Maths-
General
Easy

Question

A non - zero vector stack a with rightwards arrow on top is parallel to the line of intersection of the plane P subscript 1 end subscript determined by i with ˆ on top plus j with ˆ on top and i plus 2 j with ˆ on top and plane P subscript 2 end subscript determined by vector 2 i with ˆ on top minus j with ˆ on top and 3 i with ˆ on top plus 2 k with ˆ on top, then angle between stack a with rightwards arrow on top and vector i with ˆ on top minus 2 j with ˆ on top plus 2 k with ˆ on top is

  1. fraction numerator pi over denominator 4 end fraction    
  2. fraction numerator pi over denominator 2 end fraction    
  3. fraction numerator pi over denominator 3 end fraction    
  4. none of these    

The correct answer is: fraction numerator pi over denominator 2 end fraction

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Let a,b,c be distinct non - negative numbers and the vectors a i with ˆ on top blank plus a j with ˆ on top plus c k with ˆ on top comma i with ˆ on top plus k with ˆ on top comma i with ˆ on top plus j with ˆ on top plus b k with ˆ on top comma lie in a plane, then the quadratic equation a x to the power of 2 end exponent plus 2 c x plus b equals 0 has

Let a,b,c be distinct non - negative numbers and the vectors a i with ˆ on top blank plus a j with ˆ on top plus c k with ˆ on top comma i with ˆ on top plus k with ˆ on top comma i with ˆ on top plus j with ˆ on top plus b k with ˆ on top comma lie in a plane, then the quadratic equation a x to the power of 2 end exponent plus 2 c x plus b equals 0 has

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If stack a with rightwards arrow on top comma stack b with rightwards arrow on top and stack c with rightwards arrow on top are anythree vectors forming a linearly independent system then for all theta element of R open square brackets stack a with minus on top c o s invisible function application theta plus stack b with rightwards arrow on top s i n invisible function application theta plus stack c blank with rightwards arrow on top c o s invisible function application 2 theta comma stack a with minus on top c o s invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction plus theta close parentheses plus stack b with minus on top s i n invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction plus theta close parentheses plus stack c with rightwards arrow on top c o s invisible function application 2 open parentheses fraction numerator 2 pi over denominator 3 end fraction plus theta close parentheses close open stack a with ‾ on top c o s invisible function application open parentheses theta minus fraction numerator 2 pi over denominator 3 end fraction close parentheses plus stack b with ‾ on top s i n invisible function application open parentheses theta minus fraction numerator 2 pi over denominator 3 end fraction close parentheses plus stack c with rightwards arrow on top c o s invisible function application 2 open parentheses theta minus fraction numerator 2 pi over denominator 3 end fraction close parentheses close square brackets equals

If stack a with rightwards arrow on top comma stack b with rightwards arrow on top and stack c with rightwards arrow on top are anythree vectors forming a linearly independent system then for all theta element of R open square brackets stack a with minus on top c o s invisible function application theta plus stack b with rightwards arrow on top s i n invisible function application theta plus stack c blank with rightwards arrow on top c o s invisible function application 2 theta comma stack a with minus on top c o s invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction plus theta close parentheses plus stack b with minus on top s i n invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction plus theta close parentheses plus stack c with rightwards arrow on top c o s invisible function application 2 open parentheses fraction numerator 2 pi over denominator 3 end fraction plus theta close parentheses close open stack a with ‾ on top c o s invisible function application open parentheses theta minus fraction numerator 2 pi over denominator 3 end fraction close parentheses plus stack b with ‾ on top s i n invisible function application open parentheses theta minus fraction numerator 2 pi over denominator 3 end fraction close parentheses plus stack c with rightwards arrow on top c o s invisible function application 2 open parentheses theta minus fraction numerator 2 pi over denominator 3 end fraction close parentheses close square brackets equals

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Roasting of concentrated zinc sulphide ore is completed at the temperature of 1200 K to _______

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Reactivity of M e M g B r with the following in the decreasing order is:
i)
ii)
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Chemistry-General
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When is treated with C subscript 2 end subscript H subscript 5 end subscriptMgBr,followed by hydrolysis, the product is:

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Chemistry-General
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Consider the following halogen-containing compounds;
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Maths-General
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Maths-

At A, h>0 and alpha with ‾ on top equals fraction numerator i with ˆ on top over denominator a end fraction plus fraction numerator 4 j with ˆ on top over denominator b end fraction plus b k with ˆ on top 1192720 and beta with rightwards arrow on top equals b i with ˆ on top plus a j plus 1 over b k with ˆ on top, then the maximum value of fraction numerator 10 over denominator 5 plus stack alpha with rightwards arrow on top times beta end fraction is

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Maths-General
parallel
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The position vector of a point P is r with rightwards arrow on top equals x i with ˆ on top plus y j plus z k with ˆ on top where x comma y comma z element of N and a with ‾ on top equals i with ˆ on top plus 2 j plus k with ˆ on top. If stack r with rightwards arrow on top blank stack a with rightwards arrow on top equals 20, the number of possible position of P is

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Maths-General

For such questions, we should know different formulas of limit.

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For such questions, we should know different formulas of limits. We should try different methods to simplify the function in a way that it doesn't become zero.

L t subscript left parenthesis x rightwards arrow 3 right parenthesis invisible function application left parenthesis 4 x squared minus 17 x plus 15 right parenthesis divided by left parenthesis x squared minus x minus 6 right parenthesis

Maths-General

For such questions, we should know different formulas of limits. We should try different methods to simplify the function in a way that it doesn't become zero.

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