1
JEE Main 2022 (Online) 25th July Evening Shift
+4
-1

Let $$\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$$ and let $$\vec{b}$$ be a vector such that $$\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$$ and $$\vec{a} \cdot \vec{b}=3$$. Then the projection of $$\vec{b}$$ on the vector $$\vec{a}-\vec{b}$$ is :

A
$$\frac{2}{\sqrt{21}}$$
B
$$2 \sqrt{\frac{3}{7}}$$
C
$$\frac{2}{3} \sqrt{\frac{7}{3}}$$
D
$$\frac{2}{3}$$
2
JEE Main 2022 (Online) 25th July Morning Shift
+4
-1

Let $$\mathrm{ABC}$$ be a triangle such that $$\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}},|\overrightarrow{\mathrm{a}}|=6 \sqrt{2},|\overrightarrow{\mathrm{b}}|=2 \sqrt{3}$$ and $$\vec{b} \cdot \vec{c}=12$$. Consider the statements :

$$(\mathrm{S} 1):|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})|-|\vec{c}|=6(2 \sqrt{2}-1)$$

$$(\mathrm{S} 2): \angle \mathrm{ACB}=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$$

Then

A
both (S1) and (S2) are true
B
only (S1) is true
C
only (S2) is true
D
both (S1) and (S2) are false
3
JEE Main 2022 (Online) 30th June Morning Shift
+4
-1
Out of Syllabus

Let a vector $$\overrightarrow c$$ be coplanar with the vectors $$\overrightarrow a = - \widehat i + \widehat j + \widehat k$$ and $$\overrightarrow b = 2\widehat i + \widehat j - \widehat k$$. If the vector $$\overrightarrow c$$ also satisfies the conditions $$\overrightarrow c \,.\,\left[ {\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right] = - 42$$ and $$\left( {\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right)\,.\,\widehat k = 3$$, then the value of $$|\overrightarrow c {|^2}$$ is equal to :

A
24
B
29
C
35
D
42
4
JEE Main 2022 (Online) 29th June Evening Shift
+4
-1
Let A, B, C be three points whose position vectors respectively are

$$\overrightarrow a = \widehat i + 4\widehat j + 3\widehat k$$

$$\overrightarrow b = 2\widehat i + \alpha \widehat j + 4\widehat k,\,\alpha \in R$$

$$\overrightarrow c = 3\widehat i - 2\widehat j + 5\widehat k$$

If $$\alpha$$ is the smallest positive integer for which $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c$$ are noncollinear, then the length of the median, in $$\Delta$$ABC, through A is :

A
$${{\sqrt {82} } \over 2}$$
B
$${{\sqrt {62} } \over 2}$$
C
$${{\sqrt {69} } \over 2}$$
D
$${{\sqrt {66} } \over 2}$$
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