1
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
Let $$\overrightarrow a = 2\widehat i + {\lambda _1}\widehat j + 3\widehat k,\,\,$$   $$\overrightarrow b = 4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k,$$  and  $$\overrightarrow c = 3\widehat i + 6\widehat j + \left( {{\lambda _3} - 1} \right)\widehat k$$  be three vectors such that $$\overrightarrow b = 2\overrightarrow a$$ and $$\overrightarrow a$$ is perpendicular to $$\overrightarrow c$$. Then a possible value of $$\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right)$$ is :
A
(1, 5, 1)
B
(1, 3, 1)
C
$$\left( { - {1 \over 2},4,0} \right)$$
D
$$\left( {{1 \over 2},4, - 2} \right)$$
2
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
Let  $$\overrightarrow a = \widehat i + \widehat j + \sqrt 2 \widehat k,$$   $$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + \sqrt 2 \widehat k$$,    $$\overrightarrow c = 5\widehat i + \widehat j + \sqrt 2 \widehat k$$   be three vectors such that the projection vector of $$\overrightarrow b$$ on $$\overrightarrow a$$ is $$\overrightarrow a$$.
If   $$\overrightarrow a + \overrightarrow b$$   is perpendicular to $$\overrightarrow c$$ , then $$\left| {\overrightarrow b } \right|$$ is equal to :
A
$$\sqrt {32}$$
B
6
C
$$\sqrt {22}$$
D
4
3
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
Out of Syllabus
Let $$\overrightarrow a$$ = $$\widehat i - \widehat j$$, $$\overrightarrow b$$ = $$\widehat i + \widehat j + \widehat k$$ and $$\overrightarrow c$$

be a vector such that $$\overrightarrow a$$ × $$\overrightarrow c$$ + $$\overrightarrow b$$ = $$\overrightarrow 0$$

and $$\overrightarrow a$$ . $$\overrightarrow c$$ = 4, then |$$\overrightarrow c$$|2 is equal to :
A
8
B
$$19 \over 2$$
C
9
D
$$17 \over 2$$
4
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow c = \widehat j - \widehat k$$ and a vector $$\overrightarrow b$$ be such that $$\overrightarrow a \times \overrightarrow b = \overrightarrow c$$ and $$\overrightarrow a .\overrightarrow b = 3.$$ Then $$\left| {\overrightarrow b } \right|$$ equals :
A
$${{11} \over 3}$$
B
$${{11} \over {\sqrt 3 }}$$
C
$$\sqrt {{{11} \over 3}}$$
D
$${{\sqrt {11} } \over 3}$$
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