Find the Cartesian equation of the plane passing through the poin

Subject

Mathematics

Class

CBSE Class 12

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 Multiple Choice QuestionsLong Answer Type

21.

On a multiple choice examination with three possible answers (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?


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22.

Find the Cartesian equation of the plane passing through the points A(0, 0, 0) and B(3, -1, 2) and parallel to the line  x - 41 = y + 3-4 = z + 17


Let the equation of the plane be  ax + by + cz + d = 0       ........(i)

Since the plane passes through the point  A ( 0, 0, 0 )  and   B ( 3, -1, 2),

we have 

a x 0 + b x 0 + c x 0 + d = 0

 d = 0                 ................(ii)

Similarly for point B ( 3, -1, 2 ),    a x 3 + b x ( - 1 ) + c x 2 + d = 0

3a - b + 2c = 0           ( Using ,  d = 0 )             ............(iii)

Given equation of the line is  x - 41 = y + 3- 4 = z + 17We can also write the above equation as  x - 41 = y - ( -  3 )- 4 = z - ( -1 )7

The required plane is parellel to the above line .

Therefore,  a x 1 + b x ( - 4 ) + c x 7 = 0

 a - 4b + 7c = 0         ............(iv)

Cross multiplying equations (iii) and (iv), we obtain:

a( - 1 ) x 7 - ( - 4 ) x 2 = b2 x 1 - 3 x 7 = c3 x ( -4 ) - 1 x ( - 1 ) a- 7 + 8 = b2 - 21 = c- 12 + 1 a1 = b- 19 = c- 11 = k a = k,  b = - 19 k,  c = - 11 k

Substituting the values of  a,  b  and c  in equation ( 1 ), we obtain the equation of plane as: 

kx - 19ky - 11kz + d = 0

 k ( x - 19y - 11z )  = 0              ..........( From equation (ii) )

 x - 19y - 11z  = 0

So, the equation of the required plane is  x - 19y - 11z .


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23.

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are   2a + b   and   a - 3b  respectively, externally in the ratio 1:2. Also, show that P is the midpoint of the line segment R.


24.

Evaluate: 0π x1 + sinx  dx


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25.

Evaluate:  ex  sin4x - 41 - cos4x dx


26.

Evaluate:  1 - x2x  1 - 2x  dx


27.

Find the particular solution of the differential equation satisfying the given conditions: x2 dy + (xy + y2 )dx = 0; y = 1 when x = 1.


28.

Find the general solution of the differential equation,

x log x dydx + y = 2x log x


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29.

Find the particular solution of the differential equation satisfying the given conditions:

dydx = y tan x,    given that   y = 1  when   x= 0.


30.

Evaluate 13  3 x2 + 2 x  dx  as limit of sums.


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