7 Most FAQ’s of Direction Cosines and Direction Ratios Chapter in Inter 1st Year Maths-1B (TS/AP)

Given $$l + m + n = 0$$

$$\Rightarrow l = -(m + n) \quad \text{(1)}$$

$$l^2 + m^2 – n^2 = 0 \quad \text{(2)}$$

Solving (1) & (2) we get:

$$[-(m + n)]^2 + m^2 – n^2 = 0$$

$$\Rightarrow (m^2 + n^2 + 2mn) + m^2 – n^2 = 0$$

$$\Rightarrow 2m^2 + 2mn = 0$$

$$\Rightarrow 2(m^2 + mn) = 0$$

$$\Rightarrow m^2 + mn = 0$$

$$\Rightarrow m(m + n) = 0$$

$$\Rightarrow m = 0 \text{ or } m + n = 0$$

$$\Rightarrow m = 0 \text{ or } m = -n$$

Case (i): Put m = 0 in (1) then

$$l = -(0 + n) = -n$$

$$l = -n$$

Now, $$l : m : n = -n : 0 : n = -1 : 0 : 1$$

So, direction ratios of $$L1 = (a1, b1, c1) = (-1, 0, 1) \quad \text{(3)}$$

Case (ii): Put m = -n in (1) then

$$l = -(-n + n) = 0$$

$$l = 0$$

Now $$l : m : n = 0 : -n : n = 0 : -1 : 1$$

So, direction ratios of $$L2 = (a2, b2, c2) = (0, -1, 1) \quad \text{(4)}$$

If θ is the angle between the lines, then from (3), (4) we get

$$\cos \theta = \left| \frac{a1 \cdot a2 + b1 \cdot b2 + c1 \cdot c2}{\sqrt{(a1^2 + b1^2 + c1^2)(a2^2 + b2^2 + c2^2)}} \right|$$

$$= \left| \frac{(-1)(0) + (0)(-1) + (1)(1)}{\sqrt{((-1)^2 + 0^2 + 1^2)(0^2 + (-1)^2 + 1^2)}} \right|$$

$$= \frac{1}{\sqrt{2 \cdot 2}} = \frac{1}{\sqrt{4}} = \frac{1}{2} = \cos 60^\circ$$

$$\Rightarrow \theta = 60^\circ$$

Hence, the angle between the lines is $$60^\circ$$

Given $$3l + m + 5n = 0$$

$$\Rightarrow m = -3l – 5n \quad \text{(1)}$$

$$6mn – 2nl + 5lm = 0 \quad \text{(2)}$$

Solving (1) & (2) we get:

$$6n(-3l – 5n) – 2nl + 5l(-3l – 5n) = 0$$

$$\Rightarrow -18ln – 30n^2 – 2ln – 15l^2 – 25ln = 0$$

$$\Rightarrow -15l^2 – 45ln – 30n^2 = 0$$

$$\Rightarrow -15(l^2 + 3ln + 2n^2) = 0$$

$$\Rightarrow l^2 + 3ln + 2n^2 = 0$$

$$\Rightarrow l^2 + ln + 2ln + 2n^2 = 0$$

$$\Rightarrow l(l + n) + 2n(l + n) = 0$$

$$\Rightarrow (l + n)(l + 2n) = 0$$

$$\Rightarrow l = -n \text{ or } l = -2n$$

Case (i) : Put l = -n in (1) then

$$m = -3(-n) – 5n = 3n – 5n = -2n$$

$$m = -2n$$

Now $$l : m : n = -n : -2n : n$$

$$= -1 : -2 : 1 = 1 : 2 : -1$$

So, direction ratios of $$L1 = (a1, b1, c1) = (1, 2, -1)$$

Case (ii) : Put l = -2n in (1) then

$$m = -3(-2n) – 5n = 6n – 5n = n$$

$$m = n$$

Now $$l : m : n = -2n : n : n$$

$$= -2 : 1 : 1 = 2 : -1 : -1$$

So, direction ratios of $$L2 = (a2, b2, c2) = (2, -1, -1)$$

If θ is the angle between the lines, then from (3), (4) we get:

$$\cos \theta = \left| \frac{a1 \cdot a2 + b1 \cdot b2 + c1 \cdot c2}{\sqrt{(a1^2 + b1^2 + c1^2)(a2^2 + b2^2 + c2^2)}} \right|$$

$$= \left| \frac{1 \cdot 2 + 2 \cdot (-1) + (-1) \cdot (-1)}{\sqrt{(1^2 + 2^2 + (-1)^2)(2^2 + (-1)^2 + (-1)^2)}} \right|$$

$$= \left| 2 – 2 + 1 \right|/\sqrt{6 \cdot 6} = 1/\sqrt{36} = 1/6$$

$$\cos \theta = 1/6$$

$$\Rightarrow \theta = \cos^{-1}(1/6)$$

LAQ-3 : Find the direction cosines of two lines which are connected by the relations l – 5m + 3n = 0 and 7l2 + 5m2 – 3n2 = 0

Given $$l – 5m + 3n = 0$$

$$\Rightarrow l = 5m – 3n \quad \text{(1)}$$

$$7l^2 + 5m^2 – 3n^2 = 0 \quad \text{(2)}$$

Solving (1) & (2) we get:

$$7(5m – 3n)^2 + 5m^2 – 3n^2 = 0$$

$$\Rightarrow 7(25m^2 + 9n^2 – 30mn) + 5m^2 – 3n^2 = 0$$

$$\Rightarrow 175m^2 + 63n^2 – 210mn + 5m^2 – 3n^2 = 0$$

$$\Rightarrow 180m^2 – 210mn + 60n^2 = 0$$

$$\Rightarrow 30(6m^2 – 7mn + 2n^2) = 0$$

$$\Rightarrow 6m^2 – 7mn + 2n^2 = 0$$

$$\Rightarrow 2m(3m – 2n) – n(3n – 2n) = 0$$

$$\Rightarrow (3m – 2n)(2m – n) = 0$$

$$3m – 2n = 0$$

$$\Rightarrow 3m = 2n$$

$$\Rightarrow m = \frac{2n}{3} \text{ or } 2m – n = 0$$

$$\Rightarrow 2m = n$$

$$\Rightarrow m = \frac{n}{2}$$

Case (i) : Put $$m = \frac{2n}{3}$$ in (1) then

$$l = 5\left(\frac{2n}{3}\right) – 3n = \frac{10n}{3} – 3n = \frac{n}{3}$$

$$\Rightarrow l = \frac{n}{3}$$

$$l : m : n = \frac{n}{3} : \frac{2n}{3} : n = \frac{1}{3} : \frac{2}{3} : 1 = 1 : 2 : 3$$

So, direction ratios of $$L1 = (a1, b1, c1) = (1, 2, 3)$$

On dividing by $$\sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$ we get

$$L1 = \left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)$$

Case (ii) : Put $$m = \frac{n}{2}$$ in (1) then

$$l = 5\left(\frac{n}{2}\right) – 3n = \frac{5n}{2} – 3n = -\frac{n}{2}$$

$$\Rightarrow l = -\frac{n}{2}$$

$$l : m : n = -\frac{n}{2} : \frac{n}{2} : n = -\frac{1}{2} : \frac{1}{2} : 1 = -1 : 1 : 2$$

So, direction ratios of $$L2 = (a2, b2, c2) = (-1, 1, 2)$$

On dividing by $$\sqrt{(-1)^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$ we get

$$L2 = \left(-\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right)$$

LAQ-4 : Find the direction cosines of two lines which are connected by the relations l + m + n = 0 and mn – 2nl – 2lm = 0

Given $$l + m + n = 0$$

$$\Rightarrow l = -m – n \quad \text{(1)}$$

$$mn – 2nl – 2lm = 0 \quad \text{(2)}$$

Solving (1) & (2) we get:

$$mn – 2n(-m – n) – 2m(-m – n) = 0$$

$$\Rightarrow mn + 2mn + 2n^2 + 2m^2 + 2mn = 0$$

$$\Rightarrow 2m^2 + 5mn + 2n^2 = 0$$

$$\Rightarrow 2m(m + 2n) + n(m + 2n) = 0$$

$$\Rightarrow (2m + n)(m + 2n) = 0$$

$$\Rightarrow 2m + n = 0 \text{ or } m + 2n = 0$$

So, solving for each scenario:

Case (i): If $$2m + n = 0$$

$$\Rightarrow 2m = -n$$

$$\Rightarrow m = -\frac{n}{2}$$

Put $$m = -\frac{n}{2}$$ in (1)

$$l = -\left(-\frac{n}{2}\right) – n = \frac{n}{2} – n = -\frac{n}{2}$$

$$l = -\frac{n}{2}$$

Now, the ratios $$l : m : n = -\frac{n}{2} : -\frac{n}{2} : n$$

$$= -1 : -1 : 2 = 1 : 1 : -2$$

So, direction ratios of $$L1 = (a1, b1, c1) = (1, 1, -2)L1 = (a1, b1, c1) = (1, 1, -2)$$

On dividing by $$\sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$

Direction cosines of $$L1 = \left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}\right)$$

Case (ii): If $$m + 2n = 0$$

$$\Rightarrow m = -2n$$

Put m = -2n in (1)

$$l = -(-2n) – n = 2n – n = n$$

$$l = n$$

Now, the ratios $$l : m : n = n : -2n : n$$

$$= 1 : -2 : 1$$

So, direction ratios of $$L2 = (a2, b2, c2) = (1, -2, 1)$$

On dividing by $$\sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$$

Direction cosines of $$L2 = \left(\frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$$

LAQ-5 : Show that the lines whose direction cosines are given by l + m + n = 0, 2mn + 3nl – 5lm = 0 are perpendicular to each other

Given $$l + m + n = 0$$

$$\Rightarrow l = -(m + n) \quad \text{(1)}$$

$$2mn + 3nl – 5lm = 0 \quad \text{(2)}$$

Solving (1) & (2) we get:

$$2mn – 3n(m + n) + 5m(m + n) = 0$$

$$\Rightarrow 2mn – 3mn – 3n^2 + 5m^2 + 5mn = 0$$

$$\Rightarrow 5m^2 + 4mn – 3n^2 = 0$$

$$\Rightarrow 5\left(\frac{m}{n}\right)^2 + 4\left(\frac{m}{n}\right) – 3 = 0$$

This is a quadratic equation in $$\frac{m}{n}$$ its roots are taken as $$\frac{m_1}{n_1}, \frac{m_2}{n_2}$$

Product of roots $$\frac{m_1}{n_1} \cdot \frac{m_2}{n_2} = \frac{c}{a} = -\frac{3}{5}$$

$$\Rightarrow \frac{m_1m_2}{n_1n_2} = -\frac{3}{5} \quad \text{(3)}$$

From (1), substitute $$n = -(l + m)$$ in (2)

$$-2m(l + m) – 3l(l + m) – 5lm = 0$$

$$\Rightarrow -2lm – 2m^2 – 3l^2 – 3lm – 5lm = 0$$

$$\Rightarrow 3l^2 + 10lm + 2m^2 = 0$$

$$\Rightarrow 3\left(\frac{l}{m}\right)^2 + 10\left(\frac{l}{m}\right) + 2 = 0$$

This is a quadratic equation in $$\frac{l}{m}$$ its roots are taken as $$\frac{l_1}{m_1}, \frac{l_2}{m_2}$$

Product of roots $$\frac{l_1}{m_1} \cdot \frac{l_2}{m_2} = \frac{c}{a} = \frac{2}{3}$$

$$\Rightarrow \frac{l_1l_2}{m_1m_2} = \frac{2}{3} \quad \text{(4)}$$

From equations (3) and (4) we derive:

$$\frac{l_1l_2}{2} = \frac{m_1m_2}{3} = \frac{n_1n_2}{-5} = k$$

$$\Rightarrow l_1l_2 = 2k, \quad m_1m_2 = 3k, \quad n_1n_2 = -5k$$

Summing these, we find:

$$l_1l_2 + m_1m_2 + n_1n_2 = 2k + 3k – 5k = 0$$

Hence proved that the two lines are perpendicular

LAQ-6 : Find the angle between two diagonals of a cube

Consider a cube of side a with vertices O, A, B, C, L, M, N, P where O = (0,0,0)

A, B and C are on the X-axis, Y-axis, and Z-axis respectively:

$$A = (a,0,0)$$

$$B = (0,a,0)$$

$$C = (0,0,a)$$

L, M and N are on the XY-plane, YZ-plane, and ZX-plane respectively:

$$L = (a,a,0)$$

$$M = (0,a,a)$$

$$N = (a,0,a)$$

P is in the XYZ space:

$$P = (a,a,a)$$

Consider the two diagonals OP and CL:

Direction ratios of OP are given by:

$$\text{d.r’s of OP} = (a-0, a-0, a-0) = (a, a, a) = (a1, b1, c1)$$

Direction ratios of CL are given by:

$$\text{d.r’s of CL} = (a-0, a-0, 0-a) = (a, a, -a) = (a2, b2, c2)$$

The angle between the two diagonals is calculated as follows:

$$\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{(a_1^2 + b_1^2 + c_1^2)(a_2^2 + b_2^2 + c_2^2)}}$$

$$= \frac{|a \cdot a + a \cdot a + a \cdot (-a)|}{\sqrt{(a^2 + a^2 + a^2)(a^2 + a^2 + a^2)}}$$

$$= \frac{a^2}{\sqrt{3a^2 \cdot 3a^2}} = \frac{a^2}{3a^2} = \frac{1}{3}$$

$$\cos\theta = \frac{1}{3}$$

$$\Rightarrow \theta = \cos^{-1}(1/3)$$

LAQ-7 : If a ray makes an angle α,β,γ,δ with the four diagonals of a cube then show that cos2α + cos2β + cos2γ + cos2δ = 4/3

Let one of the vertices of the cube coincide with the origin O (0, 0, 0) and the edges coincide with the coordinate axes.

Let A, B, C be the vertices of the cube on the x-axis, y-axis, z-axis so that OA = OB = OC = a

The vertices of the cube on the xy-plane, yz-plane, zx-plane be L, M, N and P be the remaining vertex in the space.

The 4 diagonals of the cube are OP, CL, NB, AM

The coordinates of the vertices are:

$$A(a, 0, 0)$$

$$B(0, a, 0)$$

$$C(0, 0, a)$$

$$L(a, a, 0)$$

$$M(0, a, a)$$

$$N(a, 0, a)$$

$$P(a, a, a)$$

The direction ratios (d.r’s) of:

$$OP = (a-0, a-0, a-0) = (a, a, a)$$

$$CL = (a-0, a-0, 0-a) = (a, a, -a)

$$NB = (0-a, a-0, 0-a) = (-a, a, -a)$$

$$AM = (0-a, a-0, a-0) = (-a, a, a)$$

Let (l, m, n) be the direction cosines (d.c’s) of the given ray, then:

$$l^2 + m^2 + n^2 = 1$$

The ray makes angles $$\alpha, \beta, \gamma, \delta$$ with OP, CL, NB, AM respectively:

$$\cos \alpha = \frac{al + am + an}{\sqrt{a^2 + a^2 + a^2}(l^2 + m^2 + n^2)} = \frac{a(l + m + n)}{\sqrt{3a^2}(1)} = \frac{a(l + m + n)}{\sqrt{3a}}$$

$$\Rightarrow \cos \alpha = \frac{l + m + n}{\sqrt{3}}$$

Similarly:

$$\cos \beta = \frac{l + m – n}{\sqrt{3}}, \cos \gamma = \frac{-l + m – n}{\sqrt{3}}, \cos \delta = \frac{-l + m + n}{\sqrt{3}}$$

$$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta$$

$$= \frac{1}{3}[(l + m + n)^2 + (l + m – n)^2 + (-l + m – n)^2 + (-l + m + n)^2]$$

$$= \frac{1}{3}[(l + m + n)^2 + (l + m – n)^2 + (l – m + n)^2 + (l – m – n)^2]$$

$$= \frac{1}{3}[2((l + m)^2 + n^2) + 2((l – m)^2 + n^2)]$$

$$= \frac{2}{3}[(l + m)^2 + (l – m)^2 + 2n^2]$$

$$= \frac{2}{3}[2(l^2 + m^2) + 2n^2]$$

$$= \frac{4}{3}(l^2 + m^2 + n^2) = \frac{4}{3}(1) = \frac{4}{3}$$