Show that the four points A, B, C and D with position vectors r

Subject

Mathematics

Class

CBSE Class 12

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 Multiple Choice QuestionsShort Answer Type

11.

Use Properties of determinants, prove that:
open vertical bar table row cell 1 plus straight a end cell cell space 1 end cell cell space 1 end cell row 1 cell 1 plus straight b end cell 1 row 1 1 cell 1 plus straight c end cell end table close vertical bar space equals space abc plus bc plus ca plus ab

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 Multiple Choice QuestionsLong Answer Type

12.

Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award x each, y each and z each for the three respective values to 3, 2 and 1 students respectively with a total award money of 1,600. School B wants to spend 2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount for one prize on each value is 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.

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13.

If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum when the angle between them is 60 degree.

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 Multiple Choice QuestionsShort Answer Type

14. If space straight f left parenthesis straight x right parenthesis space equals space integral subscript 0 superscript straight x straight t space sint space dt comma space write space the space value space of space straight f apostrophe left parenthesis straight x right parenthesis
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15.

Find the value of 'p' for which the vectors 3 straight i with hat on top plus 2 straight j with hat on top plus 9 straight k with hat on top space and space straight i with hat on top minus 2 straight p straight j with hat on top plus 3 straight k with hat on top are parallel.

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16.

If the cartesian equations of a line are fraction numerator 3 minus straight x over denominator 5 end fraction equals fraction numerator straight y plus 4 over denominator 7 end fraction equals fraction numerator 2 straight z minus 6 over denominator 4 end fraction comma write the vector equation for the line. 

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17.

If integral subscript 0 superscript straight a fraction numerator 1 over denominator 4 plus straight x squared end fraction dx equals straight pi over 8, find the value of a.

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18. If space straight a with rightwards arrow on top space and space straight b with rightwards arrow on top space are space perpendicular space vectors comma space open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar space equals space 13 space and space open vertical bar straight a with rightwards arrow on top close vertical bar space equals 5 space and space find space the space value space of space open vertical bar straight b with rightwards arrow on top close vertical bar.
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19.

Solve the differential equation left parenthesis 1 plus straight x squared right parenthesis dy over dx plus straight y equals straight e to the power of tan to the power of negative 1 end exponent straight x end exponent

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20.

Show that the four points A, B, C and D with position vectors
4 straight i with hat on top plus 5 straight j with hat on top plus straight k with hat on top comma negative straight j with hat on top minus straight k with hat on top comma space 3 straight i with hat on top plus 9 straight j with hat on top plus 4 straight k with hat on top space and space 4 left parenthesis negative straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis respectively are coplanar.


Given position vectors of four points A, B, C and D are:

OA with rightwards arrow on top space equals space 4 straight i with hat on top minus 5 straight j plus straight k
OB with rightwards arrow on top space equals negative straight j minus straight k
OC with rightwards arrow on top equals 3 straight i with hat on top plus 9 straight j plus 4 straight k
OD with rightwards arrow on top equals space 4 open parentheses negative straight i with hat on top plus straight j plus straight k close parentheses
These points are coplanar, if the vectors, AB with rightwards arrow on top comma space AC with rightwards arrow on top space and space AD with rightwards arrow on top are coplanar. 

AB with rightwards arrow on top space equals OB with rightwards arrow on top minus OA with rightwards arrow on top
equals negative straight j minus straight k minus open parentheses 4 straight i with hat on top plus 5 straight j plus straight k close parentheses equals negative 4 straight i with hat on top minus 6 straight j minus 2 straight k
AC with rightwards arrow on top equals OC with rightwards arrow on top minus OA with rightwards arrow on top
equals 3 straight i with hat on top plus 9 straight j plus 4 straight k minus open parentheses 4 straight i with hat on top plus 5 straight j plus straight k close parentheses equals negative straight i with hat on top plus 4 straight j plus 3 straight k
AB with rightwards arrow on top space equals OD with rightwards arrow on top minus OA with rightwards arrow on top
equals 4 open parentheses negative straight i with hat on top plus straight j plus straight k close parentheses minus open parentheses 4 straight i with hat on top plus 5 straight j plus straight k close parentheses equals negative 8 straight i with hat on top minus straight j plus 3 straight k

These vectors are coplanar if and only if, they can be expressed as a linear combination of other two. 
So Let

AB with rightwards arrow on top space equals space straight x AC with rightwards arrow on top space plus straight y AD with rightwards arrow on top
rightwards double arrow space minus 4 straight i with hat on top minus 6 straight j minus 2 straight k equals straight x open parentheses negative straight i with hat on top plus 4 straight j plus 3 straight k close parentheses plus straight y open parentheses negative 8 straight i with hat on top minus straight j with hat on top plus 3 straight k with hat on top close parentheses
rightwards double arrow space minus 4 straight i with hat on top minus 6 straight j with hat on top minus 2 straight k with hat on top equals space open parentheses negative straight x minus 8 straight y close parentheses straight i with hat on top plus left parenthesis 4 straight x minus straight y right parenthesis straight j with hat on top plus left parenthesis 3 straight x plus 3 straight y right parenthesis straight k with hat on top
Comparing the coefficients, we have, 

negative straight x minus 8 straight y equals negative 4 semicolon space space 4 straight x minus straight y equals negative 6 semicolon space space 3 straight x plus 3 straight y equals negative 2
Thus comma space solving space the space first space two space equations comma space space we space get
straight x equals fraction numerator negative 4 over denominator 3 end fraction space and space straight y space equals 2 over 3
These values of x and y satisfy the equation  3x + 3y = -2.
Hence the vectors are coplanar. 

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