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).
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`
APPEARS IN
संबंधित प्रश्न
Find the vector equation of the plane passing through a point having position vector `3 hat i- 2 hat j + hat k` and perpendicular to the vector `4 hat i + 3 hat j + 2 hat k`
Find the vector equation of the plane which is at a distance of 5 units from the origin and which is normal to the vector `2hati + hatj + 2hatk.`
Parametric form of the equation of the plane is `bar r=(2hati+hatk)+lambdahati+mu(hat i+2hatj+hatk)` λ and μ are parameters. Find normal to the plane and hence equation of the plane in normal form. Write its Cartesian form.
Find the vector equation of the plane passing through three points with position vectors ` hati+hatj-2hatk , 2hati-hatj+hatk and hati+2hatj+hatk` . Also find the coordinates of the point of intersection of this plane and the line `vecr=3hati-hatj-hatk lambda +(2hati-2hatj+hatk)`
Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
Find the equation of the plane which contains the line of intersection of the planes
`vecr.(hati-2hatj+3hatk)-4=0" and"`
`vecr.(-2hati+hatj+hatk)+5=0`
and whose intercept on x-axis is equal to that of on y-axis.
The x-coordinate of a point of the line joining the points P(2,2,1) and Q(5,1,-2) is 4. Find its z-coordinate
Find the Cartesian equation of the following planes:
`vecr.[(s-2t)hati + (3 - t)hatj + (2s + t)hatk] = 15`
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
2x + 3y + 4z – 12 = 0
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
3y + 4z – 6 = 0
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
x + y + z = 1
Find the vector and Cartesian equation of the planes that passes through the point (1, 0, −2) and the normal to the plane is `hati + hatj - hatk`
Find the equation of the plane through the line of intersection of `vecr*(2hati-3hatj + 4hatk) = 1`and `vecr*(veci - hatj) + 4 =0`and perpendicular to the plane `vecr*(2hati - hatj + hatk) + 8 = 0`. Hence find whether the plane thus obtained contains the line x − 1 = 2y − 4 = 3z − 12.
The Cartesian equation of the line is 2x - 3 = 3y + 1 = 5 - 6z. Find the vector equation of a line passing through (7, –5, 0) and parallel to the given line.
Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes \[\vec{r} \cdot \left( \hat{i} - \hat{j} + 2 \hat{k} \right) = 5 \text{ and } \vec{r} \cdot \left( 3 \hat{i} + \hat{j} + 2 \hat{k} \right) = 6\]
Find the vector and Cartesian forms of the equation of the plane passing through the point (1, 2, −4) and parallel to the lines \[\vec{r} = \left( \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right)\] and \[\vec{r} = \left( \hat{i} - 3 \hat{j} + 5 \hat{k} \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)\] Also, find the distance of the point (9, −8, −10) from the plane thus obtained.
Find the equation of the plane passing through the intersection of the planes `vec(r) .(hat(i) + hat(j) + hat(k)) = 1"and" vec(r) . (2 hat(i) + 3hat(j) - hat(k)) +4 = 0 `and parallel to x-axis. Hence, find the distance of the plane from x-axis.
Find the Cartesian equation of the plane, passing through the line of intersection of the planes `vecr. (2hati + 3hatj - 4hatk) + 5 = 0`and `vecr. (hati - 5hatj + 7hatk) + 2 = 0` intersecting the y-axis at (0, 3).
Find the vector and cartesian equation of the plane passing through the point (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).
Vector equation of a line which passes through a point (3, 4, 5) and parallels to the vector `2hati + 2hatj - 3hatk`.
Find the vector equation of the plane that contains the lines `vecr = (hat"i" + hat"j") + λ (hat"i" + 2hat"j" - hat"k")` and the point (–1, 3, –4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane thus obtained.
The Cartesian equation of the plane `vec"r" * (hat"i" + hat"j" - hat"k")` = 2 is ______.
The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is `vec"r" = 5hat"i" - 4hat"j" + 6hat"k" + lambda(3hat"i" + 7hat"j" + 2hat"k")`.
Find the vector and the cartesian equations of the plane containing the point `hati + 2hatj - hatk` and parallel to the lines `vecr = (hati + 2hatj + 2hatk) + s(2hati - 3hatj + 2hatk)` and `vecr = (3hati + hatj - 2hatk) + t(hati - 3hatj + hatk)`
