If the foot of the perpendicular drawn from the point (5, 1, - 3)

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 Multiple Choice QuestionsMultiple Choice Questions

131.

The straight line passing through the point (1, 0, - 2) and perpendicular to the plane x - 2y + 5z - 7 = 0 is

  • x - 11 = y0 = z - 5- 2

  • x - 15 = y- 2 = z + 21

  • x - 5- 2 = y - 1- 5 = z1

  • x - 11 = y- 2 = z + 25


132.

The equation of the plane passing through (1, 2, 3) and parallel to 3x - 2y + 4z = 5 is

  • 3x - 2y + 4z = 11

  • 3x - 2y + 4z = 0

  • 3x - 2y + 4z = 10

  • 3(x - 1) - 2(y - 2) + 4(z - 3) = 5


133.

If the straight lines x - 21 = y - 31 = z - 40 and x - 1k = y - 42 = z - 51 are copalnar then, the value of k is

  • - 3

  • 0

  • 1

  • - 2


134.

The line x - x10 = y - y11 = z - z12 is

  • perpendicular to the x-axis

  • perpendicular to the yz-plane

  • parallel to the y-axis

  • parallel to the xz-plane


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135.

The point which divides the line joining the points (1, 3, 4) and (4, 3, 1) internally in the ratio 2 : 1, is

  • (2, - 3, 3)

  • (2, 3, 3)

  • 52, 3, 52

  • (3, 3, 2)


136.

The angle between the lines x - 71 = y + 3- 5 = z3 and 2 - x- 7 = y2 = z + 51 is equal to

  • π4

  • π3

  • π2

  • π6


137.

The equation of the plane which is equidistant from the two parallel planes 2x - 2y + z + 3= 0 and 4x - 4y + 2z + 9 = 0 is

  • 8x - 8y + 4z + 15 = 0

  • 8x - 8y + 4z - 15 = 0

  • 8x - 8y + 4z + 3 = 0

  • 8x - 8y + 4z - 3 = 0


138.

The angle between the planes 3x + 4y + 5z = 3 and 4x - 3y + 5z = 9 is equal to

  • π2

  • π4

  • π6

  • π3


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139.

The vector equation of the plane through the point (2, 1, - 1) and parallel to the plane r - (i + 3j - k) = 0 is

  • r . (i + 9j + 11k) = 6

  • r . (i - 9j + 11k) = 4

  • r . (i + 3j - k) = 6

  • r . (i + 3j - k) = 4


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140.

If the foot of the perpendicular drawn from the point (5, 1, - 3) to a plane is (1, - 1, 3), then the equation of the plane is

  • 2x + y - 3z + 8 = 0

  • 2x + y + 3z + 8 = 0

  • 2x - y - 3z + 8 = 0

  • 2x - y + 3z + 8 = 0


A.

2x + y - 3z + 8 = 0

Let equation of a plane passing through (1, - 1, 3) is

a(x - 1) + b(y + 1) + c(z - 3) = 0

Now, DR's of PQ = (5 - 1, 1 + 1, - 3- 3)

                        = (4, 2, - 6)

The normal to the plane are proportional to PQ.

i.e., (a, b, c) = (4, 2, - 6) or (2, 1, - 3)

 2(x - 1) + 1(y + 1) - 3(z - 3) = 0

 2x + y - 3z + 8 = 0


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