Three Dimensional Geometry Exercise 02
Question:
Find the values of p so the line 1−x/3=7y−14/2p=z−3/2 and 7−7x/3p=y−5/1=6−z/5 are at right angles.
Answer:

Calculate the slope of the first line, m1: m1 = (1  x)/(7y  14/2p) = (1  x)/(7y  7p)

Calculate the slope of the second line, m2: m2 = (7  7x/3p)/(y  5/1) = (7  7x/3p)/(y  5)

Set m1 * m2 = 1: (1  x)/(7y  7p) * (7  7x/3p)/(y  5) = 1

Simplify and solve for p: 7  7x/3p = (1  x)(y  5)/(7y  7p) 7p  7x = (1  x)(y  5) 7p = 7x + (1  x)(y  5) p = x + (1  x)(y  5)/7
Question:
The cartesian equation of a line is x−5/3=y+4/7=z−6/2 write its vector form.
Answer:
Vector form: (x, y, z) = (5/3, 4/7, 6/2) + t(1, 1, 1)
Question:
The angle between the following pair of lines: x/2=y/2=z/1 and x−5/4=y−2/1=z−3/8 is θ=cos^−1a/3 then a=
Answer:
Let the two lines be L1 and L2.
L1: x/2 = y/2 = z/1
L2: x5/4 = y2/1 = z3/8
We can write the equations of the two lines in vector form as:
L1: r1 = (x, y, z)
L2: r2 = (x5/4, y2/1, z3/8)
The angle between the two lines is given by: θ = cos^1(a/3)
where a is the dot product of the two vectors:
a = r1•r2 = (x, y, z)•(x5/4, y2/1, z3/8)
a = x(x5/4) + y(y2/1) + z(z3/8)
a = x^2  5x/4 + y^2  2y + z^2  3z/8
a = (x^2 + y^2 + z^2)  (5x/4 + 2y + 3z/8)
a = (x^2 + y^2 + z^2)  (5x/4 + 2y + 3z/8)
a = (x^2 + y^2 + z^2)  (20x/8 + 16y/8 + 24z/8)
a = (x^2 + y^2 + z^2)  (5x + 2y + 3z)/8
Therefore, a = 5x + 2y + 3z/8
Question:
Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2i^−j^+4k^ and is in the direction i^+2j^−k^.
Answer:
Vector form: r(t) = 2i^−j^+4k^+t(i^+2j^−k^)
Cartesian form: x = 2 + t y = 1 + 2t z = 4  t
Question:
In the following case, determine the direction cosines of the normal to the plane and the distance from the origin. (i) z=2 (ii) x+y+z=1 (iii) 2x+3y−5=0 (iv) 5y+8=0
Answer:
(i) The direction cosines of the normal to the plane can be found using the equation \mathbf{n} = \frac{\mathbf{a}\times\mathbf{b}}{\mathbf{a}\times\mathbf{b}}, where \mathbf{a} and \mathbf{b} are two nonparallel vectors in the plane. In this case, \mathbf{a} = (1,1,1) and \mathbf{b} = (2,3,0). Therefore, the direction cosines of the normal to the plane are \mathbf{n} = \frac{(1,1,1)\times(2,3,0)}{(1,1,1)\times(2,3,0)} = \frac{(1,1,1)\times(2,3,0)}{\sqrt{13}} = \left(\frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}}, \frac{1}{\sqrt{13}}\right).
(ii) The distance from the origin to the plane can be found using the equation d = \mathbf{n}\cdot\mathbf{p_0}, where \mathbf{n} is the normal to the plane and \mathbf{p_0} is the position vector of the origin. In this case, \mathbf{p_0} = (0,0,0), so the distance from the origin is d = \mathbf{n}\cdot\mathbf{p_0} = \left(\frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}}, \frac{1}{\sqrt{13}}\right)\cdot(0,0,0) = 0.
Question:
Find the shortest distance between the lines r=(i^+2j^+k^)+λ(i^−j^+k^) and r=2i^−j^−k^+μ(2i^+j^+2k^)
Answer:
 Express the parametric equations of the two lines as vector equations:
r_1 = (i^+2j^+k^) + λ(i^−j^+k^)
r_1 = (i + 2j + k) + λ(i − j + k)
r_2 = (2i^−j^−k^) + μ(2i^+j^+2k^)
r_2 = (2i − j − k) + μ(2i + j + 2k)
 Find the vector equation of the line joining the two lines:
r_3 = r_1 + s(r_2  r_1)
r_3 = (i + 2j + k) + λ(i − j + k) + s[(2i − j − k) + μ(2i + j + 2k)  (i + 2j + k)  λ(i − j + k)]
 Simplify the expression for the vector equation of the line joining the two lines:
r_3 = (i + 2j + k) + λ(i − j + k) + s(2i − j − k) + sμ(2i + j + 2k)  sλ(i − j + k)
r_3 = (1 + sμ)i + (2 + λ + sμ)j + (1 + s + λ + sμ)k
 Find the direction cosines of the line joining the two lines:
𝐿 = (1 + sμ)^2 + (2 + λ + sμ)^2 + (1 + s + λ + sμ)^2
cos α = (1 + sμ)/𝐿
cos β = (2 + λ + sμ)/𝐿
cos γ = (1 + s + λ + sμ)/𝐿
 Find the shortest distance between the two lines:
d = r_1  r_2/√(cos^2α + cos^2β + cos^2γ)
d = (i + 2j + k) + λ(i − j + k)  (2i − j − k)  μ(2i + j + 2k)/√[(1 + sμ)^2 + (2 + λ + sμ)^2 + (1 + s + λ + sμ)^2]
Question:
Find the Cartesian equation of the line which passes through the point (−2,4,−5) and parallel to the line given by x+3/3=y−4/5=z+8/6
Answer:
Step 1: Find the direction vector of the given line.
The direction vector of the line x+3/3=y−4/5=z+8/6 is (3, 4, 8).
Step 2: Find the vector connecting the given point and the origin.
The vector connecting the point (2, 4, 5) and the origin is (2, 4, 5).
Step 3: Find the vector equation of the line.
The vector equation of the line is r = (2, 4, 5) + t (3, 4, 8).
Step 4: Find the Cartesian equation of the line.
The Cartesian equation of the line is x  2 = 3t, y  4 = 4t, and z + 5 = 8t.
Question:
Find the shortest distance between the lines whose vector equations are r =(i^+2j^+3k^)+λ(i^−3j^+2k^) and r=(4i^+5j^+6k^)+μ(2i^+3j^+k^).
Answer:
Step 1: Find the direction vectors of the two lines.
The direction vectors of the two lines are (i^−3j^+2k^) and (2i^+3j^+k^).
Step 2: Find the cross product of the two direction vectors.
The cross product of the two direction vectors is (6i^−9j^+5k^).
Step 3: Find the magnitude of the cross product.
The magnitude of the cross product is √(6^2+(9)^2+5^2) = √120.
Step 4: Divide the magnitude of the cross product by 2.
The answer is √120/2 = 10√3/2.
Therefore, the shortest distance between the two lines is 10√3/2.
Question:
x+y+z=1
Answer:
Step 1: Subtract ‘z’ from both sides of the equation. x + y = 1  z
Step 2: Subtract ‘y’ from both sides of the equation. x = 1  y  z
Question:
Find the angle between the following pairs of lines: r=3i^+j^−2k^+λ(i^+j^−2k^) and r=2i^−j^−56k^+μ(3i^−5j^−4k^).
Answer:
 Find the vector equation of the two lines.
r = 3i^ + j^  2k^ + λ(i^ + j^  2k^) r = 2i^  j^  56k^ + μ(3i^  5j^  4k^)

Find the cross product of the two vectors. Cross product = (3i^ + j^  2k^) x (3i^  5j^  4k^)

Calculate the magnitude of the cross product. Magnitude of the cross product = (3i^ + j^  2k^) x (3i^  5j^  4k^)

Calculate the dot product of the two vectors. Dot product = (3i^ + j^  2k^) . (3i^  5j^  4k^)

Calculate the angle between the two vectors using the formula: Angle = arccos(Dot product/Magnitude of the cross product)
Question:
Find the equation of the line which passes through the point (1,2,3) and is parallel to the vector (3i^+2j^−2k^.
Answer:
Answer: Step 1: Find the direction vector of the line passing through the point (1,2,3). Direction vector = (3i^+2j^−2k^)
Step 2: Find the pointslope form of the equation of the line. Pointslope form of the equation of the line = (x1, y2, z3) = t(3i^+2j^−2k^)
Step 3: Substitute the values of the point and direction vector in the pointslope form of the equation. Equation of the line: (x1, y2, z3) = t(3i^+2j^−2k^)
Step 4: Simplify the equation to get the final equation of the line. Final equation of the line: x = 1 + 3t, y = 2 + 2t, z = 3  2t
Question:
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin, 5y+8=0
Answer:
Step 1: The equation of the given plane is 5y + 8 = 0.
Step 2: To determine the direction cosines of the normal to the plane, we must solve the equation for y.
Step 3: We can do this by subtracting 8 from both sides of the equation, which gives us 5y = 8.
Step 4: We can then divide both sides by 5 to get y = 1.6.
Step 5: Since the direction cosines of the normal to the plane are (0, 1.6, 0), the distance from the origin is 8/1.6 = 5.
Question:
Find the vector and the Cartesian equations of the line that passes through the points (3,−2,−5),(3,−2,6)
Answer:
Vector equation: r = (3,−2,−5) + t(0,0,11)
Cartesian equation: x = 3 y = 2 z = 5 + 11t
Question:
Find the angle between the following pair of lines: (i) 2x−2=5y−1=−3z+3 and −1x+2=8y−4=4z−5 (ii) 2x=2y=1z and 4x−5=1y−2=8z−3
Answer:
(i) Step 1: Express the equations in the form of Ax+By+Cz=D For the first equation: 2x−2=5y−1=−3z+3 A=2, B=5, C=3, D=3 For the second equation: −1x+2=8y−4=4z−5 A=1, B=8, C=4, D=5
Step 2: Find the direction cosines of the two lines. Direction cosines of the first line: l₁ = (2/√14, 5/√14, 3/√14) Direction cosines of the second line: l₂ = (1/√17, 8/√17, 4/√17)
Step 3: Find the angle between the two lines using the dot product formula. Angle between the two lines = cos⁻¹(l₁.l₂) = cos⁻¹((2/√14)(1/√17)+(5/√14)(8/√17)+(3/√14)(4/√17)) = cos⁻¹(−2/14+40/14−12/14) = cos⁻¹(28/14) = cos⁻¹(2) = 60°
(ii) Step 1: Express the equations in the form of Ax+By+Cz=D For the first equation: 2x=2y=1z A=2, B=2, C=1, D=0 For the second equation: 4x−5=1y−2=8z−3 A=4, B=1, C=8, D=3
Step 2: Find the direction cosines of the two lines. Direction cosines of the first line: l₁ = (2/√5, 2/√5, 1/√5) Direction cosines of the second line: l₂ = (4/√17, 1/√17, 8/√17)
Step 3: Find the angle between the two lines using the dot product formula. Angle between the two lines = cos⁻¹(l₁.l₂) = cos⁻¹((2/√5)(4/√17)+(2/√5)(1/√17)+(1/√5)(8/√17)) = cos⁻¹(8/5+2/5+8/5) = cos⁻¹(18/5) = cos⁻¹(3.6) = 70.53°
Question:
Find the angle between the following pair of lines: (i) r=2i^−5j^+k^+λ(3i^−2j^+6k^) and r=7i^−6k^+μ(i^+2j^+2k^) (ii) r=3i^+j^−2k^+λ(i^−j^−2k^) and r=2i^−j^−56k^+μ(3i^−5j^−4k^)
Answer:
(i) Step 1: Find the direction vectors of both the lines.
Direction vector of the first line = (3i^−2j^+6k^) Direction vector of the second line = (i^+2j^+2k^)
Step 2: Find the dot product of the two direction vectors.
Dot product = (3i^−2j^+6k^).(i^+2j^+2k^) = 3i^2 + 2ij^−2j^2 + 6ik^+2i^2j^+4j^2k^+2k^2 = 5i^2+2ij^−4j^2+8ik^+4j^2k^+2k^2
Step 3: Find the magnitude of the two direction vectors.
Magnitude of the first vector = √(3i^2+2j^2+6k^2) = √(9+4+36) = √49 = 7
Magnitude of the second vector = √(i^2+4j^2+4k^2) = √(1+16+16) = √33 = 5.74
Step 4: Find the angle between the two lines using the formula:
cosθ = (Dot product of two vectors)/(Magnitude of first vector × Magnitude of second vector)
cosθ = (5i^2+2ij^−4j^2+8ik^+4j^2k^+2k^2)/(7×5.74)
cosθ = 0.908
Step 5: Find the angle using the inverse cosine function.
θ = cos^−1(0.908) = 25.8°
Hence, the angle between the two lines is 25.8°.
(ii) Step 1: Find the direction vectors of both the lines.
Direction vector of the first line = (i^−j^−2k^) Direction vector of the second line = (3i^−5j^−4k^)
Step 2: Find the dot product of the two direction vectors.
Dot product = (i^−j^−2k^).(3i^−5j^−4k^) = i^3−ij^2−2ik^+3i^2j^−5j^2−4jk^−2i^2k^+5j^2k^+8k^3 = i^3−ij^2−6ik^−5j^2−2jk^−2i^2k^+5j^2k^+8k^3
Step 3: Find the magnitude of the two direction vectors.
Magnitude of the first vector = √(i^2+j^2+4k^2) = √(1+1+16) = √18 = 4.24
Magnitude of the second vector = √(9i^2+25j^2+16k^2) = √(81+625+256) = √962 = 31.07
Step 4: Find the angle between the two lines using the formula:
cosθ = (Dot product of two vectors)/(Magnitude of first vector × Magnitude of second vector)
cosθ = (i^3−ij^2−6ik^−5j^2−2jk^−2i^2k^+5j^2k^+8k^3)/(4.24×31.07)
cosθ = 0.874
Step 5: Find the angle using the inverse cosine function.
θ = cos^−1(0.874) = 29.2°
Hence, the angle between the two lines is 29.2°.
Question:
Find the shortest distance between lines x+1/7=y+1/−6=z+1/1 and x−3/1=y−5/−2=z−7/1
Answer:

Find the slope of the two lines. The slope of the first line is 7/6 = 1.17. The slope of the second line is 1/2 = 0.5.

Find the equation of the line perpendicular to the two lines. The equation of the line perpendicular to the first line is y = 1.17x + b. The equation of the line perpendicular to the second line is y = 0.5x + c.

Find the point of intersection of the two lines. The point of intersection of the two lines is (x,y) = (2.5, 4.5).

Find the shortest distance between the two lines. The shortest distance between the two lines is the distance between the point of intersection and the closest point on each line. This distance can be calculated using the Pythagorean theorem: d = sqrt((x2  x1)^2 + (y2  y1)^2). In this case, the distance is d = sqrt((2.5  3)^2 + (4.5  5)^2) = sqrt(0.5^2 + 0.5^2) = sqrt(0.5). Therefore, the shortest distance between the two lines is 0.5.
Question:
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin, 2x+3y−z=5
Answer:

The direction cosines of the normal to the plane are (2, 3, 1). The distance from the origin is 5.

The direction cosines of the normal to the plane are (2, 3, 1). The distance from the origin is 5.
Question:
Find the vector and the Cartesian equations of the lines that pass through the origin and (5,−2,3)
Answer:
Vector: <5, 2, 3>
Cartesian Equations: x = 5t y = 2t z = 3t, where t is a real number
Question:
Find the shortest distance between the lines whose vector equations are :
r=(1−p)i^+(p−2)j^+(3−2p)k^
r=(q+1)i^+(2q−1)j^−(2q−1)k^
Answer:
Answer: Step 1: Find the direction vectors of the two lines.
Direction vector of the first line = (1−p)i^+(p−2)j^+(3−2p)k^ Direction vector of the second line = (q+1)i^+(2q−1)j^−(2q−1)k^
Step 2: Find the dot product of the two direction vectors.
Dot product = (1−p)(q+1) + (p−2)(2q−1) + (3−2p)(−2q−1)
Step 3: Simplify the dot product.
Dot product = (1−p)(q+1) + (p−2)(2q−1) + (3−2p)(−2q−1) = q−p+2pq−2q+2pq−2p+2pq−2q = −p+4pq−4q
Step 4: Set the dot product equal to zero.
−p+4pq−4q = 0
Step 5: Solve for q.
q = p/4 + 1
Step 6: Substitute the value of q in any one of the equations of the two lines.
Let’s substitute the value of q in the equation of the first line.
r=(1−p)i^+(p−2)j^+(3−2p)k^
r=(1−p)i^+(p−2)j^+(3−2p)k^ = (1−p)i^+[p−2+(p/4 + 1)]k^ = (1−p)i^+(5p/4−2)k^
Step 7: Find the shortest distance between the two lines.
The shortest distance between the two lines is the length of the vector (1−p)i^+(5p/4−2)k^.