The distance of the point (1, −5, 9) from the plane x−y+z=5

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 Multiple Choice QuestionsLong Answer Type

331.

Find the equation of the plane which contains the line of intersection of the planes 

r.  i^ + 2 j^ + 3 k  - 4 = 0,     r.  2 i^ + j^  - k  + 5 = 0  and which is perpendicular to

the plane  r.  5 i^ + 3 j^  - 6 k  + 8 = 0.


 Multiple Choice QuestionsShort Answer Type

332.

If a line has direction ratios 2,-1,-2 then what are its direction cosines?


 Multiple Choice QuestionsLong Answer Type

333.

Find the equation of the line passing through the point  (-1,3,-2)  and

perpendicular to the lines   x1 = y2 = z3   and   x + 2- 3 = y - 12 = z + 15.


334.

Find the equation of the plane determined by the point A( 3, - 1, 2 ), B( 5, 2, 4 ) and C( -1, -1, 6 ) and hence find the distance between the plane and the point P( 6, 5, 9 ).


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 Multiple Choice QuestionsMultiple Choice Questions

335.

The line passing through  - 1, π2 and perpendicular to 

3 + sinθ + 2cosθ = 4r is : 

  • 2 = 3rcosθ - 2rsinθ

  • 5 = - 23rsinθ +  4rsinθ

  • 2 = 3rcosθ +  2rsinθ

  • 5 = 23rsinθ +  4rcosθ


336.

If m1 and m2 are the roots of the equation x2 + 3 + 2x + 3 - 1 = 0,then the area of the triangle formed by the linesy = m1x, y = m2x and y = c

  • 33 - 114c2

  • 33 + 114c2

  • 11 - 332c2

  • 332c2


337.

 ABC is formed by A(1, 8, 4), B (0, - 11, 4) and C(2, - 3,1). If D is the foot of the perpendicular from A to BC. Then the coordinates of D are

  • ( - 4, 5, 2)

  • (4, 5, - 2)

  • (4,  - 5, 2)

  • (4,  - 5, - 2)


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338.

The distance of the point (1, −5, 9) from the plane x−y+z=5 measured along the line x=y=z is:

  • 3√10

  • 10√3

  • 10/√3

  • 20/3


B.

10√3

the equation of the line passing through the point (1,5-9 and parallel to x =y=z is


fraction numerator straight x minus 1 over denominator 1 end fraction space equals space fraction numerator straight y plus 5 over denominator 1 end fraction space equals space fraction numerator straight z minus 9 over denominator 1 end fraction space equals space straight lambda
Thus, any point on this line is of the form
(λ +1, λ-5 ,λ+9) 
Now, if P (λ +1, λ-5, λ+9) is the point of intersection of line and plane, then
 (λ+1) - (λ-5) +λ+9 = 5
λ +15 = 5
λ = -10
therefore coordinates of point P are (-9, -15,-1)
Hence, required distance
=equals square root of left parenthesis 1 plus 9 right parenthesis squared plus left parenthesis negative 5 plus 15 right parenthesis squared plus left parenthesis 9 plus 1 right parenthesis squared end root
equals square root of 10 squared plus space 10 squared plus 10 squared end root space equals space 10 square root of 3

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339.

If the line   lies in the plane lx +my -z = 9, then l2 +m2 is equal to 

  • 26

  • 18

  • 5

  • 2

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340.

Locus the image of the point (2,3) in the line (2x - 3y +4) + k (x-2y+3) = 0, k ε R is a 

  • straight line parallel to X - axis

  • a straight line parallel to Y- axis

  • circle of radius square root of 2

  • circle of radius square root of 2

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