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QUESTION: 1

Equation of a plane which is at a distance d from the origin and the direction cosines of the normal to the plane are l, m, n is.

Solution:

QUESTION: 2

The distance between the planes 3x – 2y + 6z + 21 = 0 and – 6x + 4y – 12z + 35 = 0 is:

Solution:

QUESTION: 3

The distance of the point (2, 3, – 5) from the plane x + 2y – 2z = 9 is:

Solution:

Length of perpendicular from (2,3,-5) to the plane x + 2y − 2z − 9 = 0.

= |(2 + 2×3 −2×(−5) − 9)|√1^{2} + 2^{2} + (−2)^{2}

= |2 + 6 + 10 − 9|/√9

= 9/3

= 3 units.

QUESTION: 4

The angle between the planes and is

Solution:

QUESTION: 5

The foot of the perpendicular drawn from the (- 1, – 3, – 5) to a plane is (2, 4, 6). The equation of the plane is:

Solution:

Since, the foot of the perpendicular to the plane is A(2,4,6). Therefore (4,2,6) is the point on the plane.

So, equation of the plane passing through the point (2,4,6) is:

a(x–2)+b(y–4)+c(z–6)=0.

Now, the direction ratios of the perpendicular line OA=2+1, 4+3, 6+5, i.e., 3,7,11

Therefore, the required plane is:

3(x–2)+7(y–4)+6(z–11)=0

i.e, 3x + 7y + 11z = 100

QUESTION: 6

A point is 5 units away from the vertical plane and 4 units away from profile plane and 3 units away from horizontal plane in 1st quadrant then the projections are drawn on paper the distance between the front view and top view of point is _____________

Solution:

Since the point is 3 units away from the horizontal plane the distance from the point to xy reference line will be 3 units. And then the point is at distance of 5 units from the vertical plane the distance from reference line and point will be 5, sum is 8.

QUESTION: 7

The distance of the plane 6x – 2y + 3z = 12 from the origin is:

Solution:

QUESTION: 8

The distance of the point from the plane is:

Solution:

QUESTION: 9

Distance of the point P (1, 1, p) from the plane whose equation is is

Solution:

QUESTION: 10

The angle between the line and the plane 2x – y + 2z + 7 = 0 is:

Solution:

(x + 1)/(-2) = (y - 2)/(3) = (z + 5)/(-6)

b = -2i + 3j - 6k

n = 2i – yj + 2k

Sinθ = (b.n)/|b||n|

Sinθ = (-2i + 3j - 6k).(2i – yj + 2k)]/{[(-2)^{2} + (3)^{2} + (-6)^{2}]^½ *[(2)^{2} + (-1)^{2} + (2)^{2}]^{½}}

Sinθ = (-4 -3 -12)/[(49)^{½} * (9)^{½}]

Sinθ = 19/(7 * 3)

Sinθ= 19/21

θ = sin^{-1}(19/21)

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