\frac{3}{2} \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$$
C
$$\frac{1}{2} \hat{\mathbf{i}}+\frac{3}{2} \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$$
D
$$\frac{1}{2} \hat{\mathbf{i}}-\frac{3}{2} \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$$
4
AP EAPCET 2021 - 20th August Morning Shift
MCQ (Single Correct Answer)
+1
-0

The position vectors of the points $$A$$ and $$B$$ with respect to $$O$$ are $$2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$$ and $$2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$$. The length of the internal bisector of $$\angle B O A$$ of $$\triangle A O B$$ is (take proportionality constant is 2)

A
$$\frac{\sqrt{136}}{9}$$
B
$$\frac{\sqrt{136}}{3}$$
C
$$\frac{20}{3}$$
D
$$\frac{25}{3}$$

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