Advertisements
Advertisements
Question
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
Solution 1
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`
Solution 2
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
RELATED QUESTIONS
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 passing through the points `hati +hatj-2hatk, hati+2hatj+hatk,2hati-hatj+hatk`. Hence find the cartesian equation of the plane.
Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane `vec r.(hati+hatj+hatk)=2`
Find the vector equation of the plane which contains the line of intersection of the planes `vecr (hati+2hatj+3hatk)-4=0` and `vec r (2hati+hatj-hatk)+5=0` which is perpendicular to the plane.`vecr(5hati+3hatj-6hatk)+8=0`
Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
Find the Cartesian equation of the following planes:
`vecr.(hati + hatj-hatk) = 2`
Find the Cartesian equation of the following planes:
`vecr.(2hati + 3hatj-4hatk) = 1`
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 cartesian form of the equation of the plane `bar r=(hati+hatj)+s(hati-hatj+2hatk)+t(hati+2hatj+hatj)`
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 image of a point having the position vector: `3hati - 2hatj + hat k` in the plane `vec r.(3hati - hat j + 4hatk) = 2`
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 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).
Vector equation of a line which passes through a point (3, 4, 5) and parallels to the vector `2hati + 2hatj - 3hatk`.
Find the vector and Cartesian equations of the plane passing through the points (2, 2 –1), (3, 4, 2) and (7, 0, 6). Also find the vector equation of a plane passing through (4, 3, 1) and parallel to the plane obtained above.
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)`
