1
JEE Main 2019 (Online) 12th April Morning Slot
+4
-1
Let $$\overrightarrow a = 3\widehat i + 2\widehat j + 2\widehat k$$ and $$\overrightarrow b = \widehat i + 2\widehat j - 2\widehat k$$ be two vectors. If a vector perpendicular to both the vectors $$\overrightarrow a + \overrightarrow b$$ and $$\overrightarrow a - \overrightarrow b$$ has the magnitude 12 then one such vector is :
A
$$4\left( {2\widehat i - 2\widehat j - \widehat k} \right)$$
B
$$4\left( { - 2\widehat i - 2\widehat j + \widehat k} \right)$$
C
$$4\left( {2\widehat i + 2\widehat j + \widehat k} \right)$$
D
$$4\left( {2\widehat i + 2\widehat j - \widehat k} \right)$$
2
JEE Main 2019 (Online) 12th April Morning Slot
+4
-1
Out of Syllabus
If the volume of parallelopiped formed by the vectors $$\widehat i + \lambda \widehat j + \widehat k$$, $$\widehat j + \lambda \widehat k$$ and $$\lambda \widehat i + \widehat k$$ is minimum, then $$\lambda$$ is equal to :
A
$$- {1 \over {\sqrt 3 }}$$
B
$${\sqrt 3 }$$
C
$$-{\sqrt 3 }$$
D
$${1 \over {\sqrt 3 }}$$
3
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
The distance of the point having position vector $$- \widehat i + 2\widehat j + 6\widehat k$$ from the straight line passing through the point (2, 3, – 4) and parallel to the vector, $$6\widehat i + 3\widehat j - 4\widehat k$$ is :
A
6
B
7
C
$$2\sqrt {13}$$
D
$$4\sqrt 3$$
4
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
Let A (3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos ($$\angle$$GOA) (O being the origin) is equal to :
A
$${1 \over {\sqrt {15} }}$$
B
$${1 \over {6\sqrt {10} }}$$
C
$${1 \over {\sqrt {30} }}$$
D
$${1 \over {2\sqrt {15} }}$$
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