# Mathematics Questions and Answers – Direction Cosines and Direction Ratios of a Line

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Direction Cosines and Direction Ratios of a Line”.

1. If a, b, c are the direction ratios of the line and l, m, n are the direction cosines of the line, then which of the following is true?
a) $$\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=μ$$
b) $$\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=μ-1$$
c) $$\frac{l}{a}=\frac{m}{c}=\frac{n}{b}=μ$$
d) $$\frac{l}{a}=\frac{n+1}{b}=\frac{n}{c}=μ$$

Explanation: For a given line, if a, b, c are the direction ratios and l, m, n are the direction cosines of the line then
a=μl, b=μm, c=μn
Or we can say that,
$$\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=μ$$, where μ is a constant.

2. If a line makes an angle of 120°, 45°, 30° with the positive x, y, z-axis respectively then find the direction cosines.
a) l=$$\frac{1}{2}, \,m=\frac{1}{\sqrt{2}}, \,n=\frac{\sqrt{3}}{2}$$
b) l=-$$\frac{1}{2}, \,m=-\frac{1}{\sqrt{2}}, \,n=-\frac{\sqrt{3}}{2}$$
c) l=-$$\frac{1}{2}, \,m=\frac{1}{\sqrt{2}}, \,n=\frac{\sqrt{3}}{2}$$
d) l=$$0, \,m=\frac{1}{\sqrt{2}}, \,n=\frac{\sqrt{3}}{2}$$

Explanation: Let l, m, n be the direction cosines of the line.
We know that, if α, β, γ are the angles that the line makes with the x, y, z- axis respectively, then
l=cos⁡α
m=cos⁡β
n=cos⁡γ
∴l=cos⁡120°, m=cos⁡45°, n=cos⁡30°
Hence, $$l=-\frac{1}{2}, \,m=\frac{1}{\sqrt{2}}, \,n=\frac{\sqrt{3}}{2}$$

3. If a line has direction ratios 2, -3, 7 then find the direction cosines.
a) l=$$\frac{2}{\sqrt{62}},m=-\frac{7}{\sqrt{62}},n=\frac{7}{\sqrt{62}}$$
b) l=$$\frac{2}{\sqrt{6}},m=-\frac{3}{\sqrt{6}},n=\frac{7}{\sqrt{6}}$$
c) l=-$$\frac{2}{\sqrt{62}},m=-\frac{3}{\sqrt{62}},n=-\frac{7}{\sqrt{62}}$$
d) l=$$\frac{2}{\sqrt{62}},m=-\frac{3}{\sqrt{62}},n=\frac{7}{\sqrt{62}}$$

Explanation: For a given line, if a, b, c are the direction ratios and l, m, n are the direction cosines of the line then
l=±$$\frac{a}{\sqrt{a^2+b^2+c^2}}$$
m=±$$\frac{b}{\sqrt{a^2+b^2+c^2}}$$
n=±$$\frac{c}{\sqrt{a^2+b^2+c^2}}$$
∴l=$$\frac{2}{\sqrt{2^2+(-3)^2+7^2}}, \,m=-\frac{3}{\sqrt{2^2+(-3)^2+7^2}}, \,n=\frac{7}{\sqrt{2^2+(-3)^2+7^2}}$$
Hence, l=$$\frac{2}{\sqrt{62}}, \,m=-\frac{3}{\sqrt{62}}, \,n=\frac{7}{\sqrt{62}}$$.
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4. Find the direction cosines of the line passing through two points (4, -5, -6) and (-1, 2, 8).
a) $$\frac{5}{270},\frac{7}{\sqrt{270}},\frac{14}{\sqrt{270}}$$
b) –$$\frac{7}{\sqrt{270}}, \frac{7}{\sqrt{270}},\frac{7}{\sqrt{270}}$$
c) –$$\frac{5}{\sqrt{270}}, \frac{7}{\sqrt{270}},\frac{14}{\sqrt{270}}$$
d) –$$\frac{5}{\sqrt{20}}, \frac{7}{\sqrt{720}},\frac{14}{\sqrt{270}}$$

Explanation: The direction cosines of two lines passing through two points is given by:
$$\frac{x_2-x_1}{PQ}, \frac{y_2-y_1}{PQ}, \frac{z_2-z_1}{PQ}$$
and $$PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$
In the given problem we have, P(4,-5,-6) and Q(-1,2,8)
∴$$PQ = \sqrt{(-1-4)^2+(2+5)^2+(8+6)^2}$$
$$=\sqrt{25+49+196}=\sqrt{270}$$
Hence, the direction ratios are $$l=\frac{(-1-4)}{\sqrt{270}}=-\frac{5}{\sqrt{270}}$$
$$m=\frac{(2+5)}{\sqrt{270}}=\frac{7}{\sqrt{270}}$$
$$n=\frac{(8+6)}{\sqrt{270}}=\frac{14}{\sqrt{270}}$$.

5. The direction ratios of the line segment joining $$P(x_1,y_1,z_1)$$ and $$Q(x_2,y_2,z_2)$$ is given by _______, ____________ and __________
a) $$x_2+x_1,y_2+y_1,z_2+z_1$$
b) $$x_2-x_1,y_2+y_1,z_2-z_1$$
c) $$x_2-x_1,y_2-y_1,z_2-z_1$$
d) $$x_2+x_1,y_2-y_1,z_2+z_1$$

Explanation: Let a.,b,c be the direction ratios of the line segment PQ.
Then, the direction ratios of the line segment joining $$P(x_1,y_1,z_1)$$ and $$Q(x_2,y_2,z_2)$$ is given by
$$a=x_2-x_1$$
$$b=y_2-y_1$$
$$c=z_2-z_1$$
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6. Find the direction cosines of the line passing through two points P(-6,7,3) and Q(3,-2,5).
a) –$$\frac{2}{\sqrt{166}},\frac{-9}{\sqrt{166}},\frac{2}{\sqrt{166}}$$
b) –$$\frac{9}{\sqrt{166}},\frac{-7}{\sqrt{166}},\frac{2}{\sqrt{166}}$$
c) –$$\frac{9}{\sqrt{66}},\frac{-9}{\sqrt{66}},\frac{2}{\sqrt{66}}$$
d) –$$\frac{9}{\sqrt{166}},\frac{-9}{\sqrt{166}},\frac{2}{\sqrt{166}}$$

Explanation: The direction cosines of two lines passing through two points is given by:
$$\frac{x_2-x_1}{PQ},\frac{y_2-y_1}{PQ},\frac{z_2-z_1}{PQ}$$
and $$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$
In the given problem we have, P(-6,7,3) and Q(3,-2,5)
∴$$PQ=\sqrt{(3+6)^2+(-2-7)^2+(5-3)^2}$$
=$$\sqrt{81+81+4}=\sqrt{166}$$
Hence, the direction ratios are $$l=\frac{-6-3}{\sqrt{166}}=-\frac{9}{\sqrt{166}}$$
m=$$\frac{-2-7}{\sqrt{166}}=\frac{-9}{\sqrt{166}}$$
n=$$\frac{5-3}{\sqrt{166}}=\frac{2}{\sqrt{166}}$$

7. If the direction ratios of a line are 5, 4, -7 respectively, then find the direction cosines.
a) $$\frac{75}{\sqrt{90}},\frac{4}{\sqrt{90}},\frac{5}{\sqrt{90}}$$
b) $$\frac{5}{\sqrt{90}},\frac{4}{\sqrt{90}},-\frac{7}{\sqrt{90}}$$
c) $$\frac{5}{\sqrt{70}},\frac{4}{\sqrt{70}},-\frac{7}{\sqrt{70}}$$
d) $$\frac{3}{\sqrt{90}},\frac{4}{\sqrt{90}},-\frac{5}{\sqrt{90}}$$

Explanation: If a,b,c are the direction ratios and l,m,n are the direction cosines respectively for a given line, then the direction cosines in terms of the direction ratios can be expressed as
l=±$$\frac{a}{\sqrt{a^2+b^2+c^2}}$$
m=±$$\frac{b}{\sqrt{a^2+b^2+c^2}}$$
n=±$$\frac{c}{\sqrt{a^2+b^2+c^2}}$$
Given that, a=5, b=4, c=-7
l=$$\frac{5}{\sqrt{(5^2+4^2+(-7)^2)}}=\frac{5}{\sqrt{(25+16+49)}}=\frac{5}{\sqrt{90}}$$
m=$$\frac{4}{\sqrt{(5^2+4^2+(-7)^2)}}=\frac{4}{\sqrt{90}}$$
n=-$$\frac{7}{\sqrt{(5^2+4^2+(-7)^2)}}=-\frac{7}{\sqrt{90}}$$

8. If a, b, c are the direction ratios of the line and l, m, n are the direction cosines of the line, then which of the following is incorrect?
a) $$\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=k$$
b) l2+m2+n2=1
c) k=±$$\frac{1}{\sqrt{(a^2+b^2+c^2)}}$$
d) l2-m2=n2-1

Explanation: Given that, a, b, c are the direction ratios of the line and l, m, n are the direction cosines of the line,
$$\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=k$$ and l2+m2+n2=1
⇒l=ak, m=bk, n=ck
(ak)2+(bk)2+(ck)2=1
k2 (a2+b2+c2)=1
k2=$$\frac{1}{a^2+b^2+c^2}$$
∴k=±$$\frac{1}{\sqrt{(a^2+b^2+c^2)}}$$
Hence, l2-m2=n2-1 is incorrect.

9. If the direction cosines of the line are $$\frac{1}{2},-\frac{\sqrt{3}}{2}$$,x respectively, then find the value of x.
a) 1
b) 0
c) $$\frac{\sqrt{3}}{2}$$
d) $$\frac{1}{2}$$

Explanation: If the direction cosines of a line are l,m,n respectively, then
l2+m2+n2=1
∴$$\frac{1}{2}^2+(\frac{\sqrt{3}}{2})^2+x^2=1$$
x2=$$1-\frac{1}{4}-\frac{3}{4}$$
x2=0
⇒x=0

10. If a line makes an angle of 60°, 150°, 45° with the positive x, y, z-axis respectively, find its direction cosines.
a) –$$\frac{1}{2},-\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}$$
b) –$$\frac{1}{2},-\frac{\sqrt{3}}{2},-\frac{1}{\sqrt{2}}$$
c) $$\frac{1}{2},-\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}$$
d) $$\frac{1}{2},\frac{\sqrt{3}}{2},-\frac{1}{\sqrt{2}}$$

Explanation: Let l, m, n be the direction cosines of the line.
We know that, if α, β, γ are the angles that the line makes with the x, y, z-axis respectively, then
l=cos⁡α=cos⁡60°=$$\frac{1}{2}$$
m=cos⁡β=cos⁡150°=-$$\frac{\sqrt{3}}{2}$$
n=cos⁡γ=cos⁡45°=$$\frac{1}{\sqrt{2}}$$.

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