1
MHT CET 2023 10th May Morning Shift
+2
-0

The vectors are $$\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \bar{b}=\hat{i}+\hat{j}$$. If $$\bar{c}$$ is a vector such that $$\bar{a} \cdot \bar{c}=|\bar{c}|$$ and $$|\bar{c}-\bar{a}|=2 \sqrt{2}$$, angle between $$\bar{a} \times \bar{b}$$ and $$\bar{c}$$ is $$\frac{\pi}{4}$$, then $$|(\bar{a} \times \bar{b}) \times \bar{c}|$$ is

A
3
B
$$\frac{3}{\sqrt{2}}$$
C
$$3 \sqrt{2}$$
D
1
2
MHT CET 2023 10th May Morning Shift
+2
-0

If $$\bar{a}=\hat{i}+2 \hat{j}+\hat{k}, \bar{b}=\hat{i}-\hat{j}+\hat{k}, \bar{c}=\hat{i}+\hat{j}-\hat{k}$$, then a vector in the plane of $$\bar{a}$$ and $$\bar{b}$$, whose projection on $$\overline{\mathrm{c}}$$ is $$\frac{1}{\sqrt{3}}$$, is

A
$$\hat{i}+\hat{j}-2 \hat{k}$$
B
$$3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}}$$
C
$$4 \hat{i}-\hat{j}+4 \hat{k}$$
D
$$2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$$
3
MHT CET 2023 10th May Morning Shift
+2
-0

Let $$\bar{a}, \bar{b}, \bar{c}$$ be three non-zero vectors, such that no two of them are collinear and $$(\bar{a} \times \bar{b}) \times \bar{c}=\frac{1}{3}|\bar{b}||\bar{c}| \bar{a}$$. If $$\theta$$ is the angle between the vectors $$\bar{b}$$ and $$\bar{c}$$, then the value of $$\sin \theta$$ is

A
$$\frac{2 \sqrt{2}}{3}$$
B
$$\frac{-\sqrt{2}}{3}$$
C
$$\frac{\sqrt{2}}{3}$$
D
$$\sqrt{\frac{2}{3}}$$
4
MHT CET 2023 9th May Evening Shift
+2
-0

$$\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$$, then vector $$\overline{\mathrm{r}}$$ satisfying $$\overline{\mathrm{a}} \times \overline{\mathrm{r}}=\overline{\mathrm{b}}$$ and $$\overline{\mathrm{a}} \cdot \overline{\mathrm{r}}=3$$ is

A
$$\frac{5}{3} \hat{\mathrm{i}}+\frac{2}{3} \hat{\mathrm{j}}+\frac{2}{3} \hat{\mathrm{k}}$$
B
$$-\frac{5}{3} \hat{\mathrm{i}}+\frac{2}{3} \hat{\mathrm{j}}+\frac{2}{3} \hat{\mathrm{k}}$$
C
$$\frac{5}{3} \hat{\mathrm{i}}-\frac{2}{3} \hat{\mathrm{j}}+\frac{2}{3} \hat{\mathrm{k}}$$
D
$$-\frac{5}{3} \hat{\mathrm{i}}+\frac{2}{3} \hat{\mathrm{j}}+\frac{1}{3} \hat{\mathrm{k}}$$
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