Mathematics Questions and Answers – Three Dimensional Geometry – Shortest Distance between Two Lines

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This set of Mathematics Written Test Questions and Answers for Class 12 focuses 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=\(\left |\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}-\vec{b_2}|}\right |\)
View Answer

Answer: a
Explanation: The distance between two lines l1 and l2 with the equations
\(\vec{r}=\vec{a_1}+λ\vec{b_1}\)
\(\vec{r}=\vec{a_2}+μ\vec{b_2}\)
Then, the distance between the two lines is given by the formula
d=\(\left |\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}×\vec{b_2}|}\right |\)
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2. Find the shortest distance between two lines l1 and l2 whose vector equations is given below.
\(\vec{r}=3\hat{i}-4\hat{j}+2\hat{k}+λ(4\hat{i}+\hat{j}+\hat{k})\)
\(\vec{r}=5\hat{i}+\hat{j}-\hat{k}+μ(2\hat{i}-\hat{j}-3\hat{k})\)
a) \(\frac{11}{\sqrt{12}}\)
b) \(\frac{23}{\sqrt{10}}\)
c) \(\frac{18}{\sqrt{10}}\)
d) \(\frac{10}{\sqrt{11}}\)
View Answer

Answer: c
Explanation: The distance between two skew lines is given by
d=\(\left |\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}×\vec{b_2}|}\right |\)
\(\vec{r}=3\hat{i}-4\hat{j}+2\hat{k}+λ(4\hat{i}+\hat{j}+\hat{k})\)
\(\vec{r}=5\hat{i}+\hat{j}-\hat{k}+μ(2\hat{i}-\hat{j}-3\hat{k})\)
d=\(\left |\frac{((4\hat{i}+\hat{j}+\hat{k})×(2\hat{i}-\hat{j}-\hat{k})).((3\hat{i}-4\hat{j}+2\hat{k})-(2\hat{i}-\hat{j}-\hat{k}))}{|4\hat{i}+\hat{j}+\hat{k})×(2\hat{i}-\hat{j}-3\hat{k})|}\right |\)
\((4\hat{i}+\hat{j}+\hat{k})×(2\hat{i}-\hat{j}-\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4&1&1\\2&-1&-1\end{vmatrix}\)
=\(\hat{i}(-1+1)-\hat{j}(-4-2)+\hat{k}(-4-2)\)
=\(6\hat{j}-6\hat{k}\)
d=\(\left |{6\hat{j}-6\hat{k}).(\hat{i}-3\hat{j}+3\hat{k})}{|6\hat{i}-2\hat{k}|}\right |\)
\(\left|\frac{0-18-18}{\sqrt{6^2+2^2}}\right |=\frac{36}{\sqrt{40}}=\frac{18}{\sqrt{10}}\)

3. Find the equation between the two parallel lines l1 and l2 whose equations is given below.
\(\vec{r}=3\hat{i}+2\hat{j}-\hat{k}+λ(3\hat{i}-2\hat{j}+\hat{k})\)
\(\vec{r}=2\hat{i}-\hat{j}+\hat{k}+μ(3\hat{i}-2\hat{j}+\hat{k})\)
a) \(\sqrt{\frac{172}{14}}\)
b) \(\sqrt{\frac{145}{14}}\)
c) \(\sqrt{\frac{171}{14}}\)
d) \(\sqrt{\frac{171}{134}}\)
View Answer

Answer: c
Explanation: The distance between two parallel lines is given by
d=\(\left |\frac{\vec{b}×(\vec{a_2}-\vec{a_1})}{|\vec{b}|}\right |\)
=\(\left |\frac{((3\hat{i}-2\hat{j}+\hat{k})×((3\hat{i}+2\hat{j}-\hat{k})-(2\hat{i}-\hat{j}+\hat{k})))}{|\sqrt{3^2+(-2)^2+1^2}|}\right |\)
=\(\left |\frac{(3\hat{i}-2\hat{j}+\hat{k})×(\hat{i}+3\hat{j}-2\hat{k}))}{\sqrt{14}}\right |\)
\(
(3\hat{i}-2\hat{j}+\hat{k})×(\hat{i}+3\hat{j}-2\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-2&1\\1&3&-2\end{vmatrix}\)
=\(\hat{i}(4-3)-\hat{j}(-6-1)+\hat{k}(9+2)\)
=\(\hat{i}+7\hat{j}+11\hat{k}\)
∴d=\(\frac{|\hat{i}+7\hat{j}+11\hat{k}|}{\sqrt{14}}=\frac{\sqrt{1+49+121}}{\sqrt{14}}=\sqrt{\frac{171}{14}}\).
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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) \(\frac{\sqrt{115}}{134}\)
View Answer

Answer: a
Explanation: 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=\(\left |\frac{\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}}{\sqrt{(b_1 c_2-b_2 c_1)^2+(c_1 a_2-c_2 a_1)^2+(a_1 b_2-a_2 b_1)^2}}\right |\)
d=\(\left |\frac{\begin{vmatrix}-9&5&2\\2&5&4\\3&6&7\end{vmatrix}}{\sqrt{√(35-24)^2+(12-14)^2+(12-15)^2}}\right |\)
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=\(\left|\frac{\vec{b}×(\vec{a_2}-\vec{a_1})}{|\vec{b}|}\right |\)
View Answer

Answer: d
Explanation: If l1 and l2 are two parallel lines, then they are coplanar and hence can be represented by the following equations
\(\vec{r}=\vec{a_1}+λ\vec{b}\)
\(\vec{r}=\vec{a_2}+μ\vec{b}\)
Then the distance between the lines is given by
d=\(\left|\frac{\vec{b}×(\vec{a_2}-\vec{a_1})}{|\vec{b}|}\right |\)

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6. Find the shortest distance between the following set of parallel lines.
\(\vec{r}=6\hat{i}+2\hat{j}-\hat{k}+λ(\hat{i}+2\hat{j}-4\hat{k})\)
\(\vec{r}=\hat{i}+\hat{j}+\hat{k}+μ(\hat{i}+2\hat{j}-4\hat{k})\)
a) d=\(\sqrt{\frac{324}{45}}\)
b) d=\(\sqrt{\frac{405}{21}}\)
c) d=\(\sqrt{\frac{24}{21}}\)
d) d=\(\sqrt{\frac{21}{567}}\)
View Answer

Answer: b
Explanation: 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=\(\left|\frac{(\hat{i}+2\hat{j}-4\hat{k})×(6\hat{i}+2\hat{j}-\hat{k})-(\hat{i}+\hat{j}+\hat{k})}{\sqrt{1^2+2^2+(-4)^2}}\right |\)
=\(\left |\frac{(\hat{i}+2\hat{j}-4\hat{k})×(5\hat{i}+\hat{j}-2\hat{k})}{\sqrt{21}} \right |\)
\((\hat{i}+2\hat{j}-4\hat{k})×(5\hat{i}+\hat{j}-2\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}& \hat{k}\\1&2&-4\\5&1&-2\end{vmatrix}\)
=\(\hat{i}(-4+4)-\hat{j}(-2+20)+\hat{k}(1-10)\)
=-\(18\hat{j}-9\hat{k}\)
⇒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.
\(\vec{r}=2\hat{i}+2\hat{j}-2\hat{k}+λ(3\hat{i}+2\hat{j}+5\hat{k})\)
\(\vec{r}=4 \hat{i}-\hat{j}+5\hat{k}+μ(3\hat{i}-2\hat{j}+4\hat{k})\)
a) \(\frac{57}{\sqrt{47}}\)
b) \(\frac{57}{\sqrt{77}}\)
c) \(\frac{7}{\sqrt{477}}\)
d) \(\frac{57}{\sqrt{477}}\)
View Answer

Answer: d
Explanation: 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
\(\vec{r}=2\hat{i}+2\hat{j}-2\hat{k}+λ(3\hat{i}+2\hat{j}+5\hat{k})\)
\(\vec{r}=4 \hat{i}-\hat{j}+5\hat{k}+μ(3\hat{i}-2\hat{j}+4\hat{k})\)
∴d=\(\left|\frac{((3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k}).(4\hat{i}-\hat{j}+5\hat{k})-(2\hat{i}+2\hat{j}-2\hat{k}))}{|(3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k})|}\right |\)
\((3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&2&5\\3&-2&4\end{vmatrix}\)
=\(\hat{i}(8+10)-\hat{j}(12-15)+\hat{k}(-6-6)\)
=\(18\hat{i}+3\hat{j}-12\hat{k}\)
d=\(\left|\frac{(18\hat{i}+3\hat{j}-12\hat{k}).(2\hat{i}-3\hat{j}+7\hat{k})}{\sqrt{18^2+3^2+(-12)^2}}\right |\)
d=\(\left|\frac{36-9-84}{\sqrt{477}}\right |\)=\(\frac{57}{\sqrt{477}}\).
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8. Find the shortest distance between the set of parallel lines.
\(\vec{r}=(\hat{i}+2\hat{j}-\hat{k})+λ(\hat{i}+\hat{j}+\hat{k})\)
\(\vec{r}=(3\hat{i}-\hat{j}+3\hat{k})+μ(\hat{i}+\hat{j}+\hat{k})\)
a) \(\sqrt{34}\)
b) \(\sqrt{26}\)
c) 5
d) \(\sqrt{27}\)
View Answer

Answer: b
Explanation: 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=\(\left|\frac{(\hat{i}+\hat{j}+\hat{k})×((\hat{i}+2\hat{j}-\hat{k})-(3\hat{i}-\hat{j}+3\hat{k}))}{\sqrt{1^2+1^2+1^2}}\right |\)
=\(\left|\frac{\hat{i}+\hat{j}+\hat{k})×(-2\hat{i}+3\hat{j}-4\hat{k})}{√3}\right |\)
\((\hat{i}+\hat{j}+\hat{k})×(-2\hat{i}+3\hat{j}-4\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\-2&3&-4\end{vmatrix}\)
=-\(7\hat{i}+2\hat{j}+5\hat{k}\)
d=\(\frac{|-7\hat{i}+2\hat{j}+5\hat{k}|}{√3}\)
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) \(\frac{6}{\sqrt{14}}\)
View Answer

Answer: d
Explanation: The above equations can also be expressed as
\(\vec{r}=(\hat{i}-3\hat{j}+\hat{k})+p(-\hat{i}+\hat{j}+\hat{k})\)
\(\vec{r}=(-\hat{i}-3\hat{j}+\hat{k})+q(\hat{i}-2\hat{j}+\hat{k})\)
The distance between the two lines is given by
d=\(\left|\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}×\vec{b_2}|}\right |\)
=\(\left|\frac{((-\hat{i}+\hat{j}+\hat{k})×(\hat{i}-2\hat{j}+\hat{k})).(-\hat{i}-3\hat{j}+\hat{k})-(\hat{i}-3\hat{j}+\hat{k})}{|-\hat{i}+\hat{j}+\hat{k})×(\hat{i}-2\hat{j}+\hat{k}|}\right |\)
\((-\hat{i}+\hat{j}+\hat{k})×(\hat{i}-2\hat{j}+\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&1&1\\1&-2&1\end{vmatrix}\)
=\(\hat{i}(1+2)-\hat{j}(-1-1)+\hat{k}(2-1)\)
=\(3\hat{i}+2\hat{j}+\hat{k}\)
d=\(\left|\frac{(3\hat{i}+2\hat{j}+\hat{k}).(-2\hat{i})}{\sqrt{3^2+2^2+1^2}}\right |=\frac{6}{\sqrt{14}}\)
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10. Find the shortest distance between the lines whose equations are given below.
\(\vec{r}=(1-λ) \hat{i}+(1+2λ) \hat{j}+λ\hat{k}\)
\(\vec{r}=(\hat{i}-3\hat{j}-\hat{k})+μ(2\hat{i}+\hat{j}+2\hat{k})\)
a) \(\frac{11}{50}\)
b) \(\frac{21}{\sqrt{50}}\)
c) \(\frac{11}{\sqrt{50}}\)
d) \(\frac{51}{\sqrt{30}}\)
View Answer

Answer: c
Explanation: The equations can also be written as:
\(\vec{r}=(\hat{i}+\hat{j})+λ(-\hat{i}+2\hat{j}+\hat{k})\)
\(\vec{r}=(\hat{i}-3\hat{j}-\hat{k})+μ(2\hat{i}+\hat{j}+2\hat{k})\)
The distance of two skew lines is given by
d= \(\left|\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}×\vec{b_2}|}\right |\)
\(\vec{b_1}×\vec{b_2}=(-\hat{i}+2\hat{j}+\hat{k})×(2\hat{i}+\hat{j}+2\hat{k})\)
=\(\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&2&1\\2&1&2\end{vmatrix}\)
=\(\hat{i}(4-1)-\hat{j}(-2-2)+\hat{k}(-1-4)\)
=\(3\hat{i}+4\hat{j}-5\hat{k}\)
d=\(\left| \frac{(3\hat{i}+4\hat{j}-5\hat{k}).(-4\hat{j}-\hat{k})}{\sqrt{3^2+4^2+(-5)^2}}\right |\)
=\(\left |\frac{-16+5}{√50}\right |\)=\(\frac{11}{\sqrt{50}}\).

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