Mathematics Written Test Questions and Answers for Class 12 on “Three Dimensional Geometry – Shortest Distance between Two Lines”.
1. Which of the below given is the correct formula for the distance between two skew lines l1 and l2?
a) d=(left |frac{(vec{b_1}×vec{b_2}).(a_2-a_1)}{|vec{b_1}×vec{b_2}|}right |)
b) 2d=(left |frac{(vec{b_1}-vec{b_2}).(a_2-a_1)}{|vec{b_1}-vec{b_2}|}right |)
c) d=(left |frac{(vec{b_1}×vec{b_2}).(a_2.a_1)}{3|vec{b_1}×vec{b_2}|}right |)
d) d2=(left |frac{(vec{b_1}×vec{b_2}).(a_2-a_1)}{|vec{b_1}-vec{b_2}|}right |)
Answer: a
Clarification: The distance between two lines l1 and l2 with the equations
(vec{r}=vec{a_1}+λvec{b_1})
(vec{r}=vec{a_2}+μvec{b_2})
Then, the distance between the two lines is given by the formula
d=(left |frac{(vec{b_1}×vec{b_2}).(a_2-a_1)}{|vec{b_1}×vec{b_2}|}right |)
2. Find the shortest distance between two lines l1 and l2 whose vector equations is given below.
(vec{r}=3hat{i}-4hat{j}+2hat{k}+λ(4hat{i}+hat{j}+hat{k}))
(vec{r}=5hat{i}+hat{j}-hat{k}+μ(2hat{i}-hat{j}-3hat{k}))
a) (frac{11}{sqrt{12}})
b) (frac{23}{sqrt{10}})
c) (frac{18}{sqrt{10}})
d) (frac{10}{sqrt{11}})
Answer: c
Clarification: The distance between two skew lines is given by
d=(left |frac{(vec{b_1}×vec{b_2}).(a_2-a_1)}{|vec{b_1}×vec{b_2}|}right |)
(vec{r}=3hat{i}-4hat{j}+2hat{k}+λ(4hat{i}+hat{j}+hat{k}))
(vec{r}=5hat{i}+hat{j}-hat{k}+μ(2hat{i}-hat{j}-3hat{k}))
d=(left |frac{((4hat{i}+hat{j}+hat{k})×(2hat{i}-hat{j}-hat{k})).((3hat{i}-4hat{j}+2hat{k})-(2hat{i}-hat{j}-hat{k}))}{|4hat{i}+hat{j}+hat{k})×(2hat{i}-hat{j}-3hat{k})|}right |)
((4hat{i}+hat{j}+hat{k})×(2hat{i}-hat{j}-hat{k})=begin{vmatrix}hat{i}&hat{j}&hat{k}\4&1&1\2&-1&-1end{vmatrix})
=(hat{i}(-1+1)-hat{j}(-4-2)+hat{k}(-4-2))
=(6hat{j}-6hat{k})
d=(left |{6hat{j}-6hat{k}).(hat{i}-3hat{j}+3hat{k})}{|6hat{i}-2hat{k}|}right |)
(left|frac{0-18-18}{sqrt{6^2+2^2}}right |=frac{36}{sqrt{40}}=frac{18}{sqrt{10}})
3. Find the equation between the two parallel lines l1 and l2 whose equations is given below.
(vec{r}=3hat{i}+2hat{j}-hat{k}+λ(3hat{i}-2hat{j}+hat{k}))
(vec{r}=2hat{i}-hat{j}+hat{k}+μ(3hat{i}-2hat{j}+hat{k}))
a) (sqrt{frac{172}{14}})
b) (sqrt{frac{145}{14}})
c) (sqrt{frac{171}{14}})
d) (sqrt{frac{171}{134}})
Answer: c
Clarification: The distance between two parallel lines is given by
d=(left |frac{vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
=(left |frac{((3hat{i}-2hat{j}+hat{k})×((3hat{i}+2hat{j}-hat{k})-(2hat{i}-hat{j}+hat{k})))}{|sqrt{3^2+(-2)^2+1^2}|}right |)
=(left |frac{(3hat{i}-2hat{j}+hat{k})×(hat{i}+3hat{j}-2hat{k}))}{sqrt{14}}right |)
(
(3hat{i}-2hat{j}+hat{k})×(hat{i}+3hat{j}-2hat{k})=begin{vmatrix}hat{i}&hat{j}&hat{k}\3&-2&1\1&3&-2end{vmatrix})
=(hat{i}(4-3)-hat{j}(-6-1)+hat{k}(9+2))
=(hat{i}+7hat{j}+11hat{k})
∴d=(frac{|hat{i}+7hat{j}+11hat{k}|}{sqrt{14}}=frac{sqrt{1+49+121}}{sqrt{14}}=sqrt{frac{171}{14}}).
4. Find the shortest distance between the lines given.
l1:(frac{x-5}{2}=frac{y-2}{5}=frac{z-1}{4})
l2:(frac{x+4}{3}=frac{y-7}{6}=frac{z-3}{7})
a) (frac{115}{sqrt{134}})
b) (frac{115}{sqrt{184}})
c) (frac{115}{134})
d) (frac{sqrt{115}}{134})
Answer: a
Clarification: The shortest distance between two lines in cartesian form is given by:
l1:(frac{x-x_1}{a_1}=frac{y-y_1}{b_1}=frac{z-z_1}{c_1})
l2:(frac{x-x_2}{a_2}=frac{y-y_2}{b_2}=frac{z-z_2}{c_2})
∴d=(left |frac{begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\a_1&b_1&c_1\a_2&b_2&c_2end{vmatrix}}{sqrt{(b_1 c_2-b_2 c_1)^2+(c_1 a_2-c_2 a_1)^2+(a_1 b_2-a_2 b_1)^2}}right |)
d=(left |frac{begin{vmatrix}-9&5&2\2&5&4\3&6&7end{vmatrix}}{sqrt{√(35-24)^2+(12-14)^2+(12-15)^2}}right |)
d=(left |frac{-9(35-24)-5(14-12)+2(12-15)}{sqrt{11^2+2^2+3^2}}right |)
d=(left |frac{-99-10-6}{sqrt{134}}right |)
d=(frac{115}{sqrt{134}}).
5. Which of the following is the correct formula for the distance between the parallel lines l1 and l2?
a) d=(left|frac{vec{a_2}+vec{a_1})×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
b) d2=(left|frac{vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
c) 2d=(left|frac{vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
d) d=(left|frac{vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
Answer: d
Clarification: If l1 and l2 are two parallel lines, then they are coplanar and hence can be represented by the following equations
(vec{r}=vec{a_1}+λvec{b})
(vec{r}=vec{a_2}+μvec{b})
Then the distance between the lines is given by
d=(left|frac{vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
6. Find the shortest distance between the following set of parallel lines.
(vec{r}=6hat{i}+2hat{j}-hat{k}+λ(hat{i}+2hat{j}-4hat{k}))
(vec{r}=hat{i}+hat{j}+hat{k}+μ(hat{i}+2hat{j}-4hat{k}))
a) d=(sqrt{frac{324}{45}})
b) d=(sqrt{frac{405}{21}})
c) d=(sqrt{frac{24}{21}})
d) d=(sqrt{frac{21}{567}})
Answer: b
Clarification: The shortest distance between two parallel lines is given by:
d=(left |frac{vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
∴d=(left|frac{(hat{i}+2hat{j}-4hat{k})×(6hat{i}+2hat{j}-hat{k})-(hat{i}+hat{j}+hat{k})}{sqrt{1^2+2^2+(-4)^2}}right |)
=(left |frac{(hat{i}+2hat{j}-4hat{k})×(5hat{i}+hat{j}-2hat{k})}{sqrt{21}} right |)
((hat{i}+2hat{j}-4hat{k})×(5hat{i}+hat{j}-2hat{k})=begin{vmatrix}hat{i}&hat{j}& hat{k}\1&2&-4\5&1&-2end{vmatrix})
=(hat{i}(-4+4)-hat{j}(-2+20)+hat{k}(1-10))
=-(18hat{j}-9hat{k})
⇒d=(left|frac{sqrt{(-18)^2+(-9)^2}}{√21}right|)
d=(sqrt{frac{405}{21}})
7. Find the distance between the lines l1 and l2 with the following vector equations.
(vec{r}=2hat{i}+2hat{j}-2hat{k}+λ(3hat{i}+2hat{j}+5hat{k}))
(vec{r}=4 hat{i}-hat{j}+5hat{k}+μ(3hat{i}-2hat{j}+4hat{k}))
a) (frac{57}{sqrt{47}})
b) (frac{57}{sqrt{77}})
c) (frac{7}{sqrt{477}})
d) (frac{57}{sqrt{477}})
Answer: d
Clarification: We know that, the shortest distance between two skew lines is given by
d=(left |frac{(vec{b_1}×vec{b_2}).(a_2-a_1)}{|vec{b_1}×vec{b_2}|}right |)
The vector equations of the two lines is
(vec{r}=2hat{i}+2hat{j}-2hat{k}+λ(3hat{i}+2hat{j}+5hat{k}))
(vec{r}=4 hat{i}-hat{j}+5hat{k}+μ(3hat{i}-2hat{j}+4hat{k}))
∴d=(left|frac{((3hat{i}+2hat{j}+5hat{k})×(3hat{i}-2hat{j}+4hat{k}).(4hat{i}-hat{j}+5hat{k})-(2hat{i}+2hat{j}-2hat{k}))}{|(3hat{i}+2hat{j}+5hat{k})×(3hat{i}-2hat{j}+4hat{k})|}right |)
((3hat{i}+2hat{j}+5hat{k})×(3hat{i}-2hat{j}+4hat{k})=begin{vmatrix}hat{i}&hat{j}&hat{k}\3&2&5\3&-2&4end{vmatrix})
=(hat{i}(8+10)-hat{j}(12-15)+hat{k}(-6-6))
=(18hat{i}+3hat{j}-12hat{k})
d=(left|frac{(18hat{i}+3hat{j}-12hat{k}).(2hat{i}-3hat{j}+7hat{k})}{sqrt{18^2+3^2+(-12)^2}}right |)
d=(left|frac{36-9-84}{sqrt{477}}right |)=(frac{57}{sqrt{477}}).
8. Find the shortest distance between the set of parallel lines.
(vec{r}=(hat{i}+2hat{j}-hat{k})+λ(hat{i}+hat{j}+hat{k}))
(vec{r}=(3hat{i}-hat{j}+3hat{k})+μ(hat{i}+hat{j}+hat{k}))
a) (sqrt{34})
b) (sqrt{26})
c) 5
d) (sqrt{27})
Answer: b
Clarification: We know that, the distance between parallel lines is given by
d=(left|frac{(vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
d=(left|frac{(hat{i}+hat{j}+hat{k})×((hat{i}+2hat{j}-hat{k})-(3hat{i}-hat{j}+3hat{k}))}{sqrt{1^2+1^2+1^2}}right |)
=(left|frac{hat{i}+hat{j}+hat{k})×(-2hat{i}+3hat{j}-4hat{k})}{√3}right |)
((hat{i}+hat{j}+hat{k})×(-2hat{i}+3hat{j}-4hat{k})=begin{vmatrix}hat{i}&hat{j}&hat{k}\1&1&1\-2&3&-4end{vmatrix})
=-(7hat{i}+2hat{j}+5hat{k})
d=(frac{|-7hat{i}+2hat{j}+5hat{k}|}{√3})
d=(frac{sqrt{(-7)^2+2^2+5^2}}{√3}=sqrt{frac{78}{3}}=sqrt{26}).
9. Find the shortest distance between the lines given below.
(vec{r}=(1-p) hat{i}+(p-3) hat{j}+(1+p) hat{k})
(vec{r}=(q-1) hat{i}-(2q+3) hat{j}+(1+q)hat{k})
a) (frac{32}{sqrt{14}})
b) (frac{6}{sqrt{24}})
c) (frac{12}{sqrt{14}})
d) (frac{6}{sqrt{14}})
Answer: d
Clarification: The above equations can also be expressed as
(vec{r}=(hat{i}-3hat{j}+hat{k})+p(-hat{i}+hat{j}+hat{k}))
(vec{r}=(-hat{i}-3hat{j}+hat{k})+q(hat{i}-2hat{j}+hat{k}))
The distance between the two lines is given by
d=(left|frac{(vec{b_1}×vec{b_2}).(a_2-a_1)}{|vec{b_1}×vec{b_2}|}right |)
=(left|frac{((-hat{i}+hat{j}+hat{k})×(hat{i}-2hat{j}+hat{k})).(-hat{i}-3hat{j}+hat{k})-(hat{i}-3hat{j}+hat{k})}{|-hat{i}+hat{j}+hat{k})×(hat{i}-2hat{j}+hat{k}|}right |)
((-hat{i}+hat{j}+hat{k})×(hat{i}-2hat{j}+hat{k})=begin{vmatrix}hat{i}&hat{j}&hat{k}\-1&1&1\1&-2&1end{vmatrix})
=(hat{i}(1+2)-hat{j}(-1-1)+hat{k}(2-1))
=(3hat{i}+2hat{j}+hat{k})
d=(left|frac{(3hat{i}+2hat{j}+hat{k}).(-2hat{i})}{sqrt{3^2+2^2+1^2}}right |=frac{6}{sqrt{14}})
10. Find the shortest distance between the lines whose equations are given below.
(vec{r}=(1-λ) hat{i}+(1+2λ) hat{j}+λhat{k})
(vec{r}=(hat{i}-3hat{j}-hat{k})+μ(2hat{i}+hat{j}+2hat{k}))
a) (frac{11}{50})
b) (frac{21}{sqrt{50}})
c) (frac{11}{sqrt{50}})
d) (frac{51}{sqrt{30}})
Answer: c
Clarification: The equations can also be written as:
(vec{r}=(hat{i}+hat{j})+λ(-hat{i}+2hat{j}+hat{k}))
(vec{r}=(hat{i}-3hat{j}-hat{k})+μ(2hat{i}+hat{j}+2hat{k}))
The distance of two skew lines is given by
d= (left|frac{(vec{b_1}×vec{b_2}).(a_2-a_1)}{|vec{b_1}×vec{b_2}|}right |)
(vec{b_1}×vec{b_2}=(-hat{i}+2hat{j}+hat{k})×(2hat{i}+hat{j}+2hat{k}))
=(begin{vmatrix}hat{i}&hat{j}&hat{k}\-1&2&1\2&1&2end{vmatrix})
=(hat{i}(4-1)-hat{j}(-2-2)+hat{k}(-1-4))
=(3hat{i}+4hat{j}-5hat{k})
d=(left| frac{(3hat{i}+4hat{j}-5hat{k}).(-4hat{j}-hat{k})}{sqrt{3^2+4^2+(-5)^2}}right |)
=(left |frac{-16+5}{√50}right |)=(frac{11}{sqrt{50}}).
Mathematics Written Test Questions and Answers for Class 12,