1
JEE Main 2021 (Online) 16th March Evening Shift
MCQ (Single Correct Answer)
+4
-1
Let $$\overrightarrow a $$ = $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 3$$\widehat k$$ and $$\overrightarrow b = 2\widehat i$$ $$-$$ 3$$\widehat j$$ + 5$$\widehat k$$. If $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ $$\overrightarrow r $$,

$$\overrightarrow r $$ . $$\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right)$$ = 3 and $$\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right)$$ = $$-$$1, $$\alpha$$ $$\in$$ R, then the

value of $$\alpha$$ + $${\left| {\overrightarrow r } \right|^2}$$ is equal to :
A
13
B
11
C
9
D
15
2
JEE Main 2021 (Online) 16th March Morning Shift
MCQ (Single Correct Answer)
+4
-1
Let a vector $$\alpha \widehat i + \beta \widehat j$$ be obtained by rotating the vector $$\sqrt 3 \widehat i + \widehat j$$ by an angle 45$$^\circ$$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices ($$\alpha$$, $$\beta$$), (0, $$\beta$$) and (0, 0) is equal to :
A
$${1 \over {\sqrt 2 }}$$
B
$${1 \over 2}$$
C
1
D
2$${\sqrt 2 }$$
3
JEE Main 2021 (Online) 26th February Evening Shift
MCQ (Single Correct Answer)
+4
-1
If vectors $$\overrightarrow {{a_1}} = x\widehat i - \widehat j + \widehat k$$ and $$\overrightarrow {{a_2}} = \widehat i + y\widehat j + z\widehat k$$ are collinear, then a possible unit vector parallel to the vector $$x\widehat i + y\widehat j + z\widehat k$$ is :
A
$${1 \over {\sqrt 3 }}\left( {\widehat i - \widehat j + \widehat k} \right)$$
B
$${1 \over {\sqrt 2 }}\left( { - \widehat j + \widehat k} \right)$$
C
$${1 \over {\sqrt 2 }}\left( {\widehat i - \widehat j} \right)$$
D
$${1 \over {\sqrt 3 }}\left( {\widehat i + \widehat j - \widehat k} \right)$$
4
JEE Main 2021 (Online) 26th February Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
If $$\overrightarrow a $$ and $$\overrightarrow b $$ are perpendicular, then
$$\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)} \right)$$ is equal to :
A
$${1 \over 2}|\overrightarrow a {|^4}\overrightarrow b $$
B
$$\overrightarrow 0 $$
C
$$\overrightarrow a \times \overrightarrow b $$
D
$$|\overrightarrow a {|^4}\overrightarrow b $$
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