Advertisements
Advertisements
प्रश्न
Find the vector and cartesian equations of the plane passing throuh the points (2,5,- 3), (-2, - 3,5) and (5,3,-3). Also, find the point of intersection of this plane with the line passing through points (3, 1, 5) and (–1, –3, –1).
Advertisements
उत्तर १
Vector equation of a plane passing through
three pints A, B,C → (2, 5 ,-3 ) , (-2, -3,5) and (5, 3, -3)
Position vector of A = `2hat(i) + 5 hat(j) - 3 hat(k) = hat(a)`
Position vector of B = `2hat(i) - 3 hat(j) + 5 hat(k) = hat(b)`
Position vector of C = `5hat(i) + 3 hat(j) - 3 hat(k) = hat(c)`
Vector equation passing through `vec(a) , vec(b) , vec(c)" is" vec(r) = (x hat(i) + yhat(j) +zhat(k))` and
`(vec(r) - vec(a)) . [(vec(b) - vec(a)) xx (vec(c) - vec(a))] = 0`
`(vec(r) - (2hat(i) +5 hat(j) -3hat(k))).[(-2hat(i) - 3hat(j) +5hat(k) - 2hat(i) - 5 hat(j) + 3hat(k) ) xx (5hat(i) + 3hat(j) - 3 hat(k)-2hat(i) - 5hat(j) +3hat(k))]= 0`
` ⇒ (vec(r) - (2hat(i) + 5hat(j) - 3hat(k))).[(-4hat(i) - 8 hat(j) + 8hat(k) ) xx (3hat(i) - 2 hat(j) )]=0`
` ⇒ (vec(r) - (2hat(i) + 5hat(j) - 3hat(k))) .(8 hat(k) + 24 hat(k) + 24 hat(j) + 16 hat(i) ) = 0`
` ⇒ (vec(r) - (2hat(i) + 5hat(j) - 3hat(k))). (16hat(i) + 24 hat(j) + 32 hat(k) ) =0`
This is the required vector equation of the plane.
For cartesian equation:
`|(x-2, y-5, z+3) , (-4, -8, 8) , (3, -2, 0)|=0`
⇒(x - 2) (+16) - (y-5) (-24) + (z+3) (8+24) =0
⇒ (16x - 32 + 24y -120 + 32z + 96 ) =0
⇒ 16x + 24y + 32z = 120+ 32 -96
= 56
⇒ 16x + 24y + 32z = 56
This is the cartesian equation.
Line going two points (3,1,5) and (-1,-3,-1)
Difference between two points (-4,- 4, -6)
So x = 3 - 4t
y = 1 - 4t
z = 5 - 6t
and plane 16x + 24 y + 32z = 56
Intersection points
16(3 - 4t) + 24(1 + 4t) +32 (5 + 6t) =56
48 - 64t + 24 - 96t +160 -192t = 56
⇒ 232 - 56 =192t + 64t + 96t
⇒ 176 = 352t
⇒ `t = 1/2`
`x = 3 - 4(1/2) = 3 - 2 = 1`
`y = 1- 4 (1/2) = 1 -2 = -1`
`z = 5 -6 (1/2) = 5 -3 =2`
उत्तर २
Vector equation of a plane passing through
three pints A, B,C → (2, 5 ,-3 ) , (-2, -3,5) and (5, 3, -3)
Position vector of A = `2hat(i) + 5 hat(j) - 3 hat(k) = hat(a)`
Position vector of B = `2hat(i) - 3 hat(j) + 5 hat(k) = hat(b)`
Position vector of C = `5hat(i) + 3 hat(j) - 3 hat(k) = hat(c)`
Vector equation passing through `vec(a) , vec(b) , vec(c)" is" vec(r) = (x hat(i) + yhat(j) +zhat(k))` and
`(vec(r) - vec(a)) . [(vec(b) - vec(a)) xx (vec(c) - vec(a))] = 0`
`(vec(r) - (2hat(i) +5 hat(j) -3hat(k))).[(-2hat(i) - 3hat(j) +5hat(k) - 2hat(i) - 5 hat(j) + 3hat(k) ) xx (5hat(i) + 3hat(j) - 3 hat(k)-2hat(i) - 5hat(j) +3hat(k))]= 0`
` ⇒ (vec(r) - (2hat(i) + 5hat(j) - 3hat(k))).[(-4hat(i) - 8 hat(j) + 8hat(k) ) xx (3hat(i) - 2 hat(j) )]=0`
` ⇒ (vec(r) - (2hat(i) + 5hat(j) - 3hat(k))) .(8 hat(k) + 24 hat(k) + 24 hat(j) + 16 hat(i) ) = 0`
` ⇒ (vec(r) - (2hat(i) + 5hat(j) - 3hat(k))). (16hat(i) + 24 hat(j) + 32 hat(k) ) =0`
This is the required vector equation of the plane.
For cartesian equation:
`|(x-2, y-5, z+3) , (-4, -8, 8) , (3, -2, 0)|=0`
⇒(x - 2) (+16) - (y-5) (-24) + (z+3) (8+24) =0
⇒ (16x - 32 + 24y -120 + 32z + 96 ) =0
⇒ 16x + 24y + 32z = 120+ 32 -96
= 56
⇒ 16x + 24y + 32z = 56
This is the cartesian equation.
Line going two points (3,1,5) and (-1,-3,-1)
Difference between two points (-4,- 4, -6)
So x = 3 - 4t
y = 1 - 4t
z = 5 - 6t
and plane 16x + 24 y + 32z = 56
Intersection points
16(3 - 4t) + 24(1 + 4t) +32 (5 + 6t) =56
48 - 64t + 24 - 96t +160 -192t = 56
⇒ 232 - 56 =192t + 64t + 96t
⇒ 176 = 352t
⇒ `t = 1/2`
`x = 3 - 4(1/2) = 3 - 2 = 1`
`y = 1- 4 (1/2) = 1 -2 = -1`
`z = 5 -6 (1/2) = 5 -3 =2`
