1
MHT CET 2021 23rd September Evening Shift
+2
-0

If $$\bar{a}=2 \hat{i}-\hat{j}+\hat{k}, \bar{b}=\hat{i}+2 \hat{j}-3 \hat{k}$$ and $$\bar{c}=3 \hat{i}+\lambda \hat{j}+5 \hat{k}$$ are coplanar, then $$\lambda$$ is the root of the equation

A
$$\mathrm{x}^2+3 \mathrm{x}=6$$
B
$$x^2+2 x=4$$
C
$$x^2+3 x=4$$
D
$$x^2+2 x=6$$
2
MHT CET 2021 23th September Morning Shift
+2
-0

If $$\hat{a}$$ is a unit vector such that $$(\bar{x}-\hat{a}) \cdot(\bar{x}+\hat{a})=8$$, then $$|\bar{x}|=$$

A
$$\pm 3$$
B
$$2 \sqrt{2}$$
C
3
D
$$\pm \sqrt{7}$$
3
MHT CET 2021 23th September Morning Shift
+2
-0

Let $$\vec{v}=2 \hat{i}+2 \hat{j}-\hat{k}$$ and $$\bar{w}=\hat{i}+3 \hat{k}$$. If $$\bar{u}$$ is a unit vector, then the maximum value of the scalar triple product $$[\bar{u} \bar{v} \bar{w}]$$ is

A
$$\sqrt{6}$$
B
$$\sqrt{10}$$
C
$$\sqrt{13}$$
D
$$\sqrt{89}$$
4
MHT CET 2021 23th September Morning Shift
+2
-0

If $$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overline{\mathrm{b}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}$$ and $$\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$$ is perpendicular to $$\overline{\mathrm{c}}$$, then $$\lambda=$$

A
5
B
2
C
3
D
4
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