Find the coordinates of the foot of the perpendicular drawn from

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 Multiple Choice QuestionsMultiple Choice Questions

301.

If direction cosines of a vector of magnitude 3 are 23, - 93, 23, then vector is

  • i^ - 2j^ + 2k^

  • 2i^ + j^ + 2k^

  • i^ + 2j^ + 2k^

  • None of these


302.

Equation of line passing through the point (2, 3, 1) and parallel to the line of intersection of the planes x - 2y - z + 5 = 0 and x + y + 3z = 6 is

  • x - 25 = y - 34 = z - 13

  • x - 25 = y - 3- 4 = z - 13

  • x - 24 = y - 33 = z - 12

  • x - 2- 5 = y - 3- 4 = z - 13


303.

Foot of perpendicular drawn from the origin to the plane 2x - 3y + 4z = 29 is

  • (7, - 1, 3)

  • (5, - 1, 4)

  • (5, - 2, 3)

  • (2, - 3, 4)


304.

Suppose a + b + c = 0, a = 3, b = 5, c = 7, then the angle between a and b is

  • π

  • π2

  • π3

  • π4


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305.

The vector equation of the plane, which is at a distance of 314, from the origin and the normal from the origin is 2i^ - 3j^ + k^ is

  • r . 2i^ - 3j^ + k^ = 3

  • r . i^ + j^ + k^ = 9

  • r . i^ + 2j^ = 3

  • r . 2i^ + k^ = 3


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306.

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 5y + 8 = 0.

  • 0, - 185, 2

  • 0, 85, 2

  • 825, 0, 0

  • 0, - 85, 0


D.

0, - 85, 0

Let the coordinates of foot of the perpendicular drawnfrom origin is α, β, γ to the plane 5y + 8 = 0 α - 00 = β - 05 = γ - 00 = 0 + 0 + 0 + 802 + 52 + 02 α0 = β5 = γ0 = - 825 α = 0, β = - 85, μ = 0Hence, the coordinates of the foot of perpendicular is0, - 85, 0.


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307.

If cosα, cosβ, cosγ are the direction cosines of a vector a, then cos2α + cos2β + cos2γ is equal to

  • 2

  • 3

  • - 1

  • 0


308.

If a and b are unit vectors, then angle between a and b for 3a - b to be unit vectori

  • 45°

  • 60°

  • 90°

  • 30°


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309.

The plane 2x - 3y + 6z - 11 = 0 makes an angle sin-1(α) with X-axis, the value of α is equal to

  • 23

  • 27

  • 32

  • 37


310.

The perpendicular distance of the point P(6, 7, 8) from XY-plane is

  • 6

  • 7

  • 5

  • 8


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