Question 1
A line makes angles 90°, 135° and 45° with the x-, y- and z-axes. Find its direction cosines.
Steps only
Direction cosines are the cosines of the three given angles.
cos90°=0, cos135°=−√2/2 and cos45°=√2/2.
Answer
(0, −1/√2, 1/√2).
CLASS 12 · MATHEMATICS PART II · CHAPTER 11
Three Dimensional Geometry
Worked solutions for direction cosines, lines in space, angles between lines and shortest distances between skew lines.
27 worked answersAll 2 exercise sets
What you will learn
- Find direction ratios and direction cosines from points or angles.
- Write vector and Cartesian equations of lines in space.
- Calculate angles between lines and shortest distances between skew lines.
27 worked answers · short steps first, explanations when you need them
Exercise 11.1
Textbook pages 381
Question 2
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Steps only
Let all three direction cosines be l.
They satisfy l²+l²+l²=1, so l=±1/√3.
Answer
The direction cosines are (1/√3,1/√3,1/√3) or (−1/√3,−1/√3,−1/√3), according to the chosen direction.
Question 3
Find the direction cosines of a line with direction ratios −18, 12, −4.
Steps only
The length of the direction-ratio vector is √(324+144+16)=22.
Divide each direction ratio by 22.
Answer
(−9/11, 6/11, −2/11).
Question 4
Show that (2,3,4), (−1,−2,1) and (5,8,7) are collinear.
Steps only
From the first point, the displacement to the second is (−3,−5,−3).
The displacement to the third is (3,5,3), which is −1 times the first displacement.
Answer
The three points are collinear because the two displacement vectors are parallel.
Question 5
Find the direction cosines of the sides AB, BC and CA of the triangle with A=(3,5,−4), B=(−1,1,2), C=(−5,−5,−2).
Steps only
AB=B−A=(−4,−4,6), length 2√17.
BC=C−B=(−4,−6,−4), length 2√17.
CA=A−C=(8,10,−2), length 2√42. Divide each vector by its length.
Answer
AB: (−2/√17,−2/√17,3/√17); BC: (−2/√17,−3/√17,−2/√17); CA: (4/√42,5/√42,−1/√42).
Exercise 11.2
Textbook pages 389–390
Question 1
Show that the three lines with direction cosines (12,−3,−4)/13, (4,12,3)/13 and (3,−4,12)/13 are mutually perpendicular.
Steps only
Dot the first and second triples: 48−36−12=0.
Dot the first and third: 36+12−48=0.
Dot the second and third: 12−48+36=0.
Answer
Every pair has dot product zero, so all three pairs of lines are perpendicular.
Question 2
Show that the line through (1,−1,2),(3,4,−2) is perpendicular to the line through (0,3,2),(3,5,6).
Steps only
Direction vectors are (2,5,−4) and (3,2,4).
Their dot product is 2×3+5×2+(−4)×4=6+10−16=0.
Answer
The lines are perpendicular.
Question 3
Show that the line through (4,7,8),(2,3,4) is parallel to the line through (−1,−2,1),(1,2,5).
Steps only
Direction vectors are (−2,−4,−4) and (2,4,4).
The second is −1 times the first.
Answer
The lines are parallel.
Question 4
Find the equation of the line through (1,2,3) parallel to 3i+2j−2k.
Steps only
Use point (1,2,3) and direction (3,2,−2).
The parameter form is (x,y,z)=(1,2,3)+t(3,2,−2).
Answer
Vector form: r=i+2j+3k+t(3i+2j−2k). Cartesian form: (x−1)/3=(y−2)/2=(z−3)/(−2).
Question 5
Find vector and Cartesian equations of the line through position vector 2i−j+4k in direction i+2j−k.
Steps only
The point is (2,−1,4) and direction ratios are (1,2,−1).
Substitute these into the point-direction forms.
Answer
r=2i−j+4k+t(i+2j−k); (x−2)/1=(y+1)/2=(z−4)/(−1).
Question 6
Find the Cartesian equation of the line through (−2,4,−5) parallel to (x+3)/3=(y−4)/5=(z+8)/6.
Steps only
The given line has direction ratios (3,5,6).
Use the new point (−2,4,−5) with the same direction ratios.
Answer
(x+2)/3=(y−4)/5=(z+5)/6.
Question 7
Write the vector form of (x−5)/3=(y+4)/7=(z−6)/2.
Steps only
The line passes through (5,−4,6) and has direction ratios (3,7,2).
Answer
r=5i−4j+6k+t(3i+7j+2k).
Question 8(i)
Find the angle between r=2i−5j+k+λ(3i+2j+6k) and r=7i−6k+μ(i+2j+2k).
Steps only
Direction vectors are (3,2,6) and (1,2,2).
Their dot product is 19; their lengths are 7 and 3.
Answer
θ=cos⁻¹(19/21).
Question 8(ii)
Find the angle between r=3i+j−2k+λ(i−j−2k) and r=2i−j−5k+μ(3i−5j−4k).
Steps only
Direction vectors are (1,−1,−2) and (3,−5,−4).
Their dot product is 16; their lengths are √6 and √50.
Answer
θ=cos⁻¹(8/(5√3)).
Question 9(i)
Find the angle between (x−2)/2=(y−1)/5=(z+3)/(−3) and (x+2)/(−1)=(y−4)/8=(z−5)/4.
Steps only
Direction vectors are (2,5,−3) and (−1,8,4).
The dot product is 26; the lengths are √38 and 9.
Answer
θ=cos⁻¹(26/(9√38)).
Question 9(ii)
Find the angle between x/2=y/2=z/1 and (x−5)/4=(y−2)/1=(z−3)/8.
Steps only
The direction vectors are (2,2,1) and (4,1,8).
Their dot product is 18; their lengths are 3 and 9.
Answer
θ=cos⁻¹(2/3).
Question 10
Find p so the lines (1−x)/3=(7y−14)/(2p)=(z−3)/2 and (7−7x)/(3p)=(y−5)/1=(6−z)/5 are perpendicular.
Steps only
Using each common parameter, direction vectors are (−3,2p/7,2) and (−3p/7,1,−5).
Set their dot product to zero: 9p/7+2p/7−10=0.
Answer
p=70/11.
Question 11
Show that (x−5)/7=(y+2)/(−5)=z/1 and x/1=y/2=z/3 are perpendicular.
Steps only
Their direction vectors are (7,−5,1) and (1,2,3).
The dot product is 7−10+3=0.
Answer
The lines are perpendicular.
Question 12
Find the shortest distance between r=(i+2j+k)+λ(i−j+k) and r=2i−j−k+μ(2i+j+2k).
Steps only
Take points a₁=(1,2,1), a₂=(2,−1,−1); directions b₁=(1,−1,1), b₂=(2,1,2).
b₁×b₂=(−3,0,3), whose length is 3√2.
(a₂−a₁)·(b₁×b₂)=(1,−3,−2)·(−3,0,3)=−9.
Answer
Distance=|−9|/(3√2)=3/√2.
Question 13
Find the shortest distance between (x+1)/7=(y+1)/(−6)=(z+1)/1 and (x−3)/1=(y−5)/(−2)=(z−7)/1.
Steps only
Points are a₁=(−1,−1,−1), a₂=(3,5,7); directions are b₁=(7,−6,1), b₂=(1,−2,1).
b₁×b₂=(−4,−6,−8), length=10√5.
a₂−a₁=(4,6,8); the absolute dot product with the cross product is 92.
Answer
Distance=92/(10√5)=46/(5√5).
Question 14
Find the shortest distance between r=(i+2j+3k)+λ(i−3j+2k) and r=4i+5j+6k+μ(2i+3j+k).
Steps only
Points are a₁=(1,2,3), a₂=(4,5,6); directions b₁=(1,−3,2), b₂=(2,3,1).
b₁×b₂=(−9,3,9), length=3√19.
a₂−a₁=(3,3,3); its dot product with the cross product is 9.
Answer
Distance=9/(3√19)=3/√19.
Question 15
Find the shortest distance between r=(1−t)i+(t−2)j+(3−2t)k and r=(s+1)i+(2s−1)j−(2s+1)k.
Steps only
At t=0 and s=0, points are a₁=(1,−2,3), a₂=(1,−1,−1); directions are b₁=(−1,1,−2), b₂=(1,2,−2).
b₁×b₂=(6,−4,−3), length=√61.
a₂−a₁=(0,1,−4), dot product=8.
Answer
The shortest distance is 8/√61.
Miscellaneous Exercise on Chapter 11
Textbook pages 390–391
Question 1
Find the angle between direction ratios (a,b,c) and (b−c,c−a,a−b).
Steps only
The dot product is a(b−c)+b(c−a)+c(a−b)=0.
Thus the direction vectors are perpendicular, provided the second is nonzero.
Answer
The angle is 90° when a,b,c are not all equal. If a=b=c, the second triple is (0,0,0), which is not a line direction, so the angle is undefined.
Question 2
Find the equation of a line parallel to the x-axis and passing through the origin.
Steps only
The x-axis direction vector is (1,0,0).
Start at the origin and move any real multiple of this vector.
Answer
Vector form r=t i. Cartesian form y=0, z=0 (with x free).
Question 3
If (x−1)/(−3)=(y−2)/(2k)=(z−3)/2 and (x−1)/(3k)=(y−1)/1=(z−6)/(−5) are perpendicular, find k.
Steps only
Direction vectors are (−3,2k,2) and (3k,1,−5).
Set their dot product to zero: −9k+2k−10=0.
Answer
k=−10/7.
Question 4
Find the shortest distance between r=6i+2j+2k+λ(i−2j+2k) and r=−4i−k+μ(3i−2j−2k).
Steps only
Points are a₁=(6,2,2), a₂=(−4,0,−1); directions b₁=(1,−2,2), b₂=(3,−2,−2).
b₁×b₂=(8,8,4), length=12.
a₂−a₁=(−10,−2,−3); its dot product with the cross product is −108.
Answer
The distance is 108/12=9.
Question 5
Find the vector equation of the line through (1,2,−4) perpendicular to both lines with directions (3,−16,7) and (3,8,−5).
Steps only
A direction perpendicular to both is their cross product: (3,−16,7)×(3,8,−5)=(24,36,72).
This is a nonzero multiple of (2,3,6).
Answer
r=i+2j−4k+t(2i+3j+6k).