1
AIEEE 2003
+4
-1
The vectors $$\overrightarrow {AB} = 3\widehat i + 4\widehat k\,\,\& \,\,\overrightarrow {AC} = 5\widehat i - 2\widehat j + 4\widehat k$$ are the sides of triangle $$ABC.$$ The length of the median through $$A$$ is :
A
$$\sqrt {288}$$
B
$$\sqrt {18}$$
C
$$\sqrt {72}$$
D
$$\sqrt {33}$$
2
AIEEE 2003
+4
-1
$$\overrightarrow a \,,\overrightarrow b \,,\overrightarrow c$$ are $$3$$ vectors, such that

$$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$$ , $$\left| {\overrightarrow a } \right| = 1\,\,\,\left| {\overrightarrow b } \right| = 2,\,\,\,\left| {\overrightarrow c } \right| = 3,$$,

then $${\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a }$$ is equal to :
A
$$1$$
B
$$0$$
C
$$-7$$
D
$$7$$
3
AIEEE 2003
+4
-1
A tetrahedron has vertices at $$O(0,0,0), A(1,2,1) B(2,1,3)$$ and $$C(-1,1,2).$$ Then the angle between the faces $$OAB$$ and $$ABC$$ will be :
A
$${90^ \circ }$$
B
$${\cos ^{ - 1}}\left( {{{19} \over {35}}} \right)$$
C
$${\cos ^{ - 1}}\left( {{{17} \over {31}}} \right)$$
D
$${30^ \circ }$$
4
AIEEE 2003
+4
-1
If $$\left| {\matrix{ a & {{a^2}} & {1 + {a^3}} \cr b & {{b^2}} & {1 + {b^3}} \cr c & {{c^2}} & {1 + {c^3}} \cr } } \right| = 0$$ and vectors $$\left( {1,a,{a^2}} \right),\,\,$$

$$\left( {1,b,{b^2}} \right)$$ and $$\left( {1,c,{c^2}} \right)\,$$ are non-coplanar, then the product $$abc$$ equals :
A
$$0$$
B
$$2$$
C
$$-1$$
D
$$1$$
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