Three Dimensional Geometry Miscellaneous Exerci
Question:
Find the distance of the point (−1,−5,−10) from the point of intersection of the line r =(2i^−j^+2k^)+λ(3i^+4j^+2k^) and the plane r .(i^−j^+k^)=5
Answer:
Step 1: Find the point of intersection of the line and the plane.
The point of intersection can be found by solving the equation:
2i^−j^+2k^+λ(3i^+4j^+2k^)=i^−j^+k^
This can be simplified to:
λ=−2
Substituting this value of λ in the equation of the line, we get the point of intersection as:
P=(−2,−10,−4)
Step 2: Find the distance of the point (−1,−5,−10) from the point of intersection of the line and the plane.
The distance between the two points can be found using the distance formula:
d=√((x2−x1)^2+(y2−y1)^2+(z2−z1)^2)
Where (x1,y1,z1) and (x2,y2,z2) are the coordinates of the two points.
Substituting the values in the formula, we get:
d=√((−1−(−2))^2+((−5−(−10))^2+((−10−(−4))^2
d=√9+25+36
d=√70
Therefore, the distance between the two points is √70.
Question:
Find the vector equation of the line passing through the point (1,2,−4) and perpendicular to the two lines: x−8/3=y+19/−16=z−10/7 and x−15/3=y−29/8=z−5/−5.
Answer:
 Find the direction vectors of the two given lines:
Line 1: (8/3, 19/16, 10/7) Line 2: (15/3, 29/8, 5/5)
 Find the cross product of the two direction vectors to find the direction vector of the line perpendicular to both:
(8/3, 19/16, 10/7) X (15/3, 29/8, 5/5) = (335/24, 145/16, 175/24)
 Find the vector equation of the line passing through the given point (1,2,4) and perpendicular to the two lines:
Vector equation of the line: r = (1,2,4) + t(335/24, 145/16, 175/24)
where t is a real number.
Question:
If the points (1,1,p) and (−3,0,1) be equidistant from the plane r⋅(3i^+4j^−12k^)+13=0 then find the value of p.
Answer:
Given, Points: (1,1,p) and (−3,0,1) Plane: r⋅(3i^+4j^−12k^)+13=0
Step 1: Find the distance of the two points from the plane.
Distance of point (1,1,p) from the plane = r⋅(3,4,12)+13/√(3^2+4^2+(12)^2)
Distance of point (−3,0,1) from the plane = r⋅(3,0,12)+13/√(3^2+4^2+(12)^2)
Step 2: Set the two distances equal to each other and solve for p.
r⋅(3,4,12)+13/√(3^2+4^2+(12)^2) = r⋅(3,0,12)+13/√(3^2+4^2+(12)^2)
r⋅(3,4,12)+13 = r⋅(3,0,12)+13
3r+4(12r)+13 = 3r+13
12r = 0
r = 0
Substituting the value of r in the equation for point (1,1,p),
0⋅(3,4,12)+13/√(3^2+4^2+(12)^2) = 13/√(3^2+4^2+(12)^2)
Therefore, p = 13/√(3^2+4^2+(12)^2)
Question:
If the coordinates of the points A,B,C,D be (1,2,3),(4,5,7),(−4,3,−6) and (2,9,2) respectively, then find the angle between the lines AB and CD
Answer:
Step 1: Calculate the vector AB and CD.
AB = (41, 52, 73) = (3, 3, 4)
CD = (2(4), 93, 2(6)) = (6, 6, 8)
Step 2: Calculate the magnitude of the vectors.
AB = √(3^2 + 3^2 + 4^2) = √34
CD = √(6^2 + 6^2 + 8^2) = √100
Step 3: Calculate the dot product of the vectors.
AB.CD = (36) + (36) + (4*8) = 84
Step 4: Calculate the angle between AB and CD.
θ = cos1(AB.CD/ABCD)
θ = cos1(84/√34*√100)
θ = cos1(84/√3400)
θ = cos1(0.24)
θ = 73.9°
Question:
If the lines x−1/−3=y−2/2k=2z−3 and 3kx−1=1y−1=−5z−6 are perpendicular, find the value of k
Answer:
Step 1: Write the two equations in slopeintercept form.
x−1/−3=y−2/2k => y = 3x  1 + 2k
3kx−1=1y−1=−5z−6 => y = 1x + 1  5z + 6
Step 2: Set the slopes of the two equations equal to each other and solve for k.
3x  1 + 2k = 1x + 1  5z + 6 => 4x + 2k = 5z + 7 => 2k = 4x + 5z  7 => k = 2x + 2.5z  3.5
Question:
Find the coordinates of the point where the line through (5,1,6) and (3,4,1) crosses the ZXplane
Answer:
Answer: Step 1: Find the slope of the line using the coordinates of the given points. Slope = (41)/(35) = 3/2
Step 2: Find the yintercept of the line. yintercept = 1  (5*(3/2)) = 7/2
Step 3: Substitute x=0 in the equation of the line to find the corresponding ycoordinate. y = (7/2)  (3/2)*0 = 7/2
Step 4: The coordinates of the point where the line crosses the ZXplane is (0,7/2,0).
Question:
If O be the origin and the coordinates of P be (1,2,−3), then find the equation of the plane passing through P and perpendicular to OP.
Answer:
Answer: Step 1: Find the vector equation of the line OP:
OP = (1,2,3)  (0,0,0) = (1,2,3)
Step 2: Find the direction vector of the line OP:
OP = (1,2,3)
Step 3: Find the normal vector of the plane perpendicular to OP:
Normal vector of the plane = (1,2,3)
Step 4: Find the equation of the plane:
Equation of the plane = (1,2,3) . (x1, y2, z+3) = 0
Hence, the equation of the plane passing through P and perpendicular to OP is (1,2,3).(x1, y2, z+3) = 0.
Question:
Find the equation of the plane which contains the line of intersection of the planes.r⋅(i^+2j^+3k^)−4=0,r⋅(2i^+j^−k^)+5=0 and which is perpendicular to the plane r⋅(5i^+3j^−6k^)+8=0.
Answer:
 Find the vector equation of the line of intersection of the planes.
The vector equation of the line of intersection of the planes is given by: r = (i + 2j + 3k)t  (2i + j  k)s
 Find the normal vector of the plane which contains the line of intersection.
The normal vector of the plane which contains the line of intersection is given by: n = (i + 2j + 3k) × (2i + j  k) = (2j  3k + i)  (3i + 2j  k) = 4i + 5j + 4k
 Find the equation of the plane which contains the line of intersection and is perpendicular to the plane r⋅(5i^+3j^−6k^)+8=0.
The equation of the plane which contains the line of intersection and is perpendicular to the plane r⋅(5i^+3j^−6k^)+8=0 is given by: n⋅r = 4i + 5j + 4k⋅r = 0 or, 4r⋅i + 5r⋅j + 4r⋅k = 0
Question:
Find the vector equation of the line passing through (1,2,3) and parallel to the planes r.(i^−j^+2k^)=5 and r.(3i^+j^+k^)=6.
Answer:
Step 1: Find the direction vector of the line passing through (1,2,3).
This can be done by finding the vector from (1,2,3) to any other point on the line. Let’s take (3,4,5) as the other point.
Therefore, the direction vector of the line is (31, 42, 53) = (2,2,2).
Step 2: Find the normal vectors of the planes r.(i^−j^+2k^)=5 and r.(3i^+j^+k^)=6.
The normal vector of the first plane is (1,1,2).
The normal vector of the second plane is (3,1,1).
Step 3: Check if the direction vector of the line is parallel to the normal vectors of the planes.
The direction vector of the line is (2,2,2).
The normal vector of the first plane is (1,1,2).
The normal vector of the second plane is (3,1,1).
The dot product of the direction vector and the normal vectors of the planes is (21) + (2(1)) + (22) = 4 and (23) + (21) + (21) = 8, respectively.
Since the dot product of the direction vector and the normal vectors of the planes is not 0, the direction vector is parallel to the normal vectors of the planes.
Step 4: Find the vector equation of the line.
The vector equation of the line passing through (1,2,3) and parallel to the planes r.(i^−j^+2k^)=5 and r.(3i^+j^+k^)=6 is
r = (1,2,3) + t(2,2,2)
Question:
Find the equation plane passing (a,b,c) and parallel to the plane r⋅(i^+j^+k^)=2.
Answer:
Answer:

Find the normal vector of the given plane: r⋅(i^+j^+k^)=2 r⋅(i^+j^+k^)2=0 r⋅(i^+j^+k^)2=0 => r=(2,2,2)

Find the vector equation of the line passing through the point (a,b,c): v= (a,b,c) + λ(2,2,2)

Find the equation of the plane passing through the point (a,b,c) and parallel to the given plane: ax+by+cz=d 2a+2b+2c=d => d=2a+2b+2c
Question:
Find the equation of the plane passing through the line of intersection of the planes r.(i^+j^+k^)=1 and r.(2i^+3j^−k^)+4=0 and parallel to xaxis.
Answer:
 Find the normal vector to the line of intersection:
The normal vector is given by the cross product of the two plane normals:
n = (i^ + j^ + k^) x (2i^ + 3j^  k^) = (1, 3, 2)
 Find the normal vector to the plane parallel to the xaxis:
The normal vector to the plane parallel to the xaxis is (1, 0, 0).
 Find the equation of the plane:
The equation of the plane is given by the dot product of the normal vectors:
(1, 3, 2) . (1, 0, 0) = 1
Therefore, the equation of the plane passing through the line of intersection of the planes r.(i^+j^+k^)=1 and r.(2i^+3j^−k^)+4=0 and parallel to the xaxis is:
1 = 0
Question:
Find the equation of the plane passing through the point (−1,3,2) and perpendicular to each of the planes x+2y+3z=5 and 3x+3y+z=0.
Answer:
Step 1: Determine the normal vector of the given planes.
Normal vector of the plane x+2y+3z=5: (1,2,3)
Normal vector of the plane 3x+3y+z=0: (3,3,1)
Step 2: Find the cross product of the two normal vectors.
Cross product = (1,2,3) x (3,3,1) = (5,6,3)
Step 3: Use the pointnormal form to find the equation of the plane.
Pointnormal form: ax + by + cz = d
Let (a,b,c) be the normal vector of the desired plane, and (x₀,y₀,z₀) be the given point.
a = 5, b = 6, c = 3, x₀ = 1, y₀ = 3, z₀ = 2
Substitute these values in the pointnormal form:
(5)(1) + (6)(3) + (3)(2) = d
d = 1
Therefore, the equation of the plane is:
5x + 6y  3z = 1
Question:
Find the angle between the lines whose direction ratios are a,b,c and b−c,c−a,a−b
Answer:
 First calculate the direction cosines of the two lines,
Let the direction ratios of the first line be a, b and c and for the second line be b  c, c  a and a  b.
Therefore, the direction cosines of the first line are a/√(a2 + b2 + c2), b/√(a2 + b2 + c2) and c/√(a2 + b2 + c2).
Similarly, the direction cosines of the second line are (b  c)/√((b  c)2 + (c  a)2 + (a  b)2), (c  a)/√((b  c)2 + (c  a)2 + (a  b)2) and (a  b)/√((b  c)2 + (c  a)2 + (a  b)2).
 Now, use the dot product formula to find the angle between the two lines.
The dot product formula is given by,
cos θ = (a1b1 + a2b2 + a3b3)/(√(a12 + a22 + a32) * √(b12 + b22 + b32))
Therefore, the angle between the two lines is given by,
θ = cos1[(a/√(a2 + b2 + c2) * (b  c)/√((b  c)2 + (c  a)2 + (a  b)2) + b/√(a2 + b2 + c2) * (c  a)/√((b  c)2 + (c  a)2 + (a  b)2) + c/√(a2 + b2 + c2) * (a  b)/√((b  c)2 + (c  a)2 + (a  b)2)]
Question:
Find the coordinates of the point where the line through (3,−4,−5) and (2,−3,1) crosses the plane 2x+y+z=7
Answer:
Step 1: Find the direction vector of the line.
The direction vector of the line is given by: v = (2  3, 3  (4), 1  (5)) = (−1, 1, 6)
Step 2: Find the equation of the line using the pointslope form.
The equation of the line is given by: l: (x, y, z) = (3, 4, 5) + t(−1, 1, 6)
Step 3: Substitute the equation of the line into the equation of the plane.
2x + y + z = 7 (3  t) + (1 + t) + (5 + 6t) = 7
Step 4: Solve the equation for t.
t + 1 + 6t = 7 7t = 6 t = 6/7
Step 5: Substitute the value of t into the equation of the line.
(x, y, z) = (3, 4, 5) + (6/7)(−1, 1, 6) (x, y, z) = (3, 4, 5) + (6/7, 6/7, 42/7) (x, y, z) = (3  (6/7), 4 + 6/7, 5 + 42/7) (x, y, z) = (87/7, 10/7, 37/7)
Therefore, the coordinates of the point where the line through (3,−4,−5) and (2,−3,1) crosses the plane 2x+y+z=7 are (87/7, 10/7, 37/7).
Question:
Distance between the two planes : 2x+3y+4z=4 and 4x+6y+8z=12 is A 2 units B 4 units C 8 units D 2/√29 units
Answer:
Given, 2x + 3y + 4z = 4 4x + 6y + 8z = 12
Subtracting equation 1 from equation 2, we get 2x + 3y + 4z = 4 2x  3y  4z = 4
Simplifying, we get 0 = 0
Hence, the distance between the two planes is 0.
Answer: A 2 units
Question:
Find the vector equation of the plane passing through (1,2,3) and perpendicular to the plane r
Answer:
Answer: Step 1: Find the normal vector of the plane r.
Let the equation of the plane r be ax + by + cz = d.
The normal vector of the plane is given by (a,b,c).
Step 2: Find the normal vector of the plane passing through (1,2,3) and perpendicular to the plane r.
The normal vector of the plane passing through (1,2,3) and perpendicular to the plane r is given by (a,b,c) × (1,2,3) = (b2c, ca, ab).
Step 3: Find the vector equation of the plane passing through (1,2,3) and perpendicular to the plane r.
The vector equation of the plane passing through (1,2,3) and perpendicular to the plane r is given by (x1,y2,z3) · (b2c, ca, ab) = 0.
Question:
.(i^+2j^−5k^)+9=0
Answer:
Step 1: (i^2 + 2j^5k^) + 9 = 0
Step 2: i^2 + 2j^5k^ = 9
Step 3: i^2 = 9  2j^5k^
Step 4: i^2 = 9  2j^5k^
Step 5: i^2 = 9  2j^5(k)
Step 6: i^2 = 9  2j^5k
Question:
Find the equation of a line parallel to xaxis and passing through the origin
Answer:
Step 1: Determine the slope of the line. Since the line is parallel to the xaxis, the slope of the line is 0.
Step 2: Determine the equation of the line. The equation of a line with a slope of 0 and passing through the origin is y = 0.