1
IIT-JEE 2000 Screening
+2
-0.5
If $${x^2} + {y^2} = 1,$$ then
A
$$yy'' - 2{\left( {y'} \right)^2} + 1 = 0$$
B
$$yy'' + {\left( {y'} \right)^2} + 1 = 0$$
C
$$yy'' + {\left( {y'} \right)^2} - 1 = 0$$
D
$$yy'' + 2{\left( {y'} \right)^2} + 1 = 0$$
2
IIT-JEE 2000 Screening
+4
-1
If the vectors $$\overrightarrow a ,\overrightarrow b$$ and $$\overrightarrow c$$ form the sides $$BC,$$ $$CA$$ and $$AB$$ respectively of a triangle $$ABC,$$ then
A
$$\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a = 0$$
B
$$\overrightarrow a \times \overrightarrow b = \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a$$
C
$$\overrightarrow a .\overrightarrow b = \overrightarrow b .\overrightarrow c = \overrightarrow c .\overrightarrow a$$
D
$$\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a = \overrightarrow 0$$
3
IIT-JEE 2000 Screening
+4
-1
Let the vectors $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c$$ and $$\overrightarrow d$$ be such that
$$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow c \times \overrightarrow d } \right) = 0.$$ Let $${P_1}$$ and $${P_2}$$ be planes determined
by the pairs of vectors $$\overrightarrow a .\overrightarrow b$$ and $$\overrightarrow c .\overrightarrow d$$ respectively. Then the angle between $${P_1}$$ and $${P_2}$$ is
A
$$0$$
B
$${\pi \over 4}$$
C
$${\pi \over 3}$$
D
$${\pi \over 2}$$
4
IIT-JEE 2000 Screening
+4
-1
If $$\overrightarrow a \,,\,\overrightarrow b$$ and $$\overrightarrow c$$ are unit coplanar vectors, then the scalar triple product $$\left[ {2\overrightarrow a - \overrightarrow b ,2\overrightarrow b - \overrightarrow c ,2\overrightarrow c - \overrightarrow a } \right] =$$
A
$$0$$
B
$$1$$
C
$$- \sqrt 3$$
D
$$\sqrt 3$$
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