Maths-
General
Easy

Question

If l subscript 1 end subscript comma m subscript 1 end subscript comma n subscript 1 end subscriptand l subscript 2 end subscript comma m subscript 2 end subscript comma n subscript 2 end subscript blankare text d.c.´s end text text  of  end text stack O A with rightwards arrow on top comma stack O B with rightwards arrow on top such that  where ‘O’ is the origin, then the d.c.’s of the internal bisector of the angle  are

  1. fraction numerator l subscript 1 end subscript plus l subscript 2 end subscript over denominator 2 s i n invisible function application theta divided by 2 end fraction comma fraction numerator m subscript 1 end subscript plus m subscript 2 end subscript over denominator 2 s i n invisible function application theta divided by 2 end fraction comma fraction numerator n subscript 1 end subscript plus n subscript 2 end subscript over denominator 2 s i n invisible function application theta divided by 2 end fraction    
  2. fraction numerator l subscript 1 end subscript plus l subscript 2 end subscript over denominator 2 c o s invisible function application theta divided by 2 end fraction comma fraction numerator m subscript 1 end subscript plus m subscript 2 end subscript over denominator 2 c o s invisible function application theta divided by 2 end fraction comma fraction numerator n subscript 1 end subscript plus n subscript 2 end subscript over denominator 2 c o s invisible function application theta divided by 2 end fraction    
  3. fraction numerator l subscript 1 end subscript minus l subscript 2 end subscript over denominator 2 s i n invisible function application theta divided by 2 end fraction comma fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator 2 s i n invisible function application theta divided by 2 end fraction comma fraction numerator n subscript 1 end subscript minus n subscript 2 end subscript over denominator 2 s i n invisible function application theta divided by 2 end fraction    
  4. fraction numerator l subscript 1 end subscript minus l subscript 2 end subscript over denominator 2 c o s invisible function application theta divided by 2 end fraction comma fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator 2 c o s invisible function application theta divided by 2 end fraction comma fraction numerator n subscript 1 end subscript minus n subscript 2 end subscript over denominator 2 c o s invisible function application theta divided by 2 end fraction    

The correct answer is: fraction numerator l subscript 1 end subscript plus l subscript 2 end subscript over denominator 2 c o s invisible function application theta divided by 2 end fraction comma fraction numerator m subscript 1 end subscript plus m subscript 2 end subscript over denominator 2 c o s invisible function application theta divided by 2 end fraction comma fraction numerator n subscript 1 end subscript plus n subscript 2 end subscript over denominator 2 c o s invisible function application theta divided by 2 end fraction


    Let OA and OB be two lines with d.c’s l subscript 1 end subscript comma m subscript 1 end subscript comma n subscript 1 end subscript text  and  end text l subscript 2 end subscript comma m subscript 2 end subscript comma n subscript 2 end subscript. text  Let  end text O A equals O B equals 1. Then, the coordinates of A and B are open parentheses l subscript 1 end subscript comma m subscript 1 end subscript comma n subscript 1 end subscript close parentheses text  and  end text open parentheses l subscript 2 end subscript comma m subscript 2 end subscript comma n subscript 2 end subscript close parentheses comma blankrespectively. Let OC be the bisector of angle A O B. Then, C is the mid point of AB and so its coordinates are open parentheses fraction numerator l subscript 1 end subscript plus l subscript 2 end subscript over denominator 2 end fraction comma fraction numerator m subscript 1 end subscript plus m subscript 2 end subscript over denominator 2 end fraction comma fraction numerator n subscript 1 end subscript plus n subscript 2 end subscript over denominator 2 end fraction close parentheses.
    therefore text  d.r´s of OC are  end text fraction numerator l subscript 1 end subscript plus l subscript 2 end subscript over denominator 2 end fraction comma fraction numerator m subscript 1 end subscript plus m subscript 2 end subscript over denominator 2 end fraction comma fraction numerator n subscript 1 end subscript plus n subscript 2 end subscript over denominator 2 end fraction
    We have, O C equals square root of open parentheses fraction numerator 1 subscript 1 end subscript plus 1 subscript 2 end subscript over denominator 2 end fraction close parentheses to the power of 2 end exponent plus open parentheses fraction numerator m subscript 1 end subscript plus m subscript 2 end subscript over denominator 2 end fraction close parentheses to the power of 2 end exponent plus open parentheses fraction numerator n subscript 1 end subscript plus n subscript 2 end subscript over denominator 2 end fraction close parentheses to the power of 2 end exponent end root
    table row cell equals fraction numerator 1 over denominator 2 end fraction square root of open parentheses l subscript 1 end subscript superscript 2 end superscript plus m subscript 1 end subscript superscript 2 end superscript plus n subscript 1 end subscript superscript 2 end superscript close parentheses plus open parentheses l subscript 2 end subscript superscript 2 end superscript plus m subscript 2 end subscript superscript 2 end superscript plus n subscript 2 end subscript superscript 2 end superscript close parentheses plus 2 open parentheses l subscript 1 end subscript l subscript 2 end subscript plus m subscript 1 end subscript m subscript 2 end subscript plus n subscript 1 end subscript n subscript 2 end subscript close parentheses end root end cell row cell equals fraction numerator 1 over denominator 2 end fraction square root of 2 plus 2 c o s invisible function application theta end root open square brackets Q c o s invisible function application theta equals l subscript 1 end subscript l subscript 2 end subscript plus m subscript 1 end subscript m subscript 2 end subscript plus n subscript 1 end subscript n subscript 2 end subscript close square brackets end cell row cell equals fraction numerator 1 over denominator 2 end fraction square root of 2 left parenthesis 1 plus c o s invisible function application theta right parenthesis end root equals c o s invisible function application open parentheses fraction numerator theta over denominator 2 end fraction close parentheses end cell end table

    therefore text  d.c´s of  end text O C text  are  end text fraction numerator l subscript 1 end subscript plus l subscript 2 end subscript over denominator 2 left parenthesis O C right parenthesis end fraction comma fraction numerator m subscript 1 end subscript plus m subscript 2 end subscript over denominator 2 left parenthesis O C right parenthesis end fraction comma fraction numerator n subscript 1 end subscript plus n subscript 2 end subscript over denominator 2 left parenthesis O C right parenthesis end fraction

    Related Questions to study

    General
    Maths-

    The area bounded by the curves is square root of vertical line x vertical line end root plus square root of vertical line y vertical line end root equals square root of a and x to the power of 2 end exponent plus y to the power of 2 end exponent equals a to the power of 2 end exponent (where a greater than 0 right parenthesis is

    The area bounded by the curves is square root of vertical line x vertical line end root plus square root of vertical line y vertical line end root equals square root of a and x to the power of 2 end exponent plus y to the power of 2 end exponent equals a to the power of 2 end exponent (where a greater than 0 right parenthesis is

    Maths-General
    General
    chemistry-

    The electronic configurations of the element A,B, and C are :given below.

    The bond between and C will be:

    The electronic configurations of the element A,B, and C are :given below.

    The bond between and C will be:

    chemistry-General
    General
    chemistry-

    The electronic configurations of the element A,B, and C are :given below.

    The moleculaformulof the compound formed from Band will be:

    The electronic configurations of the element A,B, and C are :given below.

    The moleculaformulof the compound formed from Band will be:

    chemistry-General
    parallel
    General
    chemistry-

    The electronic configurations of the element A,B, and C are :given below.

    Stable form of C may be represented by the formula:

    The electronic configurations of the element A,B, and C are :given below.

    Stable form of C may be represented by the formula:

    chemistry-General
    General
    chemistry-

    The electronic configurations of the element A,B, and C are :given below.

    Stable form of A may be represented by the formula:

    The electronic configurations of the element A,B, and C are :given below.

    Stable form of A may be represented by the formula:

    chemistry-General
    General
    chemistry-

    Lewis concept of covalency of an element involvesoctetrule.Lateronitwasfoundthatmanyelementsin their compounds, e.g., B e F subscript 2 end subscript comma B F subscript 3 end subscript etc. have incompleteoctetwhereasP C l subscript 5 end subscript comma S F subscript 6 end subscript et c.haveexpandedoctet. This classicaconcept also failed inpredictingthegeometryof molecules. Modem concept of covalence was pro­ posed in terms of valence bond theory. Hybridizationconcept along with valence bond theory successfullexplained the geometryofvariousmoleculesbutfailedin many molecules ThebondangleN O subscript 2 end subscript superscript plus end superscript comma N O subscript 2 end subscriptandN O subscript 2 end subscript superscript minus end superscriptare,respectively:

    Lewis concept of covalency of an element involvesoctetrule.Lateronitwasfoundthatmanyelementsin their compounds, e.g., B e F subscript 2 end subscript comma B F subscript 3 end subscript etc. have incompleteoctetwhereasP C l subscript 5 end subscript comma S F subscript 6 end subscript et c.haveexpandedoctet. This classicaconcept also failed inpredictingthegeometryof molecules. Modem concept of covalence was pro­ posed in terms of valence bond theory. Hybridizationconcept along with valence bond theory successfullexplained the geometryofvariousmoleculesbutfailedin many molecules ThebondangleN O subscript 2 end subscript superscript plus end superscript comma N O subscript 2 end subscriptandN O subscript 2 end subscript superscript minus end superscriptare,respectively:

    chemistry-General
    parallel
    General
    chemistry-

    NO is a colorless gasandan important intermediate inthemanufactureof nitricacidby catalytic oxidationofammonia. Whichofthefollowingisaneutraloxide?</span

    NO is a colorless gasandan important intermediate inthemanufactureof nitricacidby catalytic oxidationofammonia. Whichofthefollowingisaneutraloxide?</span

    chemistry-General
    General
    chemistry-

    NO has11 valence electrons.It is impossible for all1be paired. and hence this is an odd electron molecule,and the gas is paramagnetic. Which of the following species has the smallest N-O bond length?

    NO has11 valence electrons.It is impossible for all1be paired. and hence this is an odd electron molecule,and the gas is paramagnetic. Which of the following species has the smallest N-O bond length?

    chemistry-General
    General
    chemistry-

    Many bond angles can be explained by either electronegativity or size arguments. Molecules with larger difference in electronegativity values between central and outer atoms have smaller bond ang les. As the size of outer atom increases, the angle increases. The size of central atom can also be used to determine the bond angles in the series. Which has the smallest bond angle left parenthesis X minus S minus X right parenthesis in the given molecules?

    Many bond angles can be explained by either electronegativity or size arguments. Molecules with larger difference in electronegativity values between central and outer atoms have smaller bond ang les. As the size of outer atom increases, the angle increases. The size of central atom can also be used to determine the bond angles in the series. Which has the smallest bond angle left parenthesis X minus S minus X right parenthesis in the given molecules?

    chemistry-General
    parallel
    General
    chemistry-

    Many bond angles can be explained by either electronegativity or size arguments. Molecules with larger difference in electronegativity values between central and outer atoms have smaller bond ang les. As the size of outer atom increases, the angle increases. The size of central atom can also be used to determine the bond angles in the series.

    Consider the following hydrides and arrange them in increasing order of bond ang les:
    I) N H subscript 3 end subscript
    II) P H subscript 3 end subscript
    III) A s H subscript 3 end subscript
    IV) S b H subscript 3 end subscript

    Many bond angles can be explained by either electronegativity or size arguments. Molecules with larger difference in electronegativity values between central and outer atoms have smaller bond ang les. As the size of outer atom increases, the angle increases. The size of central atom can also be used to determine the bond angles in the series.

    Consider the following hydrides and arrange them in increasing order of bond ang les:
    I) N H subscript 3 end subscript
    II) P H subscript 3 end subscript
    III) A s H subscript 3 end subscript
    IV) S b H subscript 3 end subscript

    chemistry-General
    General
    chemistry-

    Many bond angles can be explained by either electronegativity or size arguments. Molecules with larger difference in electronegativity values between central and outer atoms have smaller bond ang les. As the size of outer atom increases, the angle increases. The size of central atom can also be used to determine the bond angles in the series. Consider the following molecules:
    I) H subscript 2 end subscript O
    II) H subscript 2 end subscript text end text S
    III) H subscript 2 end subscript S e
    IV) H subscript 2 end subscript T e

    Arrange these molecules in increasing order of bond angle:

    Many bond angles can be explained by either electronegativity or size arguments. Molecules with larger difference in electronegativity values between central and outer atoms have smaller bond ang les. As the size of outer atom increases, the angle increases. The size of central atom can also be used to determine the bond angles in the series. Consider the following molecules:
    I) H subscript 2 end subscript O
    II) H subscript 2 end subscript text end text S
    III) H subscript 2 end subscript S e
    IV) H subscript 2 end subscript T e

    Arrange these molecules in increasing order of bond angle:

    chemistry-General
    General
    chemistry-

    A) 
    B) 
    C) 
    In the thiocyanate ion, S C N to the power of minus end exponent three resonance structures are consistent with the electron -dot method Structure A has only one negative formal charge on the nitrogen atom, the most electro negative atom in the ion. Structure B has a single negative charge on the S, which is less electronegative than N. Structure C has charges of -2 on N and +1 on S, consistent with the relative electronegativities of the atoms but with a larger charge and greater charge separation than the first.
    HNCs is isostructural with:

    A) 
    B) 
    C) 
    In the thiocyanate ion, S C N to the power of minus end exponent three resonance structures are consistent with the electron -dot method Structure A has only one negative formal charge on the nitrogen atom, the most electro negative atom in the ion. Structure B has a single negative charge on the S, which is less electronegative than N. Structure C has charges of -2 on N and +1 on S, consistent with the relative electronegativities of the atoms but with a larger charge and greater charge separation than the first.
    HNCs is isostructural with:

    chemistry-General
    parallel
    General
    chemistry-

    A) 
    B) 
    C) 
    In the thiocyanate ion, S C N to the power of minus end exponent three resonance structures are consistent with the electron -dot method Structure A has only one negative formal charge on the nitrogen atom, the most electro negative atom in the ion. Structure B has a single negative charge on the S, which is less electronegative than N. Structure C has charges of -2 on N and +1 on S, consistent with the relative electronegativities of the atoms but with a larger charge and greater charge separation than the first. Predict the hybridization of atom which is bonded with H in structure HNCS:

    A) 
    B) 
    C) 
    In the thiocyanate ion, S C N to the power of minus end exponent three resonance structures are consistent with the electron -dot method Structure A has only one negative formal charge on the nitrogen atom, the most electro negative atom in the ion. Structure B has a single negative charge on the S, which is less electronegative than N. Structure C has charges of -2 on N and +1 on S, consistent with the relative electronegativities of the atoms but with a larger charge and greater charge separation than the first. Predict the hybridization of atom which is bonded with H in structure HNCS:

    chemistry-General
    General
    chemistry-

    Number of pi bonds sigma bonds in the following structure is:

    Number of pi bonds sigma bonds in the following structure is:

    chemistry-General
    General
    chemistry-

    Which of the following reaction staking place in the blastfurnace during extraction of ironsendo thermic?

    Which of the following reaction staking place in the blastfurnace during extraction of ironsendo thermic?

    chemistry-General
    parallel

    card img

    With Turito Academy.

    card img

    With Turito Foundation.

    card img

    Get an Expert Advice From Turito.

    Turito Academy

    card img

    With Turito Academy.

    Test Prep

    card img

    With Turito Foundation.