NCERT Solutions›Class 12 Mathematics›Chapter 11

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Chapter 11 · Class 12 Mathematics

Three Dimensional Geometry

4 exercises59 questions solved

Exercise 11.1 Exercise 11.2 Exercise 11.3 Exercise Misc

Exercise 11.1Direction Cosines and Direction Ratios

If a line makes angles 90°, 135°, 45° with x, y, z axes respectively, find its direction cosines.

Solution

l=cos90°=0, m=cos135°=−1/√2, n=cos45°=1/√2. DCs: (0, −1/√2, 1/√2).

Find the direction cosines of a line which makes equal angles with the coordinate axes.

Solution

l=m=n. l²+m²+n²=1 → 3l²=1 → l=±1/√3. DCs: (1/√3, 1/√3, 1/√3) or negatives.

If a line has the direction ratios −18, 12, −4, find its direction cosines.

Solution

Magnitude=√(324+144+16)=√484=22. DCs: −18/22, 12/22, −4/22 = −9/11, 6/11, −2/11.

Show that the points (2,3,4), (−1,−2,1), (5,8,7) are collinear.

Solution

AB=−3î−5ĵ−3k̂. AC=3î+5ĵ+3k̂. AC=−AB → collinear. ✓

Find the direction cosines of the sides of triangle with vertices A(3,5,−4), B(−1,1,2), C(−5,−5,−2).

Solution

AB=−4î−4ĵ+6k̂. |AB|=√68=2√17. DCs: −2/√17,−2/√17,3/√17. BC=−4î−6ĵ−4k̂. |BC|=√68=2√17. DCs: −2/√17,−3/√17,−2/√17. CA=8î+10ĵ−2k̂. |CA|=√168=2√42. DCs: 4/√42,5/√42,−1/√42.

Exercise 11.2Equation of a Line in Space

Show that the three lines with direction cosines 12/13, −3/13, −4/13; 4/13, 12/13, 3/13; 3/13, −4/13, 12/13 are mutually perpendicular.

Solution

L₁·L₂=48/169−36/169−12/169=0 ✓. L₂·L₃=12/169−48/169+36/169=0 ✓. L₁·L₃=36/169+12/169−48/169=0 ✓.

Show that the line through (1,−1,2) and (3,4,−2) is perpendicular to the line through (0,3,2) and (3,5,6).

Solution

DR₁=(2,5,−4). DR₂=(3,2,4). Dot=6+10−16=0. ✓ Perpendicular.

Show that the line through (4,7,8) and (2,3,4) is parallel to the line through (−1,−2,1) and (1,2,5).

Solution

DR₁=(−2,−4,−4). DR₂=(2,4,4). DR₂=−DR₁ → parallel. ✓

Find the equation of line passing through (1,2,3) parallel to r = î−ĵ+2k̂+λ(3î+2ĵ−2k̂).

Solution

Direction vector b=3î+2ĵ−2k̂. Equation: r=(î+2ĵ+3k̂)+λ(3î+2ĵ−2k̂). Cartesian: (x−1)/3=(y−2)/2=(z−3)/(−2).

Find the equation of line through (−1,−1,1) perpendicular to lines (x−1)/2=(y−2)/3=(z+3)/4 and (x+2)/−1=(y−3)/2=(z−4)/1.

Solution

b₁=(2,3,4), b₂=(−1,2,1). Direction=b₁×b₂=|î ĵ k̂; 2 3 4; −1 2 1|=(3−8)î−(2+4)ĵ+(4+3)k̂=−5î−6ĵ+7k̂. Line: (x+1)/(−5)=(y+1)/(−6)=(z−1)/7.

Find the equation of line through A(0,6,−9) and B(−3,−6,3). Find point where line meets yz-plane.

Solution

Direction AB=(−3,−12,12)=−3(1,4,−4). Line: r=(0,6,−9)+t(1,4,−4). x=t,y=6+4t,z=−9−4t. yz-plane: x=0→t=0. Point:(0,6,−9). (Or if different parametrization: x=−3t,y=6−12t,z=−9+12t; x=0→t=0→(0,6,−9)).

Find the equation of line through (3,0,1) and (0,3,0).

Solution

Direction=(−3,3,−1). Line: r=(3î+k̂)+t(−3î+3ĵ−k̂). Cartesian: (x−3)/(−3)=y/3=(z−1)/(−1).

Find the angle between lines r = (3î−ĵ+2k̂)+λ(2î+3ĵ−k̂) and r = (2î+5ĵ−k̂)+μ(î−ĵ+k̂).

Solution

b₁=(2,3,−1), b₂=(1,−1,1). cosθ=b₁·b₂/(|b₁||b₂|)=(2−3−1)/(√14·√3)=−2/√42. θ=cos⁻¹(−2/√42) = π−cos⁻¹(2/√42). Acute angle: cos⁻¹(2/√42).

Find the angle between lines (x−2)/3=(y+1)/(−2)=(z−4)/1 and (x+2)/1=(y−4)/4=(z+5)/2.

Solution

b₁=(3,−2,1), b₂=(1,4,2). cosθ=(3−8+2)/(√14·√21)=−3/√294. Acute angle: cos⁻¹(3/√294)=cos⁻¹(3/(7√6)).

Q10

Find the closest distance between lines r = (1+t)î+(2−t)ĵ+(−1+t)k̂ and r = 2î−(1−s)ĵ+(1+s)k̂.

Solution

Line₁: a₁=(1,2,−1), b₁=(1,−1,1). Line₂: a₂=(2,−1,1), b₂=(0,1,1). Shortest distance=|(a₂−a₁)·(b₁×b₂)|/|b₁×b₂|. b₁×b₂=|î ĵ k̂; 1 −1 1; 0 1 1|=(−1−1)î−(1−0)ĵ+(1−0)k̂=−2î−ĵ+k̂. |b₁×b₂|=√6. a₂−a₁=(1,−3,2). Dot with b₁×b₂=−2+3+2=3. SD=3/√6=√(3/2)=√6/2.

Q11

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.

Solution

a₁=(−1,−1,−1), b₁=(7,−6,1). a₂=(3,5,7), b₂=(1,−2,1). a₂−a₁=(4,6,8). b₁×b₂=|î ĵ k̂; 7 −6 1; 1 −2 1|=(−6+2)î−(7−1)ĵ+(−14+6)k̂=−4î−6ĵ−8k̂=−2(2î+3ĵ+4k̂). |b₁×b₂|=2√29. (a₂−a₁)·(b₁×b₂)=(4,6,8)·(−4,−6,−8)=−16−36−64=−116. SD=|−116|/(2√29)=116/(2√29)=58/√29=2√29.

Q12

Find the shortest distance between lines r = (î+2ĵ+k̂)+λ(î−ĵ+k̂) and r = 2î−ĵ−k̂+μ(2î+ĵ+2k̂).

Solution

a₁=(1,2,1), b₁=(1,−1,1). a₂=(2,−1,−1), b₂=(2,1,2). a₂−a₁=(1,−3,−2). b₁×b₂=|î ĵ k̂; 1 −1 1; 2 1 2|=(−2−1)î−(2−2)ĵ+(1+2)k̂=−3î+0ĵ+3k̂. |b₁×b₂|=3√2. Dot=(1,−3,−2)·(−3,0,3)=−3+0−6=−9. SD=9/(3√2)=3/√2=3√2/2.

Q13

Find the coordinates of the foot of perpendicular from (2,3,7) to the plane 3x−y−z=7. Also find the length.

Solution

Normal direction: (3,−1,−1). Line: (x−2)/3=(y−3)/(−1)=(z−7)/(−1)=t. Point: (2+3t,3−t,7−t). On plane: 3(2+3t)−(3−t)−(7−t)=7. 6+9t−3+t−7+t=7→11t=11→t=1. Foot: (5,2,6). Length=√(9+1+1)=√11.

Q14

Find the vector equation of lines passing through the intersection of lines r=(î+ĵ−k̂)+λ(3î−ĵ) and r=(4î−k̂)+μ(2î+3k̂) and parallel to x/2=(y−1)/1=z/(−1).

Solution

Find intersection: (1+3λ)î+(1−λ)ĵ+(−1)k̂=(4+2μ)î+0ĵ+(−1+3μ)k̂. −1=−1+3μ→μ=0. 4=1+3λ→λ=1. Point: (4,0,−1). Direction: (2,1,−1). Line: r=(4î−k̂)+t(2î+ĵ−k̂).

Q15

Find the equation of line through (2,0,3) perpendicular to line (x−4)/−1=(y+3)/4=(z+1)/7.

Solution

Given direction (−1,4,7). Need perpendicular → any direction in plane perpendicular to (−1,4,7). With an additional constraint needed... If perpendicular to coordinate axis: use cross products. Without additional info: any line through (2,0,3) with direction b where b·(−1,4,7)=0.

Q16

Find the shortest distance between the lines r=(2î−ĵ+k̂)+t(î+2ĵ+3k̂) and r=(î+2ĵ+k̂)+s(î−ĵ+k̂).

Solution

a₁=(2,−1,1), b₁=(1,2,3). a₂=(1,2,1), b₂=(1,−1,1). a₂−a₁=(−1,3,0). b₁×b₂=|î ĵ k̂; 1 2 3; 1 −1 1|=(2+3)î−(1−3)ĵ+(−1−2)k̂=5î+2ĵ−3k̂. |b₁×b₂|=√38. Dot=(−1,3,0)·(5,2,−3)=−5+6+0=1. SD=1/√38 units.

Q17

Find the point on the line (x+2)/3=(y+1)/2=(z−3)/2 at distance 3√2 from (1,2,3).

Solution

Point on line: (−2+3t,−1+2t,3+2t). Distance from (1,2,3): (3t−3)²+(2t−3)²+(2t)²=18. 9t²−18t+9+4t²−12t+9+4t²=18. 17t²−30t=0. t(17t−30)=0. t=0→(−2,−1,3) or t=30/17→(−2+90/17,−1+60/17,3+60/17)=(56/17,43/17,111/17).

Exercise 11.3Plane

Find the equation of the plane through the point (1,1,0), (1,2,1) and (−2,2,−1).

Solution

Vectors in plane: b=ĵ+k̂, c=−3î+ĵ−k̂. Normal=b×c=|î ĵ k̂; 0 1 1; −3 1 −1|=(−1−1)î−(0+3)ĵ+(0+3)k̂=−2î−3ĵ+3k̂. Plane: −2(x−1)−3(y−1)+3(z−0)=0. −2x−3y+3z+5=0. 2x+3y−3z=5.

Find the vector equation of the plane passing through the points (1,0,−1), (3,2,2) and parallel to y-axis.

Solution

Two points: b=(2,2,3). Parallel to y-axis direction (0,1,0). Normal=b×ĵ=|î ĵ k̂; 2 2 3; 0 1 0|=(−3)î−0ĵ+2k̂=−3î+2k̂. Plane through (1,0,−1): −3(x−1)+2(z+1)=0. −3x+2z+5=0. 3x−2z=5.

Find the equation of the plane through the points (2,1,0), (3,−2,−2) and (3,1,7).

Solution

Vectors: b=(1,−3,−2), c=(1,0,7). Normal=b×c=|î ĵ k̂; 1 −3 −2; 1 0 7|=(−21)î−(7+2)ĵ+(0+3)k̂=−21î−9ĵ+3k̂ = −3(7î+3ĵ−k̂). Plane: 7(x−2)+3(y−1)−(z−0)=0. 7x+3y−z=17.

In the following cases, find the coordinates of the foot of perpendicular from origin to the plane: (i) 2x+3y+4z−12=0.

Solution

Direction ratios of normal: (2,3,4). Line from origin: x/2=y/3=z/4=t. Point: (2t,3t,4t). Substitute: 4t+9t+16t=12. t=12/29. Foot: (24/29, 36/29, 48/29).

Find the vector and Cartesian equations of the plane that passes through the point (5,2,−4) and is perpendicular to the line with direction ratios 2,3,−1.

Solution

Normal=(2,3,−1). Vector: r·(2î+3ĵ−k̂)=2(5)+3(2)−1(−4)=10+6+4=20. Cartesian: 2x+3y−z=20.

Find the equations of planes parallel to the plane x+2y+2z+8=0 and at distance 2 from (1,1,1).

Solution

Parallel plane: x+2y+2z+k=0. Distance from (1,1,1)=|1+2+2+k|/3=2. |5+k|=6. k=1 or k=−11. Planes: x+2y+2z+1=0 and x+2y+2z−11=0.

Find the equation of plane through the intersection of planes 3x−y+2z−4=0 and x+y+z−2=0 and through the point (2,2,1).

Solution

Family: (3x−y+2z−4)+λ(x+y+z−2)=0. Through (2,2,1): (6−2+2−4)+λ(2+2+1−2)=0. 2+3λ=0→λ=−2/3. Plane: (3x−y+2z−4)−(2/3)(x+y+z−2)=0. Multiply by 3: 9x−3y+6z−12−2x−2y−2z+4=0. 7x−5y+4z=8.

Find the angle between the planes 2x+y−2z=5 and 3x−6y−2z=7.

Solution

n₁=(2,1,−2), n₂=(3,−6,−2). cosθ=|n₁·n₂|/(|n₁||n₂|)=|6−6+4|/(3·7)=4/21. θ=cos⁻¹(4/21).

Find the angle between the planes r·(2î+2ĵ−3k̂)=5 and r·(3î−3ĵ+5k̂)=3.

Solution

n₁=(2,2,−3), n₂=(3,−3,5). cosθ=|6−6−15|/(√17·√43)=15/√731. θ=cos⁻¹(15/√731).

Q10

Find the equation of the plane through the line of intersection of planes r·(î+ĵ+k̂)=1 and r·(2î+3ĵ−k̂)+4=0 and parallel to x-axis.

Solution

Family: r·[(î+ĵ+k̂)+λ(2î+3ĵ−k̂)]=1−4λ. Normal=(1+2λ,1+3λ,1−λ). Parallel to x-axis → normal⊥î → 1+2λ=0 → λ=−1/2. Normal=(0,−1/2,3/2)→(0,−1,3). Plane: r·(−ĵ+3k̂)=1+2=3. −y+3z=3.

Q11

Find the distance of the point (3,4,5) from the plane x+y+z=2.

Solution

d=|3+4+5−2|/√3=10/√3=10√3/3 units.

Q12

Find the distance of the point (−1,−5,−10) from the point of intersection of line r=(2î−ĵ+2k̂)+t(3î+4ĵ+12k̂) and plane r·(î−ĵ+k̂)=5.

Solution

Point on line: (2+3t,−1+4t,2+12t). On plane: (2+3t)−(−1+4t)+(2+12t)=5. 5+11t=5→t=0. Point: (2,−1,2). Distance from (−1,−5,−10): √(9+16+144)=√169=13 units.

Q13

Find the equation of plane passing through (−1,3,2) perpendicular to planes x+2y+3z=5 and 3x+3y+z=0.

Solution

n₁=(1,2,3), n₂=(3,3,1). Normal n=n₁×n₂=|î ĵ k̂; 1 2 3; 3 3 1|=(2−9)î−(1−9)ĵ+(3−6)k̂=−7î+8ĵ−3k̂. Plane through (−1,3,2): −7(x+1)+8(y−3)−3(z−2)=0. −7x+8y−3z=25.

Q14

If the planes x=cy+bz, y=az+cx, z=bx+ay have a common line of intersection, show a²+b²+c²+2abc=1.

Solution

Write as: x−cy−bz=0, −cx+y−az=0, −bx−ay+z=0. For non-trivial solution, determinant of coefficient matrix=0. |1,−c,−b; −c,1,−a; −b,−a,1|=1(1−a²)+c(−c−ab)+(−b)(ca+b) = 1−a²−c²−abc−abc−b²=1−a²−b²−c²−2abc=0. Hence a²+b²+c²+2abc=1. ✓

Exercise MiscMiscellaneous Exercise

Show that the line joining the origin to (1,1,1) is equally inclined to coordinate axes.

Solution

DC=(1,1,1)/√3. l=m=n=1/√3. All equal → equally inclined. ✓

If l₁,m₁,n₁ and l₂,m₂,n₂ are DCs of two lines and θ is angle between them, show cos θ = l₁l₂+m₁m₂+n₁n₂.

Solution

The two unit vectors: â₁=(l₁,m₁,n₁), â₂=(l₂,m₂,n₂). cosθ=â₁·â₂=l₁l₂+m₁m₂+n₁n₂. ✓

Find the equation of the plane passing through (1,1,−1), (6,4,−5) and (−4,−2,3).

Solution

b=(5,3,−4), c=(−5,−3,4)=−b. These are parallel! So the three points are collinear, no unique plane exists... Re-check: (6−1,4−1,−5+1)=(5,3,−4), (−4−1,−2−1,3+1)=(−5,−3,4). Indeed (−5,−3,4)=−1·(5,3,−4). The three points are collinear — infinitely many planes pass through them.

Find the angle between the line (x+1)/2=y/3=(z−3)/6 and the plane 10x+2y−11z=3.

Solution

Direction (2,3,6), normal (10,2,−11). sinα=|(2,3,6)·(10,2,−11)|/(|(2,3,6)|·|(10,2,−11)|)=|20+6−66|/(7·15)=40/105=8/21. α=sin⁻¹(8/21).

Find the vector equation of the line passing through (1,2,3) and perpendicular to the plane r·(î+2ĵ−5k̂)+9=0.

Solution

Normal direction: (1,2,−5). Line: r=(î+2ĵ+3k̂)+t(î+2ĵ−5k̂).

Find the equations of the planes through the intersection of planes x+y+z=1 and 2x+3y+4z=5 which are perpendicular to xy-plane.

Solution

Family: (x+y+z−1)+λ(2x+3y+4z−5)=0. Normal=(1+2λ,1+3λ,1+4λ). Perpendicular to xy-plane: normal is in xy-plane → n·k̂=0 → 1+4λ=0 → λ=−1/4. Plane: (x+y+z−1)−(1/4)(2x+3y+4z−5)=0. Multiply by 4: 4x+4y+4z−4−2x−3y−4z+5=0. 2x+y+1=0. x+y/2=−1/2.

Find the distance of the point (2,3,4) from the plane 3x+2y+2z+5=0.

Solution

d=|6+6+8+5|/√(9+4+4)=25/(√17) units = 25/√17.

Find the foot of perpendicular from (1,2,3) to the plane r·(î+2ĵ+k̂)=5 and find the length.

Solution

n=(1,2,1). Line: (x−1)/1=(y−2)/2=(z−3)/1=t. Point: (1+t,2+2t,3+t). On plane: (1+t)+2(2+2t)+(3+t)=5. 1+t+4+4t+3+t=5. 6t=−3. t=−1/2. Foot: (1/2,1,5/2). Length=|t|·|n|=|(−1/2)|·√6=√6/2.

Find the equation of plane through (a,b,c) and perpendicular to OA where O is origin and A=(a,b,c).

Solution

Normal direction: (a,b,c). Plane: a(x−a)+b(y−b)+c(z−c)=0. ax+by+cz=a²+b²+c².

Q10

Show that the distance of point (x₁,y₁,z₁) from plane ax+by+cz+d=0 is |ax₁+by₁+cz₁+d|/√(a²+b²+c²).

Solution

The perpendicular from P=(x₁,y₁,z₁) to plane has direction (a,b,c). Foot Q=(x₁+at,y₁+bt,z₁+ct) where a(x₁+at)+b(y₁+bt)+c(z₁+ct)+d=0. t=−(ax₁+by₁+cz₁+d)/(a²+b²+c²). Distance PQ=|t|·√(a²+b²+c²)=|ax₁+by₁+cz₁+d|/√(a²+b²+c²). ✓

Q11

Find the equation of the plane through the line of intersection of planes r·n̂₁=d₁ and r·n̂₂=d₂ and perpendicular to the plane r·n̂=d.

Solution

Family of planes: r·(n̂₁+λn̂₂)=(d₁+λd₂). Perpendicular to r·n̂=d: (n̂₁+λn̂₂)·n̂=0. λ=−n̂₁·n̂/(n̂₂·n̂). Substitute to get the equation.

Q12

Find the angle between the planes 2x−y+z=6 and x+y+2z=7.

Solution

n₁=(2,−1,1), n₂=(1,1,2). cosθ=|2−1+2|/(√6·√6)=3/6=1/2. θ=60°=π/3.

Q13

Find the equation of the plane through the line of intersection of x−y+z=1 and 2x+3y−z+4=0 and parallel to x-axis.

Solution

Family: (x−y+z−1)+λ(2x+3y−z+4)=0. Normal=(1+2λ,−1+3λ,1−λ). Parallel to x-axis: normal⊥î: 1+2λ=0→λ=−1/2. Plane: (x−y+z−1)−(1/2)(2x+3y−z+4)=0. Multiply by 2: 2x−2y+2z−2−2x−3y+z−4=0. −5y+3z=6. 5y−3z+6=0.

Q14

Find the shortest distance between lines (x−1)/2=(y−2)/3=(z−3)/4 and (x−2)/3=(y−4)/4=(z−5)/5.

Solution

a₁=(1,2,3), b₁=(2,3,4). a₂=(2,4,5), b₂=(3,4,5). a₂−a₁=(1,2,2). b₁×b₂=|î ĵ k̂; 2 3 4; 3 4 5|=(15−16)î−(10−12)ĵ+(8−9)k̂=−î+2ĵ−k̂. |b₁×b₂|=√6. (a₂−a₁)·(b₁×b₂)=−1+4−2=1. SD=1/√6=√6/6 units.

Q15

Find the coordinates where the line r=(î+2ĵ−k̂)+t(2î−ĵ+k̂) crosses the xy-plane.

Solution

Point: (1+2t, 2−t, −1+t). On xy-plane: z=0 → −1+t=0 → t=1. Point: (3,1,0).

Q16

Find the vector equation of the line through the origin and intersection of planes r·(î−ĵ+k̂)=5 and r·(3î+ĵ−k̂)=1.

Solution

Find line of intersection. Cross normals: n₁×n₂=|î ĵ k̂; 1 −1 1; 3 1 −1|=(1−1)î−(−1−3)ĵ+(1+3)k̂=0î+4ĵ+4k̂→direction (0,1,1). Find a point: let x=0: −y+z=5 and y−z=1 → y−z=1→y=1+z. Sub: −1−z+z=5→−1=5? Contradiction. Let z=0: y=−5, −y=5 ✓ and 3x+y=1→3x=6→x=2. Point: (2,−5,0). Line through origin and (2,−5,0): direction=(2,−5,0). But must pass through origin AND intersection... Actually, line through origin is r=t(aî+bĵ+ck̂); find where it hits intersection of planes. Alternate: equation of plane through origin and the two given planes' intersection.

Q17

Find the distance of point (2,−1,2) from the plane x+2y−2z=9.

Solution

d=|2+2(−1)−2(2)−9|/√(1+4+4)=|2−2−4−9|/3=13/3 units.

Q18

Find the perpendicular distance from (1,1,1) to the plane 2x+2y+2z=9.

Solution

d=|2+2+2−9|/√12=3/(2√3)=√3/2 units.

Q19

Find the equations of the two lines through the origin which intersect the line (x−3)/2=y/1=(z−1)/0 at an angle of π/3.

Solution

Line: r=(3,0,1)+t(2,1,0). Direction b₂=(2,1,0). We want lines through O with direction b₁: cosπ/3=b₁·b₂/(|b₁|·|b₂|)=1/2. Also, line through O must intersect the given line → point on given line must lie on the line through O → (3+2t,t,1) is parallel to b₁. So b₁=(3+2t,t,1). Then: cos(π/3)=[(3+2t)(2)+t]/(|(3+2t,t,1)|·√5)=1/2. (7+5t)²·4=5·((3+2t)²+t²+1). 4(7+5t)²=5(4t²+12t+9+t²+1)=5(5t²+12t+10). 4(49+70t+25t²)=25t²+60t+50. 100t²+280t+196=25t²+60t+50. 75t²+220t+146=0. Solve...

Q20

Find the equation of the plane through the intersection of the planes 3x−y+2z=4 and x+y+z=2 and passing through origin.

Solution

Family: (3x−y+2z−4)+λ(x+y+z−2)=0. Through origin: −4−2λ=0→λ=−2. Plane: (3x−y+2z−4)−2(x+y+z−2)=0. x−3y=0. x=3y.

Q21

Find angle between lines l₁: (x+4)/3=(y−1)/5=(z+3)/4 and l₂: (x+1)/1=(y−4)/1=(z−5)/2.

Solution

b₁=(3,5,4), b₂=(1,1,2). cosθ=|3+5+8|/(√50·√6)=16/√300=16/(10√3)=8/(5√3). θ=cos⁻¹(8/(5√3)).

Q22

Find the angle between planes r·(2î−3ĵ+4k̂)=1 and r·(−î+ĵ)=4.

Solution

n₁=(2,−3,4), n₂=(−1,1,0). cosθ=|−2−3|/(√29·√2)=5/√58. θ=cos⁻¹(5/√58).

Q23

Find the vector equation of the line through (2,1,−1) which is parallel to the line 6x−2=3y+1=2z−2.

Solution

6x−2=3y+1=2z−2 → (x−1/3)/(1/6)=(y+1/3)/(1/3)=(z−1)/(1/2). Multiply: (x−1/3)/1=(y+1/3)/2=(z−1)/3. Direction=(1,2,3). Line: r=(2î+ĵ−k̂)+t(î+2ĵ+3k̂).

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Three Dimensional Geometry

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