Three Dimensional Geometry Exercise 03
Question:
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3i+5j−6k^.
Answer:
Step 1: Find the normal vector to the plane.
Normal vector = 3i + 5j  6k
Step 2: Find the magnitude of the normal vector.
Magnitude = √(3^2 + 5^2 + (6)^2) = √(9 + 25 + 36) = √70
Step 3: Find the unit vector of the normal vector.
Unit vector = (3/√70)i + (5/√70)j + (6/√70)k
Step 4: Find the vector equation of the plane.
Vector equation of the plane = r. (3/√70)i + (5/√70)j +(6/√70)k = 7
Question:
Find the vector equation of the plane passing through the intersection of the planes r.(2i^+2j^−3k^)=7,r.(2i^+5j^+3k^)=9 and through the point (2,1,3)
Answer:
Given,
Plane 1: 2i + 2j  3k = 7
Plane 2: 2i + 5j + 3k = 9
Point (2,1,3)
Step 1: Find the normal vector of the plane passing through the intersection of the two planes.
For this, we have to find the cross product of the normal vectors of the two planes.
Normal vector of Plane 1: n1 = (2,2,3)
Normal vector of Plane 2: n2 = (2,5,3)
Cross product of n1 and n2:
n1 x n2 = (2,2,3) x (2,5,3)
=(13, 6, 2)
Hence, the normal vector of the plane passing through the intersection of the two planes is n = (13, 6, 2).
Step 2: Find the equation of the plane passing through the given point (2,1,3) and having the normal vector n = (13, 6, 2).
Let the equation of the plane be ax + by + cz = d.
We know that,
ax + by + cz = d
13x + 6y  2z = d
Substituting the coordinates of the point (2,1,3) in the above equation, we get
13(2) + 6(1)  2(3) = d
26 + 6  6 = d
26 = d
Hence, the equation of the plane passing through the given point (2,1,3) and having the normal vector n = (13, 6, 2) is
13x + 6y  2z = 26
Question:
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. 3y+4z−6=0 A (0,24/25,18/25) B (0,24/25,24/25) C (0,18/25,24/25) D None of these
Answer:
Answer: C (0,18/25,24/25)
Question:
In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 2x−y+3z−1=0 and 2x−y+3z+3=0.
Answer:
The two planes are parallel because they have the same coefficients for the x, y, and z terms.
Question:
In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 2x−2y+4z+5=0 and 3x−3y+6z−1=0.
Answer:
Step 1: The given planes are 2x−2y+4z+5=0 and 3x−3y+6z−1=0.
Step 2: To determine whether the given planes are parallel or perpendicular, we need to compare their normal vectors.
Step 3: The normal vector of the first plane is (2, 2, 4). The normal vector of the second plane is (3, 3, 6).
Step 4: We can see that the two normal vectors are not parallel or perpendicular, so the planes are neither parallel nor perpendicular.
Step 5: To find the angle between the two planes, we can use the formula for the angle between two vectors: θ = cos−1((v₁⋅v₂)/(v₁*v₂))
Step 6: Plugging in our values for the two normal vectors, we get θ = cos−1(((23)+(−2−3)+(46))/((2,−2,4)(3,−3,6)))
Step 7: Simplifying, we get θ = cos−1(20/√(4+4+16)*√(9+9+36))
Step 8: Simplifying further, we get θ = cos−1(20/√84*√54)
Step 9: Simplifying once more, we get θ = cos−1(20/6*7.48)
Step 10: Finally, we get θ = cos−1(1.9)
Step 11: Therefore, the angle between the two planes is 1.9 radians.
Question:
Find the equation of the plane the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x−y+z=0.
Answer:
Step 1: Find the normal vectors of the planes x+y+z=1 and 2x+3y+4z=5.
Normal vector of x+y+z=1: (1,1,1)
Normal vector of 2x+3y+4z=5: (2,3,4)
Step 2: Find the vector perpendicular to both normal vectors.
The vector perpendicular to both normal vectors is given by the cross product of the two normal vectors: (1,1,1) x (2,3,4) = (1,2,1).
Step 3: Find the equation of the plane that is perpendicular to the plane x−y+z=0 and passes through the line of intersection of the two given planes.
The equation of the plane is given by the form ax + by + cz = d, where (a,b,c) is the normal vector of the plane and d is the distance of the plane from the origin.
Therefore, the equation of the plane is x + 2y  z = d.
Question:
Find the Cartesian equation of the following planes: r⋅(i^+j^−k^)=2.
Answer:
Step 1: Begin by writing the equation in the form Ax + By + Cz = D, where A, B, C, and D are constants.
A(i) + B(j)  C(k) = D
Step 2: Rewrite the given equation in the standard form by isolating the variables on one side of the equation.
A(i) + B(j)  C(k)  D = 0
Step 3: Substitute the given values into the equation.
i + (1)(j)  (1)(k)  2 = 0
Step 4: Simplify the equation.
i  j + k  2 = 0
Step 5: The Cartesian equation of the given plane is i  j + k  2 = 0.
Question:
In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 2x+y+3z−2=0 and x−2y+5=0.
Answer:

Rewrite the equations in standard form: 2x + y + 3z  2 = 0 x  2y + 5 = 0

Compare the coefficients of the two equations: 2x: x = 2 y: 2y = 2y 3z: 5 = 0

Since the coefficients of x and y are not the same, the planes are not parallel.

To find the angle between the two planes, use the formula for the angle between two planes: cos θ = (a₁a₂ + b₁b₂ + c₁c₂)/(√a₁² + b₁² + c₁²) * (√a₂² + b₂² + c₂²)

Plug in the coefficients of the two equations: cos θ = (2 * 1 + 2 * (2) + 0 * 5)/(√2² + 2² + 0²) * (√1² + (2)² + 5²)

Simplify the equation: cos θ = (4)/(2 * √26)

Calculate the angle: θ = arccos(4/√26) θ ≈ 115.3°
Question:
Find the angle between the planes whose vector equations are r.(2i^+2j^−3k^)=5 and r.(3i^−3j^+5k^)=3
Answer:
 Find the normal vectors for both planes:
Normal vector for the first plane: n1 = (2i^ + 2j^  3k^)
Normal vector for the second plane: n2 = (3i^  3j^ + 5k^)
 Find the dot product of the two normal vectors:
n1 • n2 = (2i^ + 2j^  3k^) • (3i^  3j^ + 5k^)
= (23) + (23) + (3*5)
= 6  6  15
= 15
 Find the angle between the two planes using the dot product formula:
angle = cos1(15/n1n2)
= cos1(15/√(22+22+32)√(32+32+52))
= cos1(15/√58√50)
= cos1(15/√2900)
= cos1(0.00521)
= 89.99°
Question:
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 5y+8=0.
Answer:
Step 1: Find the equation of the line passing through the origin and perpendicular to the given plane.
The equation of the line passing through the origin and perpendicular to the plane 5y + 8 = 0 is x = 0.
Step 2: Substitute the equation of the line in the given equation to find the coordinates of the foot of the perpendicular.
5y + 8 = 0
5y = 8
y = 8/5
Therefore, the coordinates of the foot of the perpendicular drawn from the origin to the plane 5y + 8 = 0 are (0, 8/5).
Question:
Find the vector and Catesian equation of the plane that passes through the point (1,4,6) and the normal vector to the plane is i^−2j^+k^
Answer:
Vector equation of the plane: (x−1, y−4, z−6)⋅(i^−2j^+k^) = 0
Cartesian equation of the plane: x−1−2(y−4)+z−6 = 0
Question:
Find the vector and Catesian equation of the plane that passes through the point (1,0,−2) and the normal to the plane is i^+j^−k^
Answer:
Step 1: Find the normal vector of the plane.
Normal vector = i^+ j^  k^
Step 2: Find the vector equation of the plane.
Vector equation of the plane = (x1, y0, z+2) . (i^+ j^  k^) = 0
Step 3: Find the Cartesian equation of the plane.
Cartesian equation of the plane = x  1 + y  0  z  2 = 0
Hence, the vector and Cartesian equation of the plane that passes through the point (1,0,2) and the normal to the plane is i^+j^−k^ is given by:
Vector equation: (x1, y0, z+2) . (i^+ j^  k^) = 0
Cartesian equation: x  1 + y  0  z  2 = 0
Question:
Find the equation of the plane through the intersection of the planes 3x−y+2z−4z=0 and x+y+z−2=0 and passes through the point (2,2,1)
Answer:
Step 1: Rewrite the equations in the form of Ax + By + Cz = D
3x − y + 2z − 4 = 0 => 3x − y + 2z = 4
x + y + z − 2 = 0 => x + y + z = 2
Step 2: Find the coefficients A, B and C of the equation of the plane.
A = 3 B = 1 C = 2
Step 3: Find the value of D using the point (2,2,1).
3*2 + (1)2 + 21 = 4 => D = 4
Step 4: The equation of the plane is Ax + By + Cz = D.
3x  y + 2z = 4
Question:
Point, Plane: (0,0,0),3x−4y+12z=3.
Answer:

This equation is in the form of Ax + By + Cz = D, where A = 3, B = 4, C = 12, and D = 3.

The point (0,0,0) lies on the plane since it satisfies the equation 3x−4y+12z=3.
Question:
Find the intercepts cut off by the plane 2x+y−z=5.
Answer:
Step 1: Set x = 0, y = 0, and z = 0 in the equation to find the intercepts on the x, y, and z axes.
2(0) + 0 − 0 = 5
0 = 5
This equation is false, so the plane does not intercept the x, y, or z axes.
Step 2: Set x = 0 and solve for y and z.
2(0) + y − z = 5
y − z = 5
y = 5 + z
Step 3: Set y = 0 and solve for x and z.
2x + 0 − z = 5
2x − z = 5
x = 5 + (1/2)z
Step 4: Set z = 0 and solve for x and y.
2x + y − 0 = 5
2x + y = 5
x = (5/2) − (1/2)y
Question:
In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 7x+5y+6z+30=0 and 3x−y−10z+4=0.
Answer:
Step 1: To determine whether two planes are parallel or perpendicular, we need to compare their normal vectors.
Step 2: The normal vector for the first plane is (7, 5, 6).
Step 3: The normal vector for the second plane is (3, 1, 10).
Step 4: We can calculate the dot product of the two normal vectors to determine if the planes are parallel or perpendicular.
Step 5: The dot product of (7, 5, 6) and (3, 1, 10) is 1.
Step 6: Since the dot product is not 0, the planes are neither parallel nor perpendicular.
Step 7: To find the angle between the two planes, we can use the formula cos θ = (v1 · v2) / (v1 v2).
Step 8: Plugging in the values, we get cos θ = (1) / (√74 √111) = 0.1736.
Step 9: Finally, we can use the inverse cosine to get the angle between the two planes: θ = 107.7°.
Question:
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. x+y+z=1.
Answer:
Step 1: First, rewrite the equation as x + y + z  1 = 0.
Step 2: To find the coordinates of the foot of the perpendicular, we must find the normal vector of the given equation. This can be done by taking the partial derivatives of the equation with respect to x, y, and z.
Step 3: The normal vector is given by n = (1, 1, 1).
Step 4: To find the coordinates of the foot of the perpendicular, we must find the dot product of the normal vector and the coordinates of the origin. This can be done by taking the dot product of n and (0, 0, 0).
Step 5: The dot product of n and (0, 0, 0) is 0.
Step 6: Therefore, the coordinates of the foot of the perpendicular drawn from the origin are (0, 0, 0).
Question:
Find the equations of the planes that passes through (1,1,0),(1,2,1),(−2,2,−1)
Answer:
Answer:
Step 1: Find the direction vectors of the two lines that pass through the given points.
Let the given points be A(1,1,0), B(1,2,1) and C(2,2,1).
The direction vectors of AB and BC are given by: AB = <1,1,1> and BC = <3,0,2>
Step 2: Find the normal vector of the plane.
The normal vector of the plane is given by the cross product of the two direction vectors, i.e. n = AB x BC = <3, 3, 1>
Step 3: Find the equation of the plane.
The equation of the plane can be found using the normal vector and any point on the plane, i.e. Equation of the plane: 3(x1)  3(y1) + (z0) = 0
Therefore, the equation of the plane that passes through (1,1,0), (1,2,1) and (2,2,1) is 3(x1)  3(y1) + (z0) = 0.
Question:
Find the equations of the planes that passes through (1,1,−1),(6,4,−5),(−4,−2,3)
Answer:
Let the equation of the plane be ax+by+cz=d
 Find the direction vector of the plane, which is the vector perpendicular to the plane.
Let the direction vector be (a,b,c).
Subtract the coordinates of the first two points to get the vector (5,3,6).
 Find the equation of the line passing through the two points.
The equation of the line passing through (1,1,1) and (6,4,5) is
x = 1 + 5t y = 1 + 3t z = 1  6t
 Find the point of intersection of the line and the plane.
Substitute the values of x, y and z in the equation of the plane to get
a(1+5t)+b(1+3t)+c(16t)=d
 Solve for t.
a+bc=datbtct
5t+a+bc=datbt
5t=datbtcab+c
t=(dab+c)/5
 Substitute the value of t in the equation of the line to get the coordinates of the point of intersection.
x = 1 + 5(dab+c)/5 y = 1 + 3(dab+c)/5 z = 1  6(dab+c)/5
 Substitute the coordinates of the point of intersection in the equation of the plane.
ax+by+cz=d
a(1+5(dab+c)/5)+b(1+3(dab+c)/5)+c(16(dab+c)/5)=d
 Solve for d.
d=a+b+c
 Substitute the values of a,b and c in the equation of the plane to get the equation of the plane.
ax+by+cz=a+b+c
Question:
Find the Cartesian equation of the following planes:r⋅[(s−2t)i^+(3−t)j^+(2s+t)k^]=15.
Answer:
Step 1: Rewrite the equation in the form of Ax + By + Cz = D. A(s−2t) + B(3−t) + C(2s+t) = 15
Step 2: Solve for A, B, and C by isolating each variable. A = 15/(s−2t) B = 15/(3−t) C = 15/(2s+t)
Step 3: Substitute the values of A, B, and C into the equation to get the Cartesian equation. 15/(s−2t) (s−2t) + 15/(3−t) (3−t) + 15/(2s+t) (2s+t) = 15
Step 4: Simplify the equation. 15 + 15 + 15 = 15
Step 5: The Cartesian equation of the plane is 15 = 15.
Question:
Point, Plane: (3,−2,1),2x−y+2z+3=0.
Answer:

Determine the equation of the plane: The equation of the plane is 2x  y + 2z + 3 = 0.

Determine the coordinates of the point: The coordinates of the point are (3, 2, 1).

Determine if the point lies on the plane: Substitute the coordinates of the point into the equation of the plane.
2(3)  (2) + 2(1) + 3 = 0 6 + 2 + 3 = 11
Since 11 does not equal 0, the point does not lie on the plane.
Question:
In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 4x+8yz−8=0 and y+z−4=0.
Answer:
Answer: The given planes are neither parallel nor perpendicular. The angle between them is arctan(1/3) radians.
Question:
Find the Cartesian equation of the following planes:r⋅(2i^+3j^−4k^)=1.
Answer:

The given equation is in the form of a scalar equation.

We can convert it into a Cartesian equation by expanding the dot product.

Thus, the Cartesian equation of the given plane is 2x + 3y  4z = 1.
Question:
Point (2,3,−5), plane x+2y−2z=9
Answer:
 Define the point: Point (2,3,5)
 Define the plane: x+2y2z=9
Question:
Point, Plane: (−6,0,0),2x−3y+6z−2=0.
Answer:

The equation of the plane is given as 2x−3y+6z−2=0.

The point given is (−6,0,0).

To determine if the point is on the plane, plug the coordinates of the point into the equation of the plane and solve for the value of the equation: 2(6)3(0)+6(0)2= 122=14

Since the equation is not equal to 0, the point is not on the plane.
Question:
Find the equation of the plane with intercept 3 on the yaxis and parallel to ZOX plane.
Answer:
Answer: Step 1: The equation of the ZOX plane is x + y = 0. Step 2: To find the equation of the plane parallel to ZOX plane, we will add a constant to the equation. Step 3: Since the intercept on the yaxis is 3, the constant should be 3. Step 4: The equation of the plane with intercept 3 on the yaxis and parallel to ZOX plane is x + y = 3.
Question:
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. 2x+3y+4z−12=0 A (2924,4936,2948) B (4924,4936,4948) C (2924,2936,2948) D (4924,2936,4948)
Answer:
A. (2924, 2936, 2948)