1
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
$$1$$
2
AIEEE 2003
+4
-1
Consider points $$A, B, C$$ and $$D$$ with position

vectors $$7\widehat i - 4\widehat j + 7\widehat k,\widehat i - 6\widehat j + 10\widehat k, - \widehat i - 3\widehat j + 4\widehat k$$ and $$5\widehat i - \widehat j + 5\widehat k$$ respectively. Then $$ABCD$$ is a :
A
parallelogram but not a rhombus
B
square
C
rhombus
D
None
3
AIEEE 2003
+4
-1
Out of Syllabus
If $$\overrightarrow u \,,\overrightarrow v$$ and $$\overrightarrow w$$ are three non-coplanar vectors, then $$\,\left( {\overrightarrow u + \overrightarrow v - \overrightarrow w } \right).\left( {\overrightarrow u - \overrightarrow v } \right) \times \left( {\overrightarrow v - \overrightarrow w} \right)$$ equals :
A
$$3\overrightarrow u .\overrightarrow v \times \overrightarrow w$$
B
$$0$$
C
$$\overrightarrow u .\overrightarrow v \times \overrightarrow w$$
D
$$\overrightarrow u .\overrightarrow w \times \overrightarrow v$$
4
AIEEE 2002
+4
-1
If $$\left| {\overrightarrow a } \right| = 4,\left| {\overrightarrow b } \right| = 2$$ and the angle between $${\overrightarrow a }$$ and $${\overrightarrow b }$$ is $$\pi /6$$ then $${\left( {\overrightarrow a \times \overrightarrow b } \right)^2}$$ is equal to :
A
$$48$$
B
$$16$$
C
$$\overrightarrow a$$
D
none of these
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